A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #pi/12 #, and the triangle's area is #7 #. What is the area of the triangle's incircle?
Jayden JacksonThe area of the incircle is
The area of the triangle is The angle The angle The angle The sine rule is So, Let the height of the triangle be The area of the triangle is But, So, Therefore, The radius of the incircle is The area of the incircle is
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Violet AldrichTheThe areaThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \The area of the incThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \fracThe area of the incircleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{sThe area of the incircle ofThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{The area of the incircle of aThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2The area of the incircle of a triangleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2}The area of the incircle of a triangle canThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \The area of the incircle of a triangle can beThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdotThe area of the incircle of a triangle can be calculatedThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot rThe area of the incircle of a triangle can be calculated usingThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r \The area of the incircle of a triangle can be calculated using theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
The area of the incircle of a triangle can be calculated using the formulaThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
whereThe area of the incircle of a triangle can be calculated using the formula:
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where (The area of the incircle of a triangle can be calculated using the formula:
[The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( sThe area of the incircle of a triangle can be calculated using the formula:
[ AThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s \The area of the incircle of a triangle can be calculated using the formula:
[ A_{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeterThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\textThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangleThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle andThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and (The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( rThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r )The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \piThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) isThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi rThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radiusThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius ofThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( rThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircleThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r )The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius ofThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
ToThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To findThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
TheThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeterThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radiusThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ),The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius ofThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sumThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengthsThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths ofThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle canThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can beThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangleThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be foundThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle'sThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found usingThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sidesThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide byThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formulaThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ rThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ sThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{ABThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{sThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB +The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC +The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + ACThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
whereThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where (The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( AThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
ThenThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\textThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then,The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radiusThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius (The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( rThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} )The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) isThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r )The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the areaThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle canThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area ofThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can beThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be foundThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found usingThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle andThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and (The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formulaThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( sThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s )The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ rThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) isThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\textThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter,The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{AreaThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given byThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area ofThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ sThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{sThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
SubThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a +The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
SubstitThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b +The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
SubstitutingThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + cThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the givenThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given valuesThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values intoThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
whereThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into theseThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where (The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulasThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( aThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas,The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we canThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ),The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can findThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), (The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( bThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the areaThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ),The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), andThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangleThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and (The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle'sThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( cThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
NoteThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the sideThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note:The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: SinceThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In thisThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this caseThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the anglesThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, sinceThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles atThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since twoThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at AThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two anglesThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A andThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles areThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B areThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given toThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are bothThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ),The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both (The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies thatThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle isThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\piThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isoscelesThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaningThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ),The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning twoThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), theThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sidesThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangleThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides areThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and thereforeThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal inThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore,The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, ABThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length ofThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and ACThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as (The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC areThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x )The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equalThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the lengthThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal inThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in lengthThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the thirdThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third sideThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side asThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ ABThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as (The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( yThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = ACThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ).The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). ThusThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus,The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ sThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeterThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter isThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2ABThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ sThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BCThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x +The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = ABThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x +The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB +The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + yThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2ABThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2ACThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x +The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + yThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \textThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{AreaThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area ofThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x +The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircleThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{sThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
NowThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now,The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdotThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, usingThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the LawThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law ofThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of SThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of SinesThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2ACThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we canThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( yThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in termsThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of (The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdotThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot rThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = ACThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdotThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot rThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{yThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sinThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\piThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \leftThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\piThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\rightThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sinThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\piThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{ACThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ yThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ ACThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
SubThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
SubstitThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
SubstitutingThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting backThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back intoThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeterThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ ACThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formulaThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrtThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ sThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x +The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ rThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrtThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now,The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the areaThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ rThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area (The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( AThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\textThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrtThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} )The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) ofThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of theThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle canThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can beThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressedThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed inThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \textThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in termsThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms ofThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{AreaThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of (The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area ofThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircleThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ AThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = ACThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\textThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot rThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangleThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}}The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdotThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdotThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdotThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdotThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sinThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}} =The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}} = \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\frac{\piThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}} = \fracThe area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\frac{\pi}{The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}} = \frac{7The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\frac{\pi}{6})The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}} = \frac{7}{The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\frac{\pi}{6}) =The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}} = \frac{7}{2The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\frac{\pi}{6}) = \The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}} = \frac{7}{2}The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\frac{\pi}{6}) = \fracThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}} = \frac{7}{2} \The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\frac{\pi}{6}) = \frac{xThe area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}} = \frac{7}{2} ]The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\frac{\pi}{6}) = \frac{x^The area of the triangle's incircle can be calculated using the formula:
[ \text{Area of incircle} = \frac{s}{2} \cdot r ]
where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.
To find the semi-perimeter ( s ), sum the lengths of the triangle's sides and divide by 2.
[ s = \frac{AB + BC + AC}{2} ]
Then, the radius ( r ) of the incircle can be found using the formula:
[ r = \frac{\text{Area of triangle}}{s} ]
Substituting the given values into these formulas, we can find the area of the triangle's incircle.
Note: Since the angles at A and B are both ( \frac{\pi}{12} ), the triangle is isosceles, and therefore, AB and AC are equal in length.
[ AB = AC ]
[ s = \frac{2AB + BC}{2} = AB + BC = 2AB = 2AC ]
[ \text{Area of incircle} = \frac{s}{2} \cdot r = \frac{2AC}{2} \cdot r = AC \cdot r ]
[ 7 = \frac{AC^2 \sin \left(\frac{\pi}{6}\right)}{2} = \frac{AC^2}{4} ]
[ AC^2 = 28 ]
[ AC = \sqrt{28} ]
[ r = \frac{7}{\sqrt{28}} ]
[ r = \frac{7}{2\sqrt{7}} ]
[ \text{Area of incircle} = AC \cdot r = \sqrt{28} \cdot \frac{7}{2\sqrt{7}} = \frac{7}{2} ]The area of the incircle of a triangle can be calculated using the formula:
[ A_{\text{incircle}} = \pi r^2 ]
where ( r ) is the radius of the incircle.
The radius of the incircle can be found using the formula:
[ r = \frac{A_{\text{triangle}}}{s} ]
where ( A_{\text{triangle}} ) is the area of the triangle and ( s ) is the semiperimeter, given by:
[ s = \frac{a + b + c}{2} ]
where ( a ), ( b ), and ( c ) are the side lengths of the triangle.
In this case, since two angles are given to be ( \frac{\pi}{12} ), it implies that the triangle is isosceles, meaning two sides are equal in length. Let's denote the length of these sides as ( x ) and the length of the third side as ( y ). Thus, the semiperimeter is:
[ s = \frac{x + x + y}{2} = \frac{2x + y}{2} = x + \frac{y}{2} ]
Now, using the Law of Sines, we can express ( y ) in terms of ( x ):
[ \frac{y}{\sin(\frac{\pi}{12})} = \frac{x}{\sin(\frac{\pi}{12})} ]
[ y = x ]
Substituting back into the semiperimeter formula:
[ s = x + \frac{x}{2} = \frac{3x}{2} ]
Now, the area ( A_{\text{triangle}} ) of the triangle can be expressed in terms of ( x ):
[ A_{\text{triangle}} = \frac{1}{2} \cdot x \cdot x \cdot \sin(\frac{\pi}{6}) = \frac{x^2 \sqrt{3}}{4} ]
Given that ( A_{\text{triangle}} = 7 ), we can solve for ( x ):
[ \frac{x^2 \sqrt{3}}{4} = 7 ]
[ x^2 = \frac{28}{\sqrt{3}} ]
[ x = \sqrt{\frac{28}{\sqrt{3}}} = \sqrt{\frac{28 \sqrt{3}}{3}} = \frac{2\sqrt{21}}{\sqrt{3}} = \frac{2\sqrt{63}}{3} ]
Now, we can find the radius of the incircle:
[ r = \frac{A_{\text{triangle}}}{s} = \frac{7}{\frac{3x}{2}} = \frac{14}{3x} ]
[ r = \frac{14}{\frac{2\sqrt{63}}{3}} = \frac{21}{\sqrt{63}} = \frac{21\sqrt{3}}{9} = \frac{7\sqrt{3}}{3} ]
Finally, the area of the incircle ( A_{\text{incircle}} ):
[ A_{\text{incircle}} = \pi r^2 = \pi \left(\frac{7\sqrt{3}}{3}\right)^2 = \pi \left(\frac{49 \cdot 3}{9}\right) = \frac{49\pi}{3} ]
So, the area of the triangle's incircle is ( \frac{49\pi}{3} ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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- What is the equation of the circle with a center at #(-4 ,6 )# and a radius of #3 #?
- A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #pi/6 #, and the triangle's area is #9 #. What is the area of the triangle's incircle?
- A triangle has corners at #(5 , 2 )#, #(2 ,3 )#, and #(3 ,4 )#. What is the radius of the triangle's inscribed circle?
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