A triangle has vertices A, B, and C. Vertex A has an angle of #pi/6 #, vertex B has an angle of #(pi)/12 #, and the triangle's area is #4 #. What is the area of the triangle's incircle?

Answer 1

The area of the triangle's incircle is #1.07# sq.unit.

#/_A = pi/6= 180/6=30^0 , /_B = pi/12=180/12= 15^0#
# /_C= 180-(30+15)=135^0 ;A_t=4# We know Area ,
# A_t= 1/2*b*c*sin A or b*c=(2*4)/sin 30 = 16 #, similarly ,
#a*c=(2*4)/sin 15 ~~ 30.91 #, and #a*b=(2*4)/sin 135~~ 11.31 #
#(a*b)*(b*c)*(c.a)=(a b c)^2= (16*30.91*11.31) # or
#a b c=sqrt(5595.31) = 74.80 #
#a= (a b c)/(b c)=74.80/16~~4.68#
#b= (a b c)/(ac)=74.80/30.91~~2.42#
#c= (a b c)/(ab)=74.80/11.31~~6.61#
Semi perimeter : #S/2=(4.68+2.42+6.61)/2~~6.855#
Incircle radius is #r_i= A_t/(S/2) = 4/6.855~~0.58#
Incircle Area = #A_i= pi* r_i^2= pi*0.58^2 ~~1.07# sq.unit
The area of the triangle's incircle is #1.07# sq.unit [Ans]
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Answer 2

The area of the incircle of a triangle can be calculated using the formula:

[ \text{Area of incircle} = \text{semiperimeter} \times \text{inradius} ]

The semiperimeter of the triangle can be calculated as ( s = \frac{a + b + c}{2} ), where ( a ), ( b ), and ( c ) are the lengths of the sides of the triangle.

Given the angles at vertices A and B, and the area of the triangle, we can find the lengths of the sides using trigonometric relations.

Let's denote the side lengths as follows:

  • ( a ) is opposite angle A,
  • ( b ) is opposite angle B,
  • ( c ) is opposite angle C.

We can use the law of sines to find the side lengths:

[ \frac{a}{\sin(\frac{\pi}{6})} = \frac{b}{\sin(\frac{\pi}{12})} = \frac{c}{\sin(\pi - \frac{\pi}{6} - \frac{\pi}{12})} ]

Then, once the side lengths are known, we can find the semiperimeter ( s ).

Next, we need to find the inradius ( r ). The formula to calculate the inradius of a triangle is:

[ r = \frac{\text{Area}}{\text{Semiperimeter}} ]

Finally, once we have the semiperimeter and inradius, we can find the area of the incircle using the formula mentioned at the beginning.

Let's proceed with the calculations:

Given: [ \text{Angle A} = \frac{\pi}{6} ] [ \text{Angle B} = \frac{\pi}{12} ] [ \text{Area of triangle} = 4 ]

We can use trigonometric relations to find the side lengths: [ a = 2 \times \sin(\frac{\pi}{6}) ] [ b = 2 \times \sin(\frac{\pi}{12}) ]

Then, calculate the semiperimeter: [ s = \frac{a + b + c}{2} ]

Find the inradius: [ r = \frac{\text{Area}}{s} ]

Finally, calculate the area of the incircle: [ \text{Area of incircle} = s \times r ]

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Answer from HIX Tutor

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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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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