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 #18 #. What is the area of the triangle's incircle?

Answer 1

Area of the triangle's incircle is #4.81# sq.unit.

#/_A = pi/12= 180/12=15^0 , /_B = pi/6=180/6= 30^0 #
#:. /_C= 180-(30+15)=135^0 ;A_t=18#. Area of triangle,
# A_t= 1/2*b*c*sin A or b*c=(2*18)/sin 15 ~~ 139.09 #,
similarly ,#a*c=(2*18)/sin 30=72.0 #, and
#a*b=(2*18)/sin 135 ~~ 50.91 #
#(a*b)*(b*c)*(c.a)=(abc)^2= (139.09*72.0*50.91) # or
#abc=sqrt((139.09*72*50.91)) = 714.03 #
#a= (abc)/(bc)=714.03/139.09~~5.13#
#b= (abc)/(ac)=714.03/72.0~~9.92#
#c= (abc)/(ab)=714.03/50.91~~14.03#
Semi perimeter : #S/2=(5.13+9.92+14.03)/2~~14.54#
Incircle radius is #r_i= A_t/(S/2) = 18/14.54~~1.24#
Incircle Area = #A_i= pi* r_i^2= pi*1.24^2 ~~4.81# sq.unit.
Area of the triangle's incircle is #4.81# sq.unit. [Ans]
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Answer 2

To find the area of the triangle's incircle, you can use the formula:

[ \text{Area of incircle} = \text{Semiperimeter of triangle} \times \text{Inradius} ]

The semiperimeter (( s )) of the triangle is calculated as the sum of the lengths of its sides divided by 2. The inradius (( r )) can be found using the formula:

[ r = \frac{A}{s} ]

Where ( A ) is the area of the triangle.

First, calculate the lengths of the sides of the triangle using the given angles and the Law of Sines. Then, find the semiperimeter and the inradius, and finally, use these values to find the area of the incircle.

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Answer from HIX Tutor

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