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

1 Answer
Binayaka C.
Jul 3, 2017

The area of triangle's incircle is #6.45# sq.unit.

Explanation:

#/_A = pi/6= 180/6=30^0 , /_B = pi/12=180/12= 15^0 , /_C= 180-(30+15)=135^0 , A_t=24#
We now Area ,# A_t= 1/2*b*c*sinA or b*c=(2*24)/sin30 = 96 #, similarly ,#a*c=(2*24)/sin15 = 185.46 #, and #a*b=(2*24)/sin135 = 67.88 #

#(a*b)*(b*c)*(c.a)=(abc)^2= (96*185.46*67.88) # or
#abc=sqrt((96*185.46*67.88)) = 1099.35 #

#a= (abc)/(bc)=1099.35/96~~11.45#
#b= (abc)/(ac)=1099.35/185.46~~5.93#
#c= (abc)/(ab)=1099.35/67.88~~16.19#

Semi perimeter : #S/2=(11.45+5.93+16.20)/2=16.79#
Incircle radius is #r_i= A_t/(S/2) = 24/16.79=1.43#
Incircla Area = #A_i= pi* r_i^2= pi*1.43^2 =6.45# sq.unit.

The area of triangle's incircle is #6.45# sq.unit. [Ans]

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