A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/6 #, and the triangle's area is #72 #. What is the area of the triangle's incircle?

1 Answer
Binayaka C.
Feb 12, 2017

Area of triangle's incircle is #35.04(2dp)# sq.unit.

Explanation:

#/_A= pi/2=90^0 ; /_B= pi/6=30^0 ; /_C =180-(90+30)=60^0#

This is a #30-60-90# triangle. So if#AC=x# then #AB=x*sqrt3; ;BC=2x#, Area of triangle is #1/2*AB*AC =72 or x*sqrt3*x=72*2 or x^2=144/sqrt3 :.x = 9.12(2dp) ; 2x=18.24(2dp) ; sqrt3x=15.79#
Sides of triangle are # 9.12,15.79 &18.24#unit
Perimeter of triangle is #P= 9.12+15.79 +18.24 =43.15 #unit
Area of triangle is #A_t=72#sq.unit
Radius of triangle's incircle is #r=(2*A_t)/P=(2*72)/43.15= 3.34(2dp)#unit
Area of triangle's incircle is #A_r= pi* *r^2=pi*3.34^2 = 35.04(2dp)#sq.unit. [Ans]

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