Question

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?

Answer 1

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]