Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/2`, vertex B has an angle of `( pi)/4`, and the triangle's area is `21`. What is the area of the triangle's incircle?

Answer 1

The area of triangle's incircle is `11.3(1dp)` sq.unit

Explanation:

The three angles are `/_A= 180/2=90^0 ; /_B=180/4=45^0 ; /_C=180-(90+45)=45^0` This is a right angled isocelles triangle. The sides opposite to the angles`/_B and /_C` are equal and their included angle is `90^0`. So the area of the triangle`21= 1/2*b^2 :. b =sqrt 42=6.48=c :.a=sqrt(6.48^2+6.48^2)=sqrt 84=9.17` So semi perimeter of the triangle `s= ((6.48+6.48+9.17)/2) =11.06` Hence the radius of the incircle is `r=A_t/s=21/11.06=1.9 ; A_t` is area of triangle. Hence the area of triangles incircle is `pi*r^2 = pi* 1.9^2 =11.3(1dp)`sq.unit [Ans]