Question

A triangle has vertices A, B, and C. Vertex A has an angle of `pi/12`, vertex B has an angle of `(5pi)/12`, and the triangle's area is `21`. What is the area of the triangle's incircle?

Answer 1

Area of triangle's incircle is `6.7(1dp)` sq.unit

Explanation:

`/_A = 180/12=15^0 ; /_B= 58180/12=75^0 ; /_C=180-(75+15)=90^0`
Area of triangle `A_r=21`(given). Let a,b,c be the sides of triangle.We know `(a*b*sinC)/2=A_r or a*b=42/sin90=42`similarly
`b*c=42/sin15=162.28` and `c*a=42/sin75=43.48:. ab*bc*ca=(abc)^2=42*162.28*43.48 or abc=544.38 :.` `a=544.38/162.28=3.355 ;b =544.38/43.48=12.52 ; c= 544.38/42=12.96`So semi perimeter`S=(3.355+12.52+12.96)/2 =14.42` Incircle radius is`A_r/S= 21/14.42=1.46` Area of triangle's incircle is `pi*1.46^2=6.7(1dp)` sq.unit [Ans]