Question

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?

Answer 1

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]