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 `4`. What is the area of the triangle's incircle?

Answer 1

The area of the triangle's incircle is `1.07` 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=4` We know Area ,

`A_t= 1/2*b*c*sin A or b*c=(2*4)/sin 30 = 16`, similarly ,

`a*c=(2*4)/sin 15 ~~ 30.91`, and `a*b=(2*4)/sin 135~~ 11.31`

`(a*b)*(b*c)*(c.a)=(a b c)^2= (16*30.91*11.31)` or

`a b c=sqrt(5595.31) = 74.80`

`a= (a b c)/(b c)=74.80/16~~4.68`

`b= (a b c)/(ac)=74.80/30.91~~2.42`

`c= (a b c)/(ab)=74.80/11.31~~6.61`

Semi perimeter : `S/2=(4.68+2.42+6.61)/2~~6.855`

Incircle radius is `r_i= A_t/(S/2) = 4/6.855~~0.58`

Incircle Area = `A_i= pi* r_i^2= pi*0.58^2 ~~1.07` sq.unit

The area of the triangle's incircle is `1.07` sq.unit [Ans]