A triangle has vertices A, B, and C. Vertex A has an angle of $pi/8$ , vertex B has an angle of $(pi)/12$ , and the triangle's area is $12$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-02-21T12:35:54

The area of triangl's incircle is $2.9$ sq.unit.

$/_A = pi/8= 180/8=22.5^0 , /_B = pi/12=180/12= 15^0$

$:./_C= 180-(22.5+15)=142.5^0 ; A_t=12$

Area , $A_t= 1/2*b*c*sinA or b*c=(2*12)/sin22.5 ~~ 62.72$ ,

similarly , $a*c=(2*12)/sin15 =92.73$ , and

$a*b=(2*12)/sin142.5 ~~ 39.42$

$(a*b)*(b*c)*(c.a)=(abc)^2= (39.42*62.72*92.73)$ or

$abc=sqrt((39.42*62.72*92.73)) ~~ 478.82$

$a= (abc)/(bc)=478.82/62.72~~7.63$

$b= (abc)/(ac)=478.82/92.73~~5.16$

$c= (abc)/(ab)=478.82/39.42~~12.15$

Semi perimeter : $S/2=(7.63+5.16+12.15)/2~~12.47$

Incircle radius is $r_i= A_t/(S/2) = 12/12.47~~0.96$

Incircle Area = $A_i= pi* r_i^2= pi*0.96^2 ~~ 2.9$ sq.unit.

The area of triangl's incircle is $2.9$ sq.unit. [Ans]

A triangle has vertices A, B, and C. Vertex A has an angle of pi/8 pi/8 , vertex B has an angle of (pi)/12 (pi)/12 , and the triangle's area is 12 12 . What is the area of the triangle's incircle?