A triangle has vertices A, B, and C. Vertex A has an angle of $pi/12$ , vertex B has an angle of $pi/6$ , and the triangle's area is $18$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-07-30T11:27:29

Area of the triangle's incircle is $4.81$ sq.unit.

$/_A = pi/12= 180/12=15^0 , /_B = pi/6=180/6= 30^0$

$:. /_C= 180-(30+15)=135^0 ;A_t=18$ . Area of triangle,

$A_t= 1/2*b*c*sin A or b*c=(2*18)/sin 15 ~~ 139.09$ ,

similarly , $a*c=(2*18)/sin 30=72.0$ , and

$a*b=(2*18)/sin 135 ~~ 50.91$

$(a*b)*(b*c)*(c.a)=(abc)^2= (139.09*72.0*50.91)$ or

$abc=sqrt((139.09*72*50.91)) = 714.03$

$a= (abc)/(bc)=714.03/139.09~~5.13$

$b= (abc)/(ac)=714.03/72.0~~9.92$

$c= (abc)/(ab)=714.03/50.91~~14.03$

Semi perimeter : $S/2=(5.13+9.92+14.03)/2~~14.54$

Incircle radius is $r_i= A_t/(S/2) = 18/14.54~~1.24$

Incircle Area = $A_i= pi* r_i^2= pi*1.24^2 ~~4.81$ sq.unit.

Area of the triangle's incircle is $4.81$ sq.unit. [Ans]

A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 pi/12 , vertex B has an angle of pi/6 pi/6 , and the triangle's area is 18 18 . What is the area of the triangle's incircle?