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?

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Answer 1 · 2018-06-07T11:29:07

The area of the triangle's incircle is $1.07$ sq.unit.

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

$/_C= 180-(30+15)=135^0 ;A_t=4$ 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]

A triangle has vertices A, B, and C. Vertex A has an angle of pi/6 pi/6 , vertex B has an angle of (pi)/12 (pi)/12 , and the triangle's area is 4 4 . What is the area of the triangle's incircle?