A triangle has vertices A, B, and C. Vertex A has an angle of $\frac{π}{12}$ $pi/12$ , vertex B has an angle of $\frac{5 π}{12}$ $(5pi)/12$ , and the triangle's area is $24$ $24$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-08-05T10:26:54

The area of the triangle's incircle is $7.63$ $7.63$ sq.unit

$∠ A = \frac{π}{12} = \frac{180}{12} = 15^{0}, ∠ B = \frac{5 π}{12} = 75^{0}$ $/_A = pi/12= 180/12=15^0 , /_B = (5 pi)/12= 75^0$

$:. /_C= 180-(75+15)=90^0 ; A_t=24$

Area , $A_t= 1/2*b*c*sin A or b*c=(2*24)/sin 15~~ 185.46$ ,

similarly , $a*c=(2*24)/sin 75 =49.69$ , and

$a*b=(2*24)/sin 90 = 48.0$

$(a*b)*(b*c)*(c.a)=(abc)^2= (185.46*49.69*48)$ or

$abc=sqrt( 185.46*49.69*48) ~~ 665.11$

$a= (abc)/(bc)=665.11/185.46~~ 3.59$ unit

$b= (abc)/(ac)=665.11/49.69~~ 13.39$ unit

$c= (abc)/(ab)=665.11/48~~ 13.86$ unit

Semi perimeter : $S/2=(3.59+13.39+13.86)/2~~15.42$

Incircle radius is $r_i= A_t/(S/2) = 24/15.42~~1.56$ unit

Incircle Area = $A_i= pi* r_i^2= pi*1.56^2 ~~7.63$ sq.unit

The area of the triangle's incircle is $7.63$ sq.unit [Ans]

A triangle has vertices A, B, and C. Vertex A has an angle of pi/12 π12pi/12 , vertex B has an angle of (5pi)/12 5π12(5pi)/12 , and the triangle's area is 24 2424 . What is the area of the triangle's incircle?