A triangle has vertices A, B, and C. Vertex A has an angle of $pi/8$ , vertex B has an angle of $( pi)/6$ , and the triangle's area is $22$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-02-02T11:42:29

The area of incircle of triangle is $7.47$ sq.unit.

$/_A = pi/8= 180/8=22.5^0 , /_B = pi/6=180/6= 30^0$

$/_C= 180-(22.5+30)=127.5^0: A_t=22$

We know Area , $A_t= 1/2*b*c*sinA$ or

$b*c=(2*22)/sin22.5 ~~114.98$ , similarly , $a*c=(2*22)/sin30$

$:.a*c = 88.0; a*b=(2*22)/sin127.5 ~~ 55.46$

$(a*b)*(b*c)*(c.a)=(abc)^2= (55.46*114.98*88.0)$

$=561165.55 :. abc =sqrt(561165.55)=749.11$

$a= (abc)/(bc)=749.11/114.98~~6.52$

$b= (abc)/(ac)=749.11/88.00~~8.51$

$c= (abc)/(ab)=749.11/55.46~~13.51$

Semi perimeter : $S/2=(6.52+8.51+13.51)/2 ~~14.27$

Incircle radius is $r_i= A_t/(S/2) = 22.0/14.27 ~~1.54$

Incircle Area = $A_i= pi* r_i^2= pi*1.54^2 ~~ 7.47$ sq.unit.

The area of incircle of triangle is $7.47$ 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)/6 ( pi)/6 , and the triangle's area is 22 22 . What is the area of the triangle's incircle?