A triangle has vertices A, B, and C. Vertex A has an angle of $pi/2$ , 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 · 2018-01-26T14:54:33

The area of the triangle's incircle is $8.75$ sq.unit.

$/_A = pi/2= 180/2=90^0 , /_B = pi/6=180/6= 30^0$

$:. /_C= 180-(90+30)=60^0 ; A_t=18$

We now Area , $A_t= 1/2*b*c*sinA or b*c=(2*18)/sin90 = 36$ ,

similarly , $a*c=(2*18)/sin30 = 72$ , and

$a*b=(2*18)/sin60 ~~41.57;(a*b)*(b*c)*(c.a)=(abc)^2$

$= (36*72*41.57) :.abc=sqrt((36*72*41.57))=328.25$

$a= (abc)/(bc)=328.25/36~~9.12$ unit

$b= (abc)/(ac)=328.25/72~~4.56$ unit

$c= (abc)/(ab)=328.25/41.57~~7.90$ unit

Semi perimeter : $S/2=(9.12+4.56+7.90)/2=10.79$ unit

Incircle radius is $r_i= A_t/(S/2) = 18/10.79=1.67$ unit

Incircla Area = $A_i= pi* r_i^2= pi*1.67^2 =8.75$ sq.unit

The area of the triangle's incircle is $8.75$ sq.unit [Ans]

A triangle has vertices A, B, and C. Vertex A has an angle of pi/2 pi/2 , 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?