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 $24$ . What is the area of the triangle's incircle?

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Answer 1 · 2018-02-25T11:20:59

The area of the triangle's incircle is $11.66$ sq.unit.

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

$/_C= 180-(90+30)=60^0 ; Delta_a=24$

This is a right triangle of $(30,60,90)$

In right triangle $(30,60,90)$ ,if base is $b=x$ , hypotenuse is

$a=2x$ and perpendicular is $c=sqrt3*x$

Area of the triangle , $A_t= 1/2*b*c=24 or 24= 1/2*x*sqrt3*x$

or $x^2=48/sqrt3 or x =sqrt(16*sqrt3) ~~ 27.71 :. b ~~ 5.264$

unit $c= sqrt3*x ~~ 9.12$ unit , $a=2x~~ 10.53$ unit

Semi perimeter is $S/2=(5.264+9.12+10.53)/2 ~~ 12.46$

Incircle radius is $r_i= A_t/(S/2) = 24/12.46~~1.93$

Incircle Area = $A_i= pi* r_i^2= pi*1.93^2 ~~11 .66$ sq.unit

The area of the triangle's incircle is $11.66$ 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 24 24 . What is the area of the triangle's incircle?