A triangle has vertices A, B, and C. Vertex A has an angle of $pi/2$ , vertex B has an angle of $( pi)/8$ , and the triangle's area is $3$ . What is the area of the triangle's incircle?

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Answer 1 · 2016-11-10T16:50:37

Area of incircle is $1.251$

It is a right angled triangle whose angles are $pi/2$ , $pi/8$ and third angle would be $pi-pi/8-pi/2=(3pi)/8$

Now let hypotenuse be $h$ , then other two sides are $hsin(pi/8)$ and $hsin((3pi)/8)$ and area of triangle $Delta$ is $3$ . So we have

$1/2h^2sin((3pi)/8)sin(pi/8)=1/2h^2xx0.9239xx0.3827=3$

or $h=sqrt((3xx2)/(0.9239xx0.3827))=sqrt16.9695=4.12$

and sides are $4.12xx0.9239=3.81$ and $4.12xx0.3827=1.58$

and half perimeter $s$ is given by $s=1/2xx(4.12+3.81+1.58)=1/2xx9.51=4.755$

As radius of incircle is $Delta/s=3/4.755=0.631$

and its area is $0.631^2pi=3.1416xx0.631^2=1.251$

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)/8 ( pi)/8 , and the triangle's area is 3 3 . What is the area of the triangle's incircle?