A triangle has vertices A, B, and C. Vertex A has an angle of pi/2 , vertex B has an angle of ( pi)/4 , and the triangle's area is 3 . What is the area of the triangle's incircle?

1 Answer
Narad T.
Jul 6, 2017

The area of the incircle is =1.62u^2

Explanation:

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The area of the triangle is A=3

The angle hatA=1/2pi

The angle hatB=1/4pi

The angle hatC=pi-(1/2pi+1/4pi)=1/4pi

The sine rule is

a/(sin hat (A))=b/sin hat (B)=c/sin hat (C)=k

So,

a=ksin hatA

b=ksin hatB

c=ksin hatC

Let the height of the triangle be =h from the vertex A to the opposite side BC

The area of the triangle is

A=1/2a*h

But,

h=csin hatB

So,

A=1/2ksin hatA*csin hatB=1/2ksin hatA*ksin hatC*sin hatB

A=1/2k^2*sin hatA*sin hatB*sin hatC

k^2=(2A)/(sin hatA*sin hatB*sin hatC)

k=sqrt((2A)/(sin hatA*sin hatB*sin hatC))

=sqrt(6/(sin(1/2pi)*sin(1/4pi)*sin(1/4pi)))

=3.46

Therefore,

a=3.46sin(1/2pi)=3.46

b=3.46sin(1/4pi)=2.45

c=3.46sin(1/4pi)=2.45

The radius of the incircle is =r

1/2*r*(a+b+c)=A

r=(2A)/(a+b+c)

=6/(8.36)=0.72

The area of the incircle is

area=pi*r^2=pi*0.72^2=1.62u^2

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