A triangle has vertices A, B, and C. Vertex A has an angle of pi/8 , vertex B has an angle of (pi)/12 , and the triangle's area is 6 . What is the area of the triangle's incircle?

1 Answer
Narad T.
Jul 18, 2017

The area of the incircle is =1.45u^2

Explanation:

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The area of the triangle is A=6

The angle hatA=1/8pi

The angle hatB=1/12pi

The angle hatC=pi-(1/8pi+1/12pi)=19/24pi

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(12/(sin(1/8pi)*sin(1/12pi)*sin(19/24pi)))

=14.11

Therefore,

a=14.11sin(1/8pi)=5.40

b=14.11sin(1/12pi)=3.65

c=14.11sin(19/12pi)=8.59

The radius of the incircle is =r

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

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

=12/(17.64)=0.68

The area of the incircle is

area=pi*r^2=pi*0.68^2=1.45u^2

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