A triangle has vertices A, B, and C. Vertex A has an angle of pi/8 , vertex B has an angle of ( 3pi)/8 , and the triangle's area is 16 . What is the area of the triangle's incircle?

1 Answer
Narad T.
Jun 13, 2017

The area of the incircle is =6.52u^2

Explanation:

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The area of the triangle is A=16

The angle hatA=1/8pi

The angle hatB=3/8pi

The angle hatC=pi-(1/8pi+3/8pi)=4/8pi=pi/2

The sine rule is

a/sinA=b/sinB=c/sinC=k

So,

a=ksinA

b=ksinB

c=ksinC

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=csinB

So,

A=1/2ksinA*csinB=1/2ksinA*ksinC*sinB

A=1/2k^2*sinA*sinB*sinC

k^2=(2A)/(sinA*sinB*sinC)

k=sqrt((2A)/(sinA*sinB*sinC))

=sqrt(32/(sin(pi/8)*sin(3/8pi)*sin(1/2pi)))

=9.81

Therefore,

a=9.81sin(1/8pi)=3.64

b=9.81sin(3/8pi)=9.06

c=9.81sin(1/2pi)=9.51

The radius of the incircle is =r

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

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

=32/(22.2)=1.44

The area of the incircle is

area=pi*r^2=pi*1.44^2=6.52u^2

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