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:

enter image source here

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#

Related questions
See all questions in Area of Inscribed Triangle
Impact of this question
1201 views around the world
You can reuse this answer
Creative Commons License