A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #( pi)/6 #, and the triangle's area is #3 #. What is the area of the triangle's incircle?

1 Answer
Narad T. · Shwetank Mauria
Jun 26, 2017

The area of the incircle is #=1.02# square units.

Explanation:

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The area of the triangle is #A=3#

The angle #hatA=1/8pi#

The angle #hatB=1/6pi#

The angle #hatC=pi-(1/8pi+1/6pi)=17/24pi#

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(6/(sin(pi/8)*sin(1/6pi)*sin(17/24pi)))#

#=6.29#

Therefore,

#a=6.29sin(1/8pi)=2.41#

#b=6.29sin(1/6pi)=3.15#

#c=6.29sin(17/24pi)=4.99#

Observe in the figure above that joining incenter with the three vertices #A,B# and #C# results in three triangles, whose bases are #a,b# and #c# and height is #r#, the radius of incenter and therefore the sum of the area of these triangles is area of #DeltaABC#.

Hence we have, #1/2*r*(a+b+c)=A#

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

#=6/(10.55)=0.57#

The area of the incircle is

#area=pi*r^2=pi*0.57^2=1.02# square units.

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