A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(3pi)/4 #, and the triangle's area is #5 #. What is the area of the triangle's incircle?

1 Answer
Narad T.
Jul 15, 2017

The area of the incircle is #=1.34u^2#

Explanation:

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The area of the triangle is #A=5#

The angle #hatA=1/12pi#

The angle #hatB=3/4pi#

The angle #hatC=pi-(1/12pi+3/4pi)=1/6pi#

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(10/(sin(1/12pi)*sin(3/4pi)*sin(1/6pi)))#

#=10.45#

Therefore,

#a=10.45sin(1/12pi)=2.71#

#b=10.45sin(3/4pi)=7.39#

#c=10.45sin(1/6pi)=5.23#

The radius of the incircle is #=r#

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

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

#=10/(15.33)=0.65#

The area of the incircle is

#area=pi*r^2=pi*0.65^2=1.34u^2#

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