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
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#