A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/3 #, and the triangle's area is #18 #. What is the area of the triangle's incircle?

1 Answer
Narad T.
Jun 30, 2017

The area of the incircle is #=8.74u^2#

Explanation:

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The area of the triangle is #A=18#

The angle #hatA=1/2pi#

The angle #hatB=1/3pi#

The angle #hatC=pi-(1/2pi+1/3pi)=1/6pi#

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(36/(sin(1/2pi)*sin(1/3pi)*sin(1/6pi)))#

#=9.12#

Therefore,

#a=9.12sin(1/2pi)=9.12#

#b=9.12sin(1/3pi)=7.90#

#c=9.12sin(1/6pi)=4.56#

The radius of the incircle is #=r#

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

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

#=36/(21.58)=1.67#

The area of the incircle is

#area=pi*r^2=pi*1.67^2=8.74u^2#

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