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 #9 #. What is the area of the triangle's incircle?

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Answer 1 · 2016-03-14T12:53:57

Inscribed circle Area#=4.37405" "#square units

Solve for the sides of the triangle using the given Area#=9#
and angles #A=pi/2# and #B=pi/3#.

Use the following formulas for Area:

Area#=1/2*a*b*sin C#

Area#=1/2*b*c*sin A#

Area#=1/2*a*c*sin B#

so that we have

#9=1/2*a*b*sin (pi/6)#
#9=1/2*b*c*sin (pi/2)#
#9=1/2*a*c*sin (pi/3)#

Simultaneous solution using these equations result to
#a=2*root4 108#
#b=3*root4 12#
#c=root4 108#

solve half of the perimeter #s#

#s=(a+b+c)/2=7.62738#

Using these sides a,b,c,and s of the triangle, solve for radius of the incribed circle

#r=sqrt(((s-a)(s-b)(s-c))/s)#

#r=1.17996#

Now, compute the Area of the inscribed circle

Area#=pir^2#
Area#=pi(1.17996)^2#
Area#=4.37405" "#square units

God bless....I hope the explanation is useful.