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?

1 Answer

Inscribed circle Area=4.37405" "square units

Explanation:

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.