A triangle has vertices A, B, and C. Vertex A has an angle of #pi/8 #, vertex B has an angle of #( pi)/4 #, and the triangle's area is #18 #. What is the area of the triangle's incircle?

1 Answer
Ananda Dasgupta
Jul 23, 2016

#~~6.97#

Explanation:

The third angle of the triangle is #/_C = pi -pi/4-pi/8={5 pi}/8#

Sine law gives us

#a/{sin (pi/8)} = b/{sin(pi/4)}=c/{sin({5 pi}/8)} = k ( "say")#

The area is given by

#A = 1/2 b c sin/_A = 1/2 k^2 sin/_A sin/_B sin/_C = 1/2 k^2 sin(pi/8) sin(pi/4)sin({5pi}/8) = 1/2 k^2 sin(pi/8) sin(pi/4) cos(pi/4) = 1/4 k^2 sin^2(pi/4) = 1/8 k^2#

and thus #k^2 = 8 xx 18 = 144# or #k=12#.

Now, the radius of the incircle can be easily found from

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

which one can derive simply by joining each of the vertices with the incenter. So, we have

#1/2 k(sin(pi/8) +sin(pi/4)+sin({5pi}/8))r = 18#

and so

#r = 6/(sin(pi/8) +sin(pi/4)+sin({5pi}/8)) ~~ 1.49 #

Thus the area of the incircle is

#pi r^2 ~~ 6.97#

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