A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( 3pi)/8 #, and the triangle's area is #16 #. What is the area of the triangle's incircle?

1 Answer
Ananda Dasgupta
Jul 17, 2016

6.681 (to four significant figures)

Explanation:

This is a right angled triangle with #a# as the hypotenuse. The other two sides obey
#b = a sin/_ B#
#c = a cos/_B#

The area #A# of the triangle is given by
#A = 1/2 bc = 1/2 a^2 sin/_B cos/_B#

Alternatively, by adding up the areas of the three triangles formed by joining the vertices to the in-center (each of which have a common height - namely, the radius #r# of the incircle), we get

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

Equating the two expressions for the area, we get

#r = {bc}/(a+b+c)= a {sin/_B cos/_B }/{1+sin/_B + cos/_B}#

So that the area of the in-circle is

#pi r^2 = pi a^2 {sin^2/_B cos^2/_B }/(1+sin/_B + cos/_B)^2 = 2 pi A {sin/_B cos/_B }/(1+sin/_B + cos/_B)^2 = pi A {sin(2/_B ) }/(1+sin/_B + cos/_B)^2#
Substituting the numerical values, we get an area of 6.681 units

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