A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, 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
Sonnhard
May 24, 2018

#A=pi*(6/(2+sqrt(2)))^2#

Explanation:

Since #B=pi/4# we get #b=c# so we can compute #b^2/2=18# so # b=6# and by the Pythagorean Theorem we get #a=sqrt(2)*6# and our inradius is #a=36/(12+6sqrt(2))=6/(2+sqrt(2))# by the formula #r=A/s# #s=(a+b+c)/2# and #A# denotes the area.

Related questions
See all questions in Area of Inscribed Triangle
Impact of this question
1018 views around the world
You can reuse this answer
Creative Commons License