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

1 Answer
bp
Feb 24, 2017

#r= sqrt30/(2+sqrt2)#

Explanation:

Triangle ABC is a right triangle. Since one of its angle is #pi/4# it is an isosceles triangle also. If one of the equal sides is 'x' , its area would be #1/2 x*x= 1/2 x^2#

Thus #1/2 x^2 =15 -> = sqrt30#

The hypotenuse of the right triangle ABC would be #sqrt(30+30)= sqrt60#

The sum of the sides of triagle ABC would this be #sqrt30+sqrt30 +sqrt60= sqrt30 (2+sqrt2)#

If 'r' is the radius of the incircle then

#r= 2(Area of triangle)/(sum of the sides of the triangle)#
Thus #r= 30/(sqrt30 (2+sqrt2)#

#r= sqrt30/(2+sqrt2)#

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