A triangle has corners at #(5 , 2 )#, #(2 ,3 )#, and #(3 ,1 )#. What is the radius of the triangle's inscribed circle?

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Answer 1 · 2017-02-24T18:59:05

#sqrt5/(2+sqrt2)#

If A is (5,2) B is (2,3) and C is (3,1), then

Side #AB= sqrt((5-2)^2 +(2-3)^2)=sqrt10#

Side #BC= sqrt ((2-3)^2 +(3-1)^2)=sqrt5#

Side#AC= sqrt((5-3)^2 +(2-1)^2)= sqrt5#

Since#AC^2+BC^2 =AB^2#, the triangle is a right triangle. Its Area would be #1/2 sqrt5 *sqrt5= 5/2#

Sum of the #sides =sqrt5 +sqrt5 +sqrt10=sqrt5(2+sqrt2)#

radius of incircle would be #2 (Area of triangle) / (sum sides oftriangle)#

=#2* 5/2 div sqrt5 (2+sqrt2)= sqrt5/(2+sqrt2)#