+595 Triangle ABC is isosceles with AB = BC. If AC = 20 and the inradius is 5, then find the perimeter of triangle ABC.
+129847 inradius = area / semi-perimeter
Let AB = BC = x
semi-perimeter = [ 20 + 2x ] / 2 = 10 + x
area = sqrt [ (10 + x) (10 + x -20) (10 +x -x)^2] = sqrt [ (10+x) (x -10) (100) ]
5 = sqrt [ (10+x) (x -10) (100) ] / [ 10 +x]
5 [10 + x] = sqrt [ 100 ( x^2 - 100) ]
5 [ 10 + x] = 10 sqrt [ (x^2 - 100) ]
[10 + x] = 2 sqrt [x^2 -100] square both sides
x^2 + 20x + 100 = 4x^2 - 400
3x^2 - 20x - 500 = 0
x = 50/3
Perimeter = 2 (50/3) + 20 = 160 / 3
