Question 901335
Pythagorean Theorem: a^2 + b^2 = c^2
(x+2)^2 + (2x+3)^2 = (2x+5)^2  solve for x
5x^2 + 16x + 13 = 4x^2 + 20x + 25
x^2 -4x - 12 = 0
(x-6)(x+2) = 0  (Tossing out the negative solution for unit measure)
x = 6checking...
64 + 225 = 289