Suppose you want to build a rectangular sandbox where the width is 5 feet less than the length and the diagonal is 5 feet longer than the length. What are the dimensions of the sandbox?

Answer 1

If the length is x, then you must have
(x-5)^2 + x^2 = (x+5)^2

You can work through the math, but just think of the simple
3-4-5 right triangle. Scale it up by a factor of 5, and you have
15-20-25

Answer 2

Length = L.
Width = L-5 ft.
Diag. = L+5 ft.

L^2 + (L-5)^2 = (L+5)^2.
L^2 + L^2 - 10L +25 = L^2 + 10L + 25,
L^2 - 20L = 0,
Divide both sides by L:
L - 20 = 0,
L = 20 ft.
Width = L-5 = 15 ft.