# problem solving in RAE

What is the area of a rectangle whose length is 1 unit longer than its width and whose diagonal is 2 unit longer than the width?

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1. looks like Pythagoras.

width --- x
length --- x+1
diagonal --- x+2

x^2 + (x+1)^2 = (x+2)^2

by recognition:
the sides are one unit apart, the only Pythagorean triple like that is 3,4,5, since 3^2 + 4^2 = 5^2.

If you did not recognize that:
x^2 + x^2 + 2x + 1 = x^2 + 4x + 4
x^2 - 2x - 3 = 0

(x - 3)(x+1) = 0
x = 3 or x = -1, but a side can't be negative, so

x = 3
find the area using base*height

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posted by Reiny

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