2008 AMC 10A Problem 18

Let the legs be $a, b$ and the hypotenuse $c$. The data give
$$a + b + c = 32,\qquad \tfrac12 ab = 20,\qquad a^2 + b^2 = c^2.$$

Square the sum of the legs and use the other two facts:
$$(a + b)^2 = a^2 + b^2 + 2ab = c^2 + 80.$$
Since $a + b = 32 - c$, this reads
$$(32 - c)^2 = c^2 + 80 \quad\Longrightarrow\quad 1024 - 64c + c^2 = c^2 + 80.$$
The $c^2$ terms cancel, leaving $64c = 944$, so
$$c = \frac{944}{64} = \frac{59}{4}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{59}{4}}$.

Perimeter, area and the Pythagorean relation are all expressible through the symmetric quantities $a + b$ and $ab$, and $(a+b)^2 = (a^2 + b^2) + 2ab$ ties them together. Because $a^2 + b^2$ is $c^2$, squaring the perimeter equation produces an equation whose $c^2$ terms cancel, so the hypotenuse is determined linearly. Look for this whenever a right triangle is described by perimeter and area.

• Trying to find the two legs individually; the legs are irrational, but the hypotenuse falls out of a single identity without ever solving for them.
• Writing $2ab = 20$ instead of $2ab = 80$ (forgetting the $\frac12$ in the area formula), which gives $c = \frac{1004}{64}$, not on the list.