2008 AMC 10A Problem 18

Call the legs $a$ and $b$ and the hypotenuse $c$.

1. Write down the three facts:
$$a + b + c = 32, \qquad \frac{1}{2}ab = 20, \qquad a^2 + b^2 = c^2.$$
So $ab = 40$ and $a + b = 32 - c$.
2. We want to connect $a + b$ to $c$. The Pythagorean theorem is about squares, so square $a + b$:
$$(a + b)^2 = a^2 + b^2 + 2ab.$$
3. Replace $a^2 + b^2$ by $c^2$ and $2ab$ by $80$:
$$(a + b)^2 = c^2 + 80.$$
4. Replace $a + b$ by $32 - c$:
$$(32 - c)^2 = c^2 + 80.$$
5. Multiply out the left side: $1024 - 64c + c^2 = c^2 + 80$. The $c^2$ on both sides cancels:
$$1024 - 64c = 80, \qquad 64c = 944, \qquad c = \frac{944}{64} = \frac{59}{4}.$$

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

Perimeter gives the sum of the legs, area gives their product, and the Pythagorean theorem gives the sum of their squares. The identity $(a + b)^2 = a^2 + b^2 + 2ab$ ties all three together. Because the $c^2$ terms cancel, you are left with a simple equation for $c$. Remember this trick for any right triangle described by its perimeter and area.

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