2011 AMC 10B Problem 14

The perimeter is $2 \times (\text{length} + \text{width})$, so what we really need is the sum of the two sides. Call them $l$ and $w$.

1. Use the diagonal. The diagonal splits the rectangle into two right triangles with legs $l$ and $w$ and long side $25$. By the Pythagorean theorem,
$$l^2 + w^2 = 25^2 = 625.$$
2. Use the area. $lw = 168$.
3. Build the sum. Squaring a sum gives $(l + w)^2 = l^2 + 2lw + w^2$. We know every piece:
$$(l + w)^2 = 625 + 2 \times 168 = 625 + 336 = 961.$$
4. Since $31 \times 31 = 961$, we get $l + w = 31$.
5. The perimeter is $2 \times 31 = 62$.

The answer is $\boxed{\textbf{(C)}\ 62}$.

The formula $(l + w)^2 = l^2 + w^2 + 2lw$ links "sum of squares," "product," and "sum." When a problem gives you two of these, you can get the third without ever finding $l$ and $w$ themselves. Look for this whenever a rectangle's diagonal and area are both given.

Right triangles with whole-number sides and long side $25$ include $7, 24, 25$ and $15, 20, 25$. Check the area: $7 \times 24 = 168$, a match. So the sides are $7$ and $24$, and the perimeter is $2 \times (7 + 24) = 62$.