2001 AMC 10 Problem 24

1. Draw the shape. $AB$ and $CD$ both meet $AD$ at right angles, so they are parallel, and $AD = 7$ is the distance between them. $AB$ is the shorter one.
2. Call $AB = a$ and $CD = b$. The slanted side is $BC = a + b$.
3. The slanted side is the hard part, so turn it into the long side of a right triangle. From $B$, draw a line straight to $CD$, at a right angle, meeting it at $E$.
4. $ABED$ is a rectangle. So $BE = AD = 7$, and $DE = AB = a$. That leaves $EC = b - a$.
5. Triangle $BEC$ has a right angle at $E$. By the Pythagorean theorem:
$$(a + b)^2 = 7^2 + (b - a)^2.$$
6. Expand both squares:
$$a^2 + 2ab + b^2 = 49 + a^2 - 2ab + b^2.$$
7. The $a^2$ and $b^2$ appear on both sides, so they cancel: $2ab = 49 - 2ab$, which gives $4ab = 49$.
8. So $ab = \frac{49}{4} = 12.25$.

The answer is $\boxed{\textbf{(B)}\ 12.25}$.

In a trapezoid with two right angles, drawing a line from the top of the shorter side straight across makes a rectangle and a right triangle. The right triangle connects the slanted side, the height, and the difference of the two parallel sides. The condition $BC = AB + CD$ is chosen so that after expanding, everything cancels except the product $AB \cdot CD$, which is exactly what is asked.

Since the answer does not depend on the exact lengths, choose numbers that are easy. Suppose $BC = 25$. The right triangle has sides $7$ and $b - a$ with long side $25$, so $b - a = \sqrt{625 - 49} = \sqrt{576} = 24$. With $a + b = 25$ and $b - a = 24$: $b = 24.5$ and $a = 0.5$. Then $ab = 0.5 \times 24.5 = 12.25$.