2001 AMC 10 Problem 24

Since $AB$ and $CD$ are both perpendicular to $AD$, they are the parallel sides and $AD = 7$ is the height. Let $AB = a$ and $CD = b$ with $a < b$, so the slanted side is $BC = a + b$.

Draw the perpendicular from $B$ to $CD$, meeting it at $E$. Then $ABED$ is a rectangle, so $BE = 7$ and $DE = a$, leaving $EC = b - a$. In right triangle $BEC$:
$$(a + b)^2 = 7^2 + (b - a)^2.$$
Expand both squares: $a^2 + 2ab + b^2 = 49 + a^2 - 2ab + b^2$. The $a^2$ and $b^2$ terms cancel, leaving $4ab = 49$, so
$$ab = \frac{49}{4} = 12.25.$$

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

In a right trapezoid, dropping the perpendicular from the top of the shorter leg creates the right triangle that ties the slanted side, the height, and the difference of the bases together. The condition $BC = AB + CD$ is designed so that $(a+b)^2 - (a-b)^2 = 4ab$ isolates exactly the product asked for; the individual lengths are never needed and in fact are not determined.

Because only the product is fixed, pick a convenient case. Take $BC = 25$: then $b - a = \sqrt{25^2 - 7^2} = 24$, so $a = \frac{25 - 24}{2} = 0.5$ and $b = 24.5$, giving $ab = 12.25$.

• Trying to find AB and CD individually; only the product is determined, and the identity (a+b)^2 - (a-b)^2 = 4ab is what the problem is testing.
• Putting $b - a$ as the hypotenuse or using $AD = 7$ as a base instead of the height.