2001 AMC 10 Problem 24

Below is the professionally curated solution for Problem 24 of the 2001 AMC 10, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2001 AMC 10 solutions, or check the answer key.

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Concepts:trapezoid Pythagorean Theorem difference of squares

Difficulty rating: 1810

24.

In trapezoid $ABCD,$ $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD},$ with $AB+CD=BC,$ $AB\lt CD,$ and $AD=7.$ What is $AB\cdot CD?$

$12$

$12.25$

$12.5$

$12.75$

$13$

Solution:

Drop a perpendicular from $B$ to $CD,$ meeting it at $E.$ Then $BE=AD=7$ and $CE=CD-AB.$ By the Pythagorean theorem, $BC^2=BE^2+CE^2.$

Since $BC=CD+AB,$ $(CD+AB)^2-(CD-AB)^2=BE^2=49.$

The left side equals $4\cdot AB\cdot CD,$ so $AB\cdot CD=\dfrac{49}{4}=12.25.$

Thus, the correct answer is B.

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