2002 AMC 10A Problem 25

Below is the professionally curated solution for Problem 25 of the 2002 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10A solutions, or check the answer key.

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Concepts:trapezoid similarity Pythagorean Triple

Difficulty rating: 1790

25.

In trapezoid $ABCD$ with bases $\overline{AB}$ and $\overline{CD},$ we have $AB=52,$ $BC=12,$ $CD=39,$ and $DA=5.$ The area of $ABCD$ is

$182$

$195$

$210$

$234$

$260$

Solution:

Extend $DA$ and $CB$ to meet at $P.$ Since $DC\parallel AB,$ $\triangle PDC\sim\triangle PAB$ with ratio $\dfrac{39}{52}=\dfrac{3}{4}.$ From $\dfrac{PD}{PD+5}=\dfrac{3}{4}$ we get $PD=15,$ and similarly $PC=36.$

Then $PD:PC:DC=15:36:39=3\cdot(5:12:13),$ so $\angle P$ is a right angle. The area of $ABCD$ is $\dfrac{1}{2}(PA)(PB)-\dfrac{1}{2}(PD)(PC)=\dfrac{1}{2}(20)(48)-\dfrac{1}{2}(15)(36)=480-270=210.$

Thus, the correct answer is C.

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