2002 AMC 10A Problem 25

Attempt Problem 25 of the 2002 AMC 10A below, then check your answer against the professionally curated solution 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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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

Answer: C

Concepts:trapezoid similarity Pythagorean Triple

Difficulty rating: 1790

Small hint:

Extend the legs

DA

and

CB

until they meet at a point

P

Big hint:

\triangle PDC\sim\triangle PAB

with ratio

\dfrac{39}{52}=\dfrac{3}{4}

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

\begin{gathered} \dfrac{1}{2}(PA)(PB) \\ {}-\dfrac{1}{2}(PD)(PC) \\ = \dfrac{1}{2}(20)(48) \\ {}-\dfrac{1}{2}(15)(36) \\ = 480-270=210. \end{gathered}

Thus, the correct answer is C.