Difference between revisions of "2001 AMC 10 Problems/Problem 24"

Revision as of 10:11, 4 July 2013

Problem

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

$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$

Solution

If $AB=x$ and $CD=y$ , we have $BC=x+y$ .

By the Pythagorean theorem, we have $(x+y)^2=(y-x)^2+49$

Solving the equation, we get $4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25}$ .

2001 AMC 10 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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Difference between revisions of "2001 AMC 10 Problems/Problem 24"

Revision as of 10:11, 4 July 2013

Problem

Solution

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