2001 AMC 10 Problems/Problem 24

Revision as of 12:05, 2 August 2015 by Eugenis (talk | contribs) (Solution)

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 $CD=y$,then $BC=x+y$. By the Pythagorean theorem, we have $(x+y)^2=(y-x)^2+49$ Solving the equation, we get $4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25}$.

See Also

2001 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS