Art of Problem Solving

2005 AMC 12B Problems/Problem 3: Difference between revisions

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== Problem ==
== Problem ==
A rectangle with a diagonal of length <math>x</math> is twice as long as it is wide.  What is the area of the rectangle?
<math>
\mathrm{(A)}\ \frac14x^2      \qquad
\mathrm{(B)}\ \frac25x^2      \qquad
\mathrm{(C)}\ \frac12x^2      \qquad
\mathrm{(D)}\ x^2      \qquad
\mathrm{(E)}\ \frac32x^2
</math>


== Solution ==
== Solution ==

Revision as of 20:40, 17 April 2009

Problem

A rectangle with a diagonal of length $x$ is twice as long as it is wide. What is the area of the rectangle?

$\mathrm{(A)}\ \frac14x^2      \qquad \mathrm{(B)}\ \frac25x^2      \qquad \mathrm{(C)}\ \frac12x^2      \qquad \mathrm{(D)}\ x^2      \qquad \mathrm{(E)}\ \frac32x^2$

Solution

See also