Difference between revisions of "2005 AMC 10A Problems/Problem 4"

Revision as of 11:29, 4 July 2013

Problem

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

$\mathrm{(A) \ } \frac{1}{4}x^2\qquad \mathrm{(B) \ } \frac{2}{5}x^2\qquad \mathrm{(C) \ } \frac{1}{2}x^2\qquad \mathrm{(D) \ } x^2\qquad \mathrm{(E) \ } \frac{3}{2}x^2$

Solution

Let the width of the rectangle be $w$ . Then the length is $2w$

Using the Pythagorean Theorem:

$x^{2}=w^{2}+(2w)^{2}$

$x^{2}=5w^{2}$

$w=\frac{x}{\sqrt{5}}$

$2w=\frac{2x}{\sqrt{5}}$

So the area of the rectangle is $w \cdot 2w = \frac{x}{\sqrt{5}} \cdot \frac{2x}{\sqrt{5}} = \frac{2}{5}x^{2} \Longrightarrow \mathrm{(B)}$

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Difference between revisions of "2005 AMC 10A Problems/Problem 4"

Revision as of 11:29, 4 July 2013

Problem

Solution

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