Difference between revisions of "2005 AMC 12A Problems/Problem 3"

Latest revision as of 20:15, 3 July 2013

Problem

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

$(\mathrm {A}) \ \frac 14x^2 \qquad (\mathrm {B}) \ \frac 25x^2 \qquad (\mathrm {C})\ \frac 12x^2 \qquad (\mathrm {D}) \ x^2 \qquad (\mathrm {E})\ \frac 32x^2$

Solution

Let $w$ be the width, so the length is $2w$ . By the Pythagorean Theorem, $w^2 + 4w^2 = x^2 \Longrightarrow \frac{x}{\sqrt{5}} = w$ . The area of the rectangle is $(w)(2w) = 2w^2 = 2\left(\frac{x}{\sqrt{5}}\right)^2 = \frac{2}{5}x^2 \ \mathrm{(B)}$ .

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Difference between revisions of "2005 AMC 12A Problems/Problem 3"

Latest revision as of 20:15, 3 July 2013

Problem

Solution

See also