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2005 AMC 12A Problems/Problem 3

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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 \...

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)}.

See also

2005 AMC 12A (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
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