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2005 AMC 10A Problems/Problem 4

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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) \ ...

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)}

See Also

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