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

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Revision as of 14:23, 13 August 2019

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:

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

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

The area of the rectangle is $2w^2=\frac{2}{5}x^2$

$(B)$

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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