2005 AMC 10A Problem 4

Below is the professionally curated solution for Problem 4 of the 2005 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 10A solutions, or check the answer key.

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Concepts:rectangle Pythagorean Theorem area

Difficulty rating: 1100

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

$\dfrac{1}{4}x^2$

$\dfrac{2}{5}x^2$

$\dfrac{1}{2}x^2$

$x^2$

$\dfrac{3}{2}x^2$

Solution:

Let the width be $w,$ so the length is $2w.$ Then $x^2 = w^2 + (2w)^2 = 5w^2,$ giving $w^2 = \dfrac{x^2}{5}.$ The area is $w \cdot 2w = 2w^2 = \dfrac{2}{5}x^2.$

Thus, the correct answer is B.

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