2005 AMC 12A Problem 3

Attempt Problem 3 of the 2005 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2005 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

Let the width be

w,

so the length is

2w

The diagonal satisfies

x^2 = w^2 + (2w)^2

Let the width be

w.

Then the length is

2w,

and the diagonal gives

x^2 = w^2 + (2w)^2 = 5w^2.

The area is

w \cdot 2w = 2w^2 = \dfrac{2}{5}x^2.

Thus, the correct answer is B.

Problem 3 in other years