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2005 AMC 10A Problem 4

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2005 AMC 10A Problem 4AMC 10 Step by Step

Name: Date: September 13, 2026

Triangles: Area & Pythagorean · difficulty 1 of 5 · about 2 min · position P1-10

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

A)
$\frac{1}{4}x^2$
B)
$\frac{2}{5}x^2$
C)
$\frac{1}{2}x^2$
D)
$x^2$
E)
$\frac{3}{2}x^2$

Answer

Key insight

With width w and length 2w, the diagonal satisfies x^2 = 5w^2, and the area 2w^2 is two fifths of x^2.

Solution

Let the width be $w$ , so the length is $2w$ . The diagonal is the hypotenuse of a right triangle with legs $w$ and $2w$ :
$x^2 = w^2 + (2w)^2 = 5w^2, \qquad\text{so}\qquad w^2 = \frac{x^2}{5}.$
The area is length times width:
$2w \cdot w = 2w^2 = \frac{2}{5}x^2.$

The answer is $\boxed{\textbf{(B)}\ \frac{2}{5}x^2}$ .

Why this works

A rectangle's diagonal, length and width always form a right triangle, so the Pythagorean theorem links the diagonal to the sides. Once the sides are expressed through one parameter, both $x^2$ and the area become multiples of that parameter squared, and the ratio between them is the answer.

Alternative approach

Pick numbers: width $1$ , length $2$ , area $2$ , diagonal $\sqrt{5}$ . Test the choices with $x^2 = 5$ : only $\frac{2}{5} \cdot 5 = 2$ matches.

The trap

Writing the diagonal as w + 2w or forgetting to square, then matching the wrong coefficient of x^2.

Common mistakes

Writing the diagonal as w + 2w or forgetting to square, then matching the wrong coefficient of x^2.
Computing $w^2 = x^2/5$ correctly but reporting $\frac{1}{5}x^2$ or $\frac{1}{2}x^2$ after forgetting that the area is $2w^2$ .

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer