AMC 10 · 2005 · #4

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

pythagorean-theoremarea-rectangles convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theoremarea-rectanglesexponents

📏 Medium solution 💡 2 insights

Problem

A rectangle is twice as long as it is wide, and its diagonal has length x. Find the area of the rectangle, written in terms of x.

Pick an answer.

(A)

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

(B)

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

(C)

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

(D)

$x^2$

(E)

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

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

No side length is given as a number, only the ratio of the sides and the diagonal, so Tool #4 (Introduce a Variable) lets us name the width and write both sides — and the area — from that single unknown. Tool #1 (Draw a Diagram) shows why the Pythagorean theorem applies: the diagonal splits the rectangle into a right triangle whose legs are the two sides. Tool #7 (Identify Subproblems) spots the shortcut — the area is 2w², so we only need the value of w², not w itself, and the Pythagorean equation hands us w² directly.

1STEP 1

Name the two sides

Let the width be w. Twice as long means the length is 2w — every length now rides on one unknown.

width=w, length=2w

2STEP 2

Diagonal is the hypotenuse

The diagonal cuts the rectangle into right triangles with legs w and 2w and hypotenuse x, so Pythagoras gives x² = w² + (2w)².

x² = w² + (2w)²

3STEP 3

Solve for w², not w

Since (2w)² = 4w², the equation becomes x² = 5w², so w² = x²/5 — and w itself is never needed.

x² = 5w² → w² = x²/5

4STEP 4

Compute the area

Area = 2w × w = 2w², and substituting w² = x²/5 gives 2/5x², choice (B).

A = 2w² = 2·x²/5 = 2/5x² → (B)

Answer

2/5x²

Test with concrete numbers: let the width be 1 and the length be 2. Then the diagonal is x=√(1²+2²)=√(5), so x²=5, and the true area is 1 × 2 = 2. Plugging x²=5 into choice (B) gives 2/5 · 5 = 2, an exact match. The other choices miss: (A) gives 1.25, (C) gives 2.5, (D) gives 5, and (E) gives 7.5. Since area has units of length squared and x is a length, the area must be a constant times x² — which every choice respects — but only (B) produces the correct constant.

💡Key takeaway

When a shape is described only by its diagonal, name one side as a variable and use the Pythagorean theorem — you can often reach the area from w² without ever finding the side itself.

Name the two sides
Diagonal is the hypotenuse
Solve for w², not w
Compute the area

Name the two sides

Diagonal is the hypotenuse

Solve for w², not w

Compute the area

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