AMC 10 · 2005 · #4
Grade 8 geometry-2dA rectangle with a diagonal of length x is twice as long as it is wide. What is the area of the rectangle?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $x$.
Givens: The rectangle's length is twice its width; The diagonal has length $x$; Answer choices: (A) $\frac{1}{4}x^2$, (B) $\frac{2}{5}x^2$, (C) $\frac{1}{2}x^2$, (D) $x^2$, (E) $\frac{3}{2}x^2$
Unknowns: The area of the rectangle, expressed in terms of $x$
Understand
Restated: 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 $x$.
Givens: The rectangle's length is twice its width; The diagonal has length $x$; Answer choices: (A) $\frac{1}{4}x^2$, (B) $\frac{2}{5}x^2$, (C) $\frac{1}{2}x^2$, (D) $x^2$, (E) $\frac{3}{2}x^2$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
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 $2w^2$, so we only need the value of $w^2$, not $w$ itself, and the Pythagorean equation hands us $w^2$ directly.
Execute — Answer: B
6.EE.B.6 Step 1 Name the two sides
- Let the width be $w$.
- Since the rectangle is twice as long as it is wide, the length is $2w$.
- Now every measurement in the problem can be written using the single unknown $w$.
💡 Naming the width as one variable lets both sides — and the area — grow out of a single unknown.
8.G.B.7 Step 2 Diagonal is the hypotenuse
- Draw the rectangle and its diagonal.
- The diagonal cuts the rectangle into two right triangles; one of them has legs $w$ (the width) and $2w$ (the length), with the diagonal $x$ as its hypotenuse.
- By the Pythagorean theorem, the square of the hypotenuse equals the sum of the squares of the legs.
💡 A rectangle's diagonal always makes a right triangle with the two sides, so the Pythagorean theorem ties the diagonal to them.
6.EE.A.2 Step 3 Solve for $w^2$, not $w$
- Simplify the right side: $(2w)^2 = 4w^2$, so $x^2 = w^2 + 4w^2 = 5w^2$.
- Dividing both sides by $5$ gives $w^2 = \frac{x^2}{5}$.
- Notice we do not need $w$ by itself — the value of $w^2$ is all the area calculation will require.
💡 The area depends on $w^2$, so getting $w^2$ from the equation is enough — no square roots needed.
6.EE.A.2 Step 4 Compute the area
- The area of the rectangle is length times width: $2w \times w = 2w^2$.
- Substitute $w^2 = \frac{x^2}{5}$: the area is $2 \cdot \frac{x^2}{5} = \frac{2}{5}x^2$.
- So the area is $\frac{2}{5}x^2$, choice (B).
💡 Once $w^2$ is known, the area formula $2w^2$ turns straight into an expression in $x^2$.
6.EE.B.6 Let the width be $w$. Since the rectangle is twice as long as it is wide, the le 8.G.B.7 Draw the rectangle and its diagonal. The diagonal cuts the rectangle into two ri 6.EE.A.2 Simplify the right side: $(2w)^2 = 4w^2$, so $x^2 = w^2 + 4w^2 = 5w^2$. Dividing 6.EE.A.2 The area of the rectangle is length times width: $2w \times w = 2w^2$. Substitut Review
Reasonableness: Test with concrete numbers: let the width be $1$ and the length be $2$. Then the diagonal is $x=\sqrt{1^2+2^2}=\sqrt{5}$, so $x^2=5$, and the true area is $1\times 2 = 2$. Plugging $x^2=5$ into choice (B) gives $\frac{2}{5}\cdot 5 = 2$, 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 $x^2$ — which every choice respects — but only (B) produces the correct constant.
Alternative: Skip the algebra and plug in a number from the start: choose width $1$ and length $2$, so $x^2 = 1^2 + 2^2 = 5$ and the area is $2$. Then just ask which answer choice equals $2$ when $x^2 = 5$. Only $\frac{2}{5}x^2 = \frac{2}{5}\cdot 5 = 2$ works, pointing straight to (B).
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the width $w$ and writing the length as $2w$, so both sides and the area are expressed from a single unknown.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Relating the diagonal $x$ to the sides through the right triangle it forms: $x^2 = w^2 + (2w)^2$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Simplifying $(2w)^2=4w^2$ to reach $x^2=5w^2$, solving for $w^2=\frac{x^2}{5}$, and substituting into the area $2w^2$ to get $\frac{2}{5}x^2$.)
⭐ 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 $w^2$ without ever finding the side itself.
⭐ 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 $w^2$ without ever finding the side itself.
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