AMC 10 · 2008 · #2
Grade 6 geometry-2dA square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is 2:1. The ratio of the rectangle's length to its width is 2:1. What percent of the rectangle's area is inside the square?
Pick an answer.
AMC 10 2008 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 square sits inside a rectangle. The rectangle's width is twice the square's side, and the rectangle's length is twice its own width. Find what percent of the rectangle's area lies inside the square.
Givens: The rectangle's width to the square's side is in ratio $2:1$; The rectangle's length to its width is in ratio $2:1$; The square lies entirely inside the rectangle; Answer choices: (A) $12.5$, (B) $25$, (C) $50$, (D) $75$, (E) $87.5$
Unknowns: The square's area as a percent of the rectangle's area
Understand
Restated: A square sits inside a rectangle. The rectangle's width is twice the square's side, and the rectangle's length is twice its own width. Find what percent of the rectangle's area lies inside the square.
Givens: The rectangle's width to the square's side is in ratio $2:1$; The rectangle's length to its width is in ratio $2:1$; The square lies entirely inside the rectangle; Answer choices: (A) $12.5$, (B) $25$, (C) $50$, (D) $75$, (E) $87.5$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities
Because only ratios are given, the shape's size is free — so pick a name for the one length everything is compared to. Tool #4 (Introduce a Variable): let the square's side be $s$ and write every other length in terms of $s$. Tool #1 (Draw a Diagram) makes the two ratios concrete and confirms the square fits inside. Once both areas are in terms of $s$, the $s$ cancels and Tool #3 (Eliminate Possibilities) confirms the single matching choice.
Execute — Answer: A
6.RP.A.1 Step 1 Draw it and name the side
- Sketch the rectangle with the square inside it.
- Let the square's side be $s$.
- The first ratio says the width to the side is $2:1$, so the width is twice the side: width $= 2s$.
- Since the side $s$ is smaller than the width $2s$, the square really does fit inside.
💡 Naming the smallest length $s$ lets every other length be measured against it.
6.RP.A.3 Step 2 Use the second ratio for the length
- The second ratio says the length to the width is $2:1$, so the length is twice the width.
- The width is $2s$, so the length is $2 \times 2s = 4s$.
💡 A $2:1$ ratio just means doubling, so apply it a second time to the width.
4.MD.A.3 Step 3 Find both areas
- Area of a rectangle is length times width, so the rectangle's area is $4s \times 2s = 8s^2$.
- Area of a square is side times side, so the square's area is $s \times s = s^2$.
💡 Both areas come from the same $s$, so their sizes can be compared directly.
6.RP.A.3 Step 4 Turn the fraction into a percent
- The part inside the square is the square's area over the rectangle's area: $\dfrac{s^2}{8s^2}$.
- The $s^2$ cancels, leaving $\dfrac{1}{8}$.
- As a percent, $\dfrac{1}{8} = 0.125 = 12.5\%$.
- No choice larger than that fits, so the answer is (A).
💡 Because the variable cancels, the percent is fixed no matter how big the square is.
6.RP.A.1 Sketch the rectangle with the square inside it. Let the square's side be $s$. Th 6.RP.A.3 The second ratio says the length to the width is $2:1$, so the length is twice t 4.MD.A.3 Area of a rectangle is length times width, so the rectangle's area is $4s \times 6.RP.A.3 The part inside the square is the square's area over the rectangle's area: $\dfr Review
Reasonableness: The rectangle is $2 \times 4 = 8$ little squares each the size of the drawn square, so the square is exactly $1$ of those $8$ parts, i.e. $\tfrac{1}{8} = 12.5\%$. That matches. A value like $25\%$ would mean the rectangle held only $4$ such squares, but the width alone already spans $2$ squares and the length spans $4$, giving $8$ — so $12.5\%$ is the sensible answer.
Alternative: Pick a concrete number instead of a variable: let the side be $1$. Then the width is $2$, the length is $4$, the rectangle's area is $8$, and the square's area is $1$. The percent is $\tfrac{1}{8} = 12.5\%$, the same choice (A).
CCSS standards used (min grade 6)
6.RP.A.1Understand the concept of a ratio and use ratio language (Reading the $2:1$ width-to-side ratio as 'the width is twice the side' and naming the side $s$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Doubling the width to get the length, and converting the area fraction $\tfrac{1}{8}$ into the percent $12.5\%$.)4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Computing the rectangle's area as $4s \times 2s = 8s^2$ and the square's area as $s^2$.)
⭐ When a problem gives only ratios, name the smallest length and build the rest from it — the letter cancels at the end, so the width of 2 and length of 4 make 8 squares, and one of them is 12.5%.
⭐ When a problem gives only ratios, name the smallest length and build the rest from it — the letter cancels at the end, so the width of 2 and length of 4 make 8 squares, and one of them is 12.5%.
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