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2010 AMC 10A Problem 2

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2010 AMC 10A Problem 2AMC 10 Step by Step
Name:                             Date: September 13, 2026
Quadrilaterals & Polygon Areas · difficulty 1 of 5 · about 1 min · position P1-10

Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?
figure

  1. A)

    54\dfrac{5}{4}

  2. B)

    43\dfrac{4}{3}

  3. C)

    32\dfrac{3}{2}

  4. D)

    22

  5. E)

    33

Answer

B

Key insight

Four small squares of side s span the big square's side, so the rectangle is 4s wide and 4s - s = 3s tall.

Solution

Let each small square have side ss. The four small squares sit in a row along the top edge, so the large square has side 4s4s.

The rectangle fills everything below that row. Its length (horizontal) is the full side of the large square, 4s4s. Its width (vertical) is what remains after removing the row of small squares: 4s−s=3s4s - s = 3s.

Therefore the length is 4s3s=43\dfrac{4s}{3s} = \dfrac{4}{3} times the width.

The answer is (B) 43\boxed{\textbf{(B)}\ \dfrac{4}{3}}.

Why this works

Assigning one variable to the smallest repeated piece lets every other dimension be read off the figure by counting and subtracting. The ratio is independent of ss, which is why the problem never needs an actual measurement.

The trap

Reading the rectangle's width as the small square's side and answering 4 or 2, or inverting the ratio to 3/4.

Common mistakes

  • Reading the rectangle's width as the small square's side and answering 4 or 2, or inverting the ratio to 3/4.
  • Forgetting that the large figure is a square, so its height must also equal 4s4s.

Techniques

Set up the equation/formula and compute; no special trick needed