2010 AMC 10A Problem 2

Give the small squares an easy size and read every other length off the picture.

1. Say each small square has side $1$.
2. The four small squares sit side by side across the top, so the top edge of the big square is $1 + 1 + 1 + 1 = 4$ long. A square has equal sides, so the big square is $4$ by $4$.
3. The rectangle fills the rest of the big square, under the row of small squares. Its long side runs all the way across: $4$.
4. Its short side is the big square's height minus the row of small squares on top: $4 - 1 = 3$.
5. So the length is $\dfrac{4}{3}$ times the width.

Using side $s$ instead of $1$ gives $\dfrac{4s}{3s}$, the same ratio, so the choice of $1$ does not matter.

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

When a figure is built from identical pieces, give the smallest piece a size and count your way to every other length. A ratio does not depend on the actual size, so picking side $1$ is safe and makes the numbers easy.

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

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

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

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