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2017 AMC 10A Problem 3

2017 AMC 10A Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Quadrilaterals & Polygon Areas · difficulty 1 of 5 · about 2 min · position P1-10

Tamara has three rows of two $6$ -feet by $2$ -feet flower beds in her garden. The beds are separated and also surrounded by $1$ -foot-wide walkways, as shown on the diagram. What is the total area of the walkways, in square feet?

A)
$72$
B)
$78$
C)
$90$
D)
$120$
E)
$150$

Answer

Key insight

Walkway area is the whole garden minus the six beds: the garden is 15 by 10 once you add the 1-foot walkways between and around the beds.

Solution

Measure the whole garden. Left to right there are two beds of width $6$ and three walkways of width $1$ (left edge, middle, right edge): $1 + 6 + 1 + 6 + 1 = 15$ feet. Top to bottom there are three beds of height $2$ and four walkways: $1 + 2 + 1 + 2 + 1 + 2 + 1 = 10$ feet.

Total garden area: $15 \cdot 10 = 150$ square feet.

The six beds each measure $6 \times 2 = 12$ square feet, for $6 \cdot 12 = 72$ square feet of beds.

Walkways are everything else: $150 - 72 = 78$ .

The answer is $\boxed{\textbf{(B)}\ 78}$ .

Why this works

An irregular region (the walkways) is easiest to measure as a complement: a simple outer rectangle minus simple inner rectangles. The only care needed is counting the walkway strips correctly; there is always one more walkway than there are beds in each direction because the outer border counts too.

The trap

Forgetting a walkway strip when measuring the garden (using 14 by 9 or 13 by 8), or computing the bed area and stopping.

Common mistakes

Forgetting a walkway strip when measuring the garden (using 14 by 9 or 13 by 8), or computing the bed area and stopping.
Reporting $72$ (the bed area) or $150$ (the whole garden) instead of the difference.

Techniques

Set up the equation/formula and compute; no special trick needed · Cut the figure into known shapes (triangles, rectangles, sectors)