AMC 10 · 2016 · #9

Grade 6 geometry-2d
perimeterarea-rectanglessystems-of-equations convert-to-algebra ↑ Prerequisites: perimeter
📏 Medium solution 💡 2 insights
Problem
Evenly spaced posts run around a rectangle, with one side holding twice as many. Find the area.

Pick an answer.

(A)
256
(B)
336
(C)
384
(D)
448
(E)
512
How to solve
Strategy Introduce a Variable

The post counts on the two sides are linked by a 'twice as many' relationship and a fixed total, so naming the shorter-side count with a variable turns the words into one equation. A quick sketch keeps the shared corner posts from being counted twice, and once a side's post count is known the length is a small multiplication subproblem.

1STEP 1

Name the posts per side

One letter names the posts on both kinds of side.

n = posts on a shorter side, 2n = posts on a longer side
2STEP 2

Count posts around the rectangle

The corners must be corrected for twice.

2n + 2(2n) - 4 = 20
3STEP 3

Solve for n

Solving gives 4 posts on a short side.

6n - 4 = 20 → 6n = 24 → n = 4
4STEP 4

Turn posts into side lengths

Gaps, not posts, give the lengths.

3 × 4 = 12 yd, 7 × 4 = 28 yd
5STEP 5

Compute the area

Multiplying gives 336, choice (B).

12 × 28 = 336 square yards
Answer
336
Check the post count directly: 4 + 8 + 4 + 8 = 24 side-tallies, minus 4 doubly counted corners = 20 posts, exactly what Carl bought. The 12 by 28 rectangle gives 336 square yards, which is answer (B) and sits sensibly between the listed choices.
💡Key takeaway

Count the gaps, not the posts: a row of posts always has one fewer space than posts, and corners get shared by two sides.

  • Name the posts per side
  • Count posts around the rectangle
  • Solve for n
  • Turn posts into side lengths
  • Compute the area