AMC 8 · 1999 · #5

Grade 4 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Rectilinear Frame / Box Border

perimeterarea-rectanglesmulti-digit-arithmetic identify-subproblemsarea-difference ↑ Prerequisites: area-rectanglesperimetermulti-digit-arithmetic

📏 Medium solution 💡 3 insights

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Problem

A rectangular garden measuring 60 ft by 20 ft is enclosed by a fence. The same fence is rearranged to form a square. By how many square feet does the square garden exceed the rectangular garden in area?

Pick an answer.

(A)

100

(B)

200

(C)

300

(D)

400

(E)

500

AMC 8 1999 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Break into Subproblems

One sentence hides three short tasks, so Tool #7 (Break into Subproblems) keeps the work in order: (i) find the rectangle's perimeter and area, (ii) reuse that perimeter to get the square's side and area, (iii) subtract to get the increase. Tool #1 (Draw a Diagram) is the visual support — sketching the 60 × 20 rectangle next to a square shows that "same fence" means equal perimeter, which is the only link between the two shapes.

1STEP 1

Subproblem 1: the rectangle's perimeter is the reusable fence length 160 ft, and its area is 1200 ft².

P = 2(60 + 20) = 2 × 80 = 160 ft, and A_rect = 60 × 20 = 1200 ft²

2STEP 2

Subproblem 2: the same 160 ft fence makes a square of side 40 ft, so its area is 1600 ft².

s =

\frac{160}{4}

= 40 ft, and A_sq = 40 × 40 = 1600 ft²

3STEP 3

Subproblem 3: subtract the old area from the new area — the garden grows by 400 ft².

1600 - 1200 = 400 ft² → (D)

Answer

400

Quick check: 2(60+20) = 160 and 4 × 40 = 160, so the fence really does fit both gardens. The rectangle is long and thin (60 × 20) while the square is balanced (40 × 40), and for a fixed perimeter the square always gives the largest area, so the area must increase — consistent with answer (D). Numerically, 1600 - 1200 = 400. The other choices fail simple ratio checks: an increase of 100 or 200 would require the square's area to be far smaller than 40² = 1600.

💡Key takeaway

Same fence means same perimeter — and for a fixed perimeter a square holds more area than a long, thin rectangle. Three Grade 3-4 steps (perimeter, side, subtract) turn this AMC 8 problem into routine work.

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