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

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Problem

A rectangular garden 60 feet long and 20 feet wide is enclosed by a fence. To make the garden larger, while using the same fence, its shape is changed to a square. By how many square feet does this enlarge the garden?

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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Original garden: rectangle $60$ ft long, $20$ ft wide; The same fence is reused to enclose a square garden; Answer choices: (A) $100$, (B) $200$, (C) $300$, (D) $400$, (E) $500$

Unknowns: Increase in area (square feet) when the rectangle becomes a square of equal perimeter

Understand

Givens: Original garden: rectangle $60$ ft long, $20$ ft wide; The same fence is reused to enclose a square garden; Answer choices: (A) $100$, (B) $200$, (C) $300$, (D) $400$, (E) $500$

Plan

Primary tool: #7 Break into Subproblems

Secondary: #1 Draw a Diagram

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 $60 \times 20$ rectangle next to a square shows that "same fence" means equal perimeter, which is the only link between the two shapes.

Execute — Answer: D

#7 Break into Subproblems 3.MD.D.8 Step 1

Subproblem 1: find the rectangle's perimeter (the fence length) and its area.

$P = 2(60 + 20) = 2 \times 80 = 160 \text{ ft}$, and $A_{\text{rect}} = 60 \times 20 = 1200 \text{ ft}^2$

💡 Perimeter and area of a rectangle are Grade 3 formulas. The perimeter $160$ is the length of fence we get to reuse.

#7 Break into Subproblems 3.MD.C.7 Step 2

Subproblem 2: the square is built from the same $160$-ft fence, so its four equal sides total $160$.
Divide by $4$ to get one side, then square it for the area.

$s = \dfrac{160}{4} = 40 \text{ ft}$, and $A_{\text{sq}} = 40 \times 40 = 1600 \text{ ft}^2$

💡 $P = 4s$ for a square, so $s = P/4$. Area of a square is side $\times$ side, another Grade 3 standard.

#7 Break into Subproblems 4.OA.A.3 Step 3

Subproblem 3: subtract the old area from the new area to get the increase.

$$1600 - 1200 = 400 \text{ ft}^2 \;\Rightarrow\; \textbf{(D)}$$

💡 "By how many" is a comparison subtraction. The Grade 4 four-operations standard covers exactly this kind of multi-step word-problem finish.

[1] #7 3.MD.D.8 Subproblem 1: find the rectangle's perimeter (the fence length) and its area.

[2] #7 3.MD.C.7 Subproblem 2: the square is built from the same $160$-ft fence, so its four equa

[3] #7 4.OA.A.3 Subproblem 3: subtract the old area from the new area to get the increase.

Review

Reasonableness: Quick check: $2(60+20) = 160$ and $4 \times 40 = 160$, so the fence really does fit both gardens. The rectangle is long and thin ($60 \times 20$) while the square is balanced ($40 \times 40$), 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 $40^2 = 1600$.

Alternative: Tool #1 (Draw a Diagram): sketch the $60 \times 20$ rectangle on grid paper — it covers $60 \times 20 = 1200$ unit squares. Reshape the same border into a $40 \times 40$ square; that grid covers $40 \times 40 = 1600$ unit squares. The extra unit squares between the two pictures are $1600 - 1200 = 400$, again giving (D). The diagram also makes it visible that a square "plumps up" the area for the same border length.

CCSS standards used (min grade 4)

3.MD.D.8 Solve real-world and mathematical problems involving perimeters of polygons (Computing the rectangle's perimeter $P = 2(60+20) = 160$ ft (the reusable fence length).)
3.MD.C.7 Relate area to the operations of multiplication and addition (Finding the rectangle's area $60 \times 20 = 1200$ and the square's area $40 \times 40 = 1600$ as side-times-side products.)
4.OA.A.3 Solve multistep word problems with whole numbers using the four operations (Chaining perimeter, division, multiplication, and subtraction across three subproblems to land on the area increase $1600 - 1200 = 400$.)

⭐ 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.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

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