AMC 8 · 1999 · #5

Easy mode Grade 4

Problem

Imagine a rectangular garden. It is 60 feet long and 20 feet wide. A fence goes all the way around it.

Now the gardener takes down the fence and uses the same pieces to build a new fence. The new fence forms a square instead of a rectangle. The total length of fence is exactly the same.

The new square garden is bigger than the old rectangular one. How many more square feet does the new garden cover?

Diagram

Same fence, two shapes — a 60-by-20 rectangle and a 40-by-40 square, each with perimeter 160 ft.

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.

View mode:

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

Diagram

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 4)

More like this