AMC 10 · 2006 · #5

Grade 4 geometry-2d

Archetype Grid Tiling / Packing with Divisibility └ Sub-archetype Fixed Piece Set — Feasibility by Elimination

area-rectanglesspatial-visualizationoptimization guess-and-check ↑ Prerequisites: area-rectangles

📏 Medium solution 💡 2 insights

Problem

A $2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2006 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 2×3 rectangle and a 3×4 rectangle must sit inside one square without overlapping, and every rectangle side stays parallel to the square's sides. Find the smallest possible area of that square.

Givens: A 2×3 rectangle; A 3×4 rectangle; Both rectangles lie inside one square; The two rectangles do not overlap at any point; Each rectangle's sides are parallel to the square's sides

Unknowns: The smallest possible area of the square

Understand

Givens: A 2×3 rectangle; A 3×4 rectangle; Both rectangles lie inside one square; The two rectangles do not overlap at any point; Each rectangle's sides are parallel to the square's sides

Plan

Primary tool: #14 Extreme Principle

Secondary: #1 Draw a Diagram, #17 Visualize Spatial Relationships

The question asks for a smallest value, so treat it as a minimum problem. First find a lower bound the square's side must clear, then show the smallest allowed size actually holds both rectangles. A diagram and some spatial reasoning confirm that a real placement exists.

Execute — Answer: B

#17 Visualize Spatial Relationships 2.G.A.1 Step 1

Big rectangle sets a floor

The 3×4 rectangle by itself needs room.
Its longest side is 4, so any square holding it must have a side of at least 4.
That leaves the square's side as 4, 5, 6, 7, or 8, which match the answer areas 16, 25, 36, 49, and 64.
Start testing from the smallest, side 4.

$$\text{side} \ge 4$$

💡 A box cannot hold a stick longer than the box itself.

#14 Extreme Principle 4.MD.A.3 Step 2

Area rules out side 4

Add the two rectangle areas: 2×3 = 6 and 3×4 = 12, so together they cover 18 square units.
A 4×4 square has only 16 square units.
Since 18 is greater than 16, the two rectangles cannot both fit inside a 4×4 square without overlapping.

$$2\cdot 3 + 3\cdot 4 = 6 + 12 = 18 > 16 = 4\cdot 4$$

💡 Shapes that need more area than the box holds can never fit inside it.

#1 Draw a Diagram 2.G.A.1 Step 3

Side 5 actually works

Try a 5×5 square.
Lay the 3×4 rectangle along the bottom so it is 4 wide and 3 tall.
That leaves a strip 5 wide and 2 tall across the top.
Place the 2×3 rectangle in that strip so it is 3 wide and 2 tall.
Both rectangles sit inside the square with no overlap.

$$3\times 4 \text{ (bottom)} \;+\; 2\times 3 \text{ (top strip)} \subset 5\times 5$$

💡 Put one rectangle on the bottom and slide the other into the leftover strip above it.

#14 Extreme Principle 3.OA.C.7 Step 4

Smallest square found

Side 4 is impossible and side 5 works, so the smallest square has side 5.
Its area is 5×5 = 25.
The answer is (B).

$$5\times 5 = 25$$

💡 The smallest side that can hold both rectangles without overlap gives the smallest area.

[1] #17 2.G.A.1 The 3×4 rectangle by itself needs room. Its longest side is 4, so any square hol

[2] #14 4.MD.A.3 Add the two rectangle areas: 2×3 = 6 and 3×4 = 12, so together they cover 18 squ

[3] #1 2.G.A.1 Try a 5×5 square. Lay the 3×4 rectangle along the bottom so it is 4 wide and 3 t

[4] #14 3.OA.C.7 Side 4 is impossible and side 5 works, so the smallest square has side 5. Its ar

Review

Reasonableness: The answer 25 is one of the offered choices and sits just above the impossible 16. The two rectangles cover 18 square units, so any working square needs area at least 18; 25 is the smallest listed square area that clears 18 and also admits a real placement. A side of 5 comfortably contains the longest rectangle side of 4.

Alternative: Start from area instead of from the big rectangle. The rectangles cover 18 square units, so the square must have area at least 18, which rules out 16 at once. Then test the next choice, 25, by exhibiting a placement, which succeeds.

CCSS standards used (min grade 4)

2.G.A.1 Recognize and draw shapes having specified attributes (Reasoning that a 3×4 rectangle needs a container at least 4 units wide, and drawing a valid placement of both rectangles inside the square)
4.MD.A.3 Apply area and perimeter formulas for rectangles in real-world problems (Using the rectangle areas to show the combined 18 square units cannot fit in a 16-unit square)
3.OA.C.7 Fluently multiply and divide within 100 (Computing the rectangle and square areas 6, 12, 16, and 25)

⭐ First find the size too small to hold everything, then show the next size up really fits, and that next size is the smallest that works.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Big rectangle sets a floor

Area rules out side 4

Side 5 actually works

Smallest square found

Review

CCSS standards used (min grade 4)

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