AMC 10 · 2006 · #5
Grade 4 geometry-2dPick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Big rectangle sets a floor
The 3×4 rectangle's longest side is 4, so the square's side must be at least 4; the choices leave sides 4, 5, 6, 7, 8 to test.
A box cannot hold a stick longer than the box itself.
2.G.A.1Visualize Spatial RelationshipsArea rules out side 4
Their areas add to 2×3 + 3×4 = 6 + 12 = 18, but a 4×4 square offers only 16 — so side 4 cannot hold both.
Shapes that need more area than the box holds can never fit inside it.
A shape that needs more area than the box holds can never fit inside it.
▸ Why?
Anything placed inside inherits the container's limits, so its area cannot exceed the container's.
▸ Why?
A square's area grows as the square of its side, so a small side buys very little room.
Side 5 actually works
In a 5×5 square lay the 3×4 along the bottom, 4 wide and 3 tall; the leftover 5-by-2 top strip takes the 2×3 with no overlap.
Put one rectangle on the bottom and slide the other into the leftover strip above it.
2.G.A.1Draw A DiagramSmallest square found
Side 4 is impossible and side 5 works, so the smallest square has side 5 and area 5×5 = 25. The answer is (B).
The smallest side that can hold both rectangles without overlap gives the smallest area.
3.OA.C.7Extreme PrincipleFirst 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.
- Big rectangle sets a floor
- Area rules out side 4
- Side 5 actually works
- Smallest square found