AMC 10 · 2006 · #7
Grade 6 geometry-2dThe 8×18 rectangle ABCD is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is y?
Pick an answer.
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.
Toolkit + CCSS Solution
Understand
Restated: A rectangle $ABCD$ measuring $8$ by $18$ is cut along a staircase line into two identical hexagons. Those two hexagons can be slid around (no flipping needed, no overlap) so they fit together into a perfect square. In the picture, $y$ is the length of the short flat piece of the cut along the top-left edge (and the matching one on the bottom-right). Find $y$.
Givens: The rectangle is $8$ tall and $18$ wide, so it measures $8 \times 18$; One straight step cut splits it into two congruent (identical) hexagons; The two hexagons can be repositioned without overlap to form a square; $y$ is the flat top-left segment of the cut, running along the $18$-length edge from the corner; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Unknowns: The length $y$
Understand
Restated: A rectangle $ABCD$ measuring $8$ by $18$ is cut along a staircase line into two identical hexagons. Those two hexagons can be slid around (no flipping needed, no overlap) so they fit together into a perfect square. In the picture, $y$ is the length of the short flat piece of the cut along the top-left edge (and the matching one on the bottom-right). Find $y$.
Givens: The rectangle is $8$ tall and $18$ wide, so it measures $8 \times 18$; One straight step cut splits it into two congruent (identical) hexagons; The two hexagons can be repositioned without overlap to form a square; $y$ is the flat top-left segment of the cut, running along the $18$-length edge from the corner; Answer choices: (A) $6$, (B) $7$, (C) $8$, (D) $9$, (E) $10$
Plan
Primary tool: #17 Visualize Spatial Relationships
Secondary: #1 Draw a Diagram, #6 Guess and Check
The heart of the problem is imagining the two hexagons sliding together into a square, so Tool #17 (Visualize Spatial Relationships) leads: picture the pieces moving. Tool #1 (Draw a Diagram) supports it — sketch the finished square to see where $y$ lands. Tool #6 (Guess and Check) handles one small arithmetic step: find the number that, times itself, gives the square's area. The chain is short: area is preserved, so the square's side is fixed, and $y$ is a simple fraction of that side.
Execute — Answer: A
4.MD.A.3 Step 1 Area does not change
- When you cut a shape and slide the pieces without overlapping, no area is gained or lost.
- So the square must have the same area as the original rectangle.
- The rectangle is $8$ by $18$, giving an area of $8 \times 18 = 144$.
💡 Rearranging pieces is like moving puzzle parts around, the total amount of space stays exactly the same.
3.OA.A.4 Step 2 Find the square's side
- A square with area $144$ has a side length $s$ where $s \times s = 144$.
- Testing $s = 12$ gives $12 \times 12 = 144$, which fits.
- So each side of the square is $12$.
💡 A square's area is one side times itself, so ask which number squared makes $144$.
6.G.A.1 Step 3 Locate y in the square
- Slide the two hexagons together into the $12 \times 12$ square.
- The flat piece of length $y$ from the original cut ends up sitting along a side of the square, running from a corner to the exact middle of that side, because the two congruent pieces meet at the center.
- That means $y$ is half of a full side: $y = 12 \div 2 = 6$.
- So $y = 6$, which is choice (A).
💡 Two identical pieces meet at the middle, so the seam reaches exactly halfway across the square's side.
4.MD.A.3 When you cut a shape and slide the pieces without overlapping, no area is gained 3.OA.A.4 A square with area $144$ has a side length $s$ where $s \times s = 144$. Testing 6.G.A.1 Slide the two hexagons together into the $12 \times 12$ square. The flat piece o Review
Reasonableness: Check the numbers against the picture: a side of $12$ is a believable square built from an $8$-by-$18$ block, since $12$ sits between $8$ and $18$. And $y = 6$ is one third of the $18$-length edge, which matches the cut starting well before the middle. The value $6$ is the smallest choice offered and lands cleanly on a whole number, with no leftover gaps in the square.
Alternative: Track the width instead of the seam. The rectangle is $18$ wide, and the finished square is only $12$ wide, so the reposition must pull the width in by $18 - 12 = 6$. That shrink comes entirely from sliding one hexagon over by the length of its flat step, so that step length is $6$, again giving $y = 6$.
CCSS standards used (min grade 6)
4.MD.A.3Apply the area and perimeter formulas for rectangles (Computing the rectangle's area $8 \times 18 = 144$, which equals the square's area.)3.OA.A.4Determine the unknown whole number in a multiplication or division equation (Finding the square's side $s$ from $s \times s = 144$, giving $s = 12$.)6.G.A.1Find the area of polygons by composing into rectangles or decomposing into triangles and other shapes (Reasoning that the two congruent hexagons recompose into the $12 \times 12$ square and that $y$ reaches the midpoint of a side.)
⭐ Cutting and rearranging keeps the area the same, so find the square's side first, then read off where the marked length lands.
⭐ Cutting and rearranging keeps the area the same, so find the square's side first, then read off where the marked length lands.
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