AMC 10 · 2006 · #12
Grade 7 geometry-2dRolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. His preliminary drawings are shown.
Which of these arrangements give the dog the greater area to roam, and by how many square feet?
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 dog is tied to a square shed ($16$ ft per side) with an $8$-ft rope. In Arrangement I the rope is tied to the middle of one side; in Arrangement II it is tied to a point $4$ ft from a corner. Find which arrangement lets the dog roam a larger area, and by how much.
Givens: Square shed, each side $16$ ft; Rope length $8$ ft, tied to the outside of one wall; Arrangement I: rope anchored at the midpoint of a side (so $8$ ft from each corner of that side); Arrangement II: rope anchored $4$ ft from one corner (so $12$ ft from the other corner); Answer choices compare the two regions as a multiple of $\pi$: (A) I by $8\pi$, (B) I by $6\pi$, (C) II by $4\pi$, (D) II by $8\pi$, (E) II by $10\pi$
Unknowns: The roaming area for Arrangement I; The roaming area for Arrangement II; Which is larger and the difference in square feet
Understand
Restated: A dog is tied to a square shed ($16$ ft per side) with an $8$-ft rope. In Arrangement I the rope is tied to the middle of one side; in Arrangement II it is tied to a point $4$ ft from a corner. Find which arrangement lets the dog roam a larger area, and by how much.
Givens: Square shed, each side $16$ ft; Rope length $8$ ft, tied to the outside of one wall; Arrangement I: rope anchored at the midpoint of a side (so $8$ ft from each corner of that side); Arrangement II: rope anchored $4$ ft from one corner (so $12$ ft from the other corner); Answer choices compare the two regions as a multiple of $\pi$: (A) I by $8\pi$, (B) I by $6\pi$, (C) II by $4\pi$, (D) II by $8\pi$, (E) II by $10\pi$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #17 Visualize Spatial Relationships
The whole problem is about what region a rope can sweep against a wall, so Tool #1 (Draw a Diagram) is the natural lead: sketch the anchor, the wall, and the arc the taut rope traces. The key move — noticing that a rope tied near a corner bends around it — is spatial (Tool #17), and once the rope wraps, the roaming area splits into a main arc plus a smaller leftover arc, which is a clean Tool #7 (Identify Subproblems) decomposition. Each piece is a fraction of a circle, so comparing the two arrangements reduces to adding and comparing multiples of $\pi$.
Execute — Answer: C
7.G.B.4 Step 1 Arrangement I: a clean semicircle
- The rope is tied at the midpoint of a $16$-ft wall, so each corner of that wall is exactly $8$ ft away — the same as the rope length.
- The taut rope reaches each corner with nothing to spare, so it never bends around a corner.
- The dog sweeps a half-circle of radius $8$ against the wall.
💡 The wall blocks the back half of the circle, so the dog gets exactly half of a full circle of radius $8$.
7.G.B.4 Step 2 Arrangement II: the same semicircle first
- Now the rope is tied $4$ ft from one corner.
- Straight out from the wall, before any bending, the dog still sweeps a half-circle of radius $8$ along the wall — the same $32\pi$ as before.
- This is the first subproblem.
💡 Moving the anchor along the wall does not change the half-circle it can sweep against that wall — a semicircle of radius $8$ is a semicircle of radius $8$.
7.G.B.4 Step 3 The rope bends around the near corner
- The anchor is only $4$ ft from the near corner, but the rope is $8$ ft.
- So the rope reaches that corner with $8 - 4 = 4$ ft left over, then bends around the corner and pivots there.
- That leftover $4$ ft sweeps a quarter-circle of radius $4$ over the top of the shed — brand-new roaming area.
- (The other corner is $12$ ft away, farther than the $8$-ft rope, so no bending happens there.)
💡 Once the rope hugs the corner, that corner becomes a new pivot and the leftover length draws a fresh arc around it.
7.EE.A.1 Step 4 Add up Arrangement II and compare
- Arrangement II's area is the semicircle plus the wrapped quarter-circle.
- Compare it to Arrangement I.
- Because both areas are multiples of $\pi$, subtracting is just combining like terms.
💡 Both regions share the $32\pi$ semicircle, so the extra wrap-around quarter-circle ($4\pi$) is exactly how much more room Arrangement II gives.
7.G.B.4 The rope is tied at the midpoint of a $16$-ft wall, so each corner of that wall 7.G.B.4 Now the rope is tied $4$ ft from one corner. Straight out from the wall, before 7.G.B.4 The anchor is only $4$ ft from the near corner, but the rope is $8$ ft. So the r 7.EE.A.1 Arrangement II's area is the semicircle plus the wrapped quarter-circle. Compare Review
Reasonableness: Sanity-check the boundary logic. In Arrangement I the rope length ($8$) equals the distance to each corner ($8$), so the rope arrives at the corner with zero slack — no wrap, pure $32\pi$; that borderline is exactly why Arrangement I gets no bonus. In Arrangement II the anchor sits $4$ ft from the near corner, strictly less than $8$, guaranteeing a wrap, and the leftover $4$ ft is smaller than the original $8$, so the bonus arc has the smaller radius $4$ — giving a quarter-circle of only $4\pi$, not a full-size chunk. The far corner ($12$ ft) is out of reach, so there is no second bonus. Both arrangements share the identical $32\pi$ front semicircle, so the whole contest is decided by that single $4\pi$ wrap-around — Arrangement II wins by $4\pi$, matching (C). The trap answers (D) $8\pi$ and (E) $10\pi$ overcount by imagining a bigger or second bonus arc.
Alternative: Skip the shared semicircle entirely (Tool #16, change focus to the difference). Both arrangements sweep the exact same radius-$8$ semicircle against the wall — $32\pi$ each — so it cancels when you subtract. The only thing that can break the tie is corner-wrapping. Arrangement I wraps nothing (anchor is $8$ ft from each corner, rope is $8$ ft). Arrangement II wraps the near corner with $8-4=4$ ft left, adding a quarter-circle of radius $4$, area $\tfrac14\pi(4)^2 = 4\pi$. Difference $= 4\pi$ in favor of II, so (C) — reached without ever computing $32\pi$.
CCSS standards used (min grade 7)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Computing each swept region as a fraction of a circle: the radius-$8$ semicircle ($\tfrac12\pi\cdot8^2 = 32\pi$) and the wrapped radius-$4$ quarter-circle ($\tfrac14\pi\cdot4^2 = 4\pi$).)5.NF.B.4Apply and extend previous understandings of multiplication to multiply a fraction or whole number by a fraction (Taking a half ($\tfrac12$) and a quarter ($\tfrac14$) of a full circle's area to get the semicircle and quarter-circle pieces.)7.EE.A.1Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients (Combining and comparing the $\pi$-terms as like terms: $32\pi + 4\pi = 36\pi$ and $36\pi - 32\pi = 4\pi$.)
⭐ When a rope is tied closer to a corner than its own length, it bends around the corner and the leftover rope sweeps an extra arc — here that one extra quarter-circle is the whole $4\pi$ difference.
⭐ When a rope is tied closer to a corner than its own length, it bends around the corner and the leftover rope sweeps an extra arc — here that one extra quarter-circle is the whole $4\pi$ difference.
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