AMC 10 · 2002 · #19
Grade 7 geometry-2dSpot's doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach?
Pick an answer.
AMC 10 2002 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 doghouse has a regular six-sided base, each side $1$ yard. A dog is tied to one corner with a rope $2$ yards long. Find the total area of the ground outside the doghouse that the rope lets the dog cover.
Givens: The base is a regular hexagon with side length $1$ yard; Each interior angle of a regular hexagon is $120^\circ$; The rope is attached at one corner (vertex) and is $2$ yards long; Only the region outside the doghouse counts; Answer choices: (A) $2\pi/3$, (B) $2\pi$, (C) $5\pi/2$, (D) $8\pi/3$, (E) $3\pi$
Unknowns: The area of the region outside the hexagon that the dog can reach with the rope
Understand
Restated: A doghouse has a regular six-sided base, each side $1$ yard. A dog is tied to one corner with a rope $2$ yards long. Find the total area of the ground outside the doghouse that the rope lets the dog cover.
Givens: The base is a regular hexagon with side length $1$ yard; Each interior angle of a regular hexagon is $120^\circ$; The rope is attached at one corner (vertex) and is $2$ yards long; Only the region outside the doghouse counts; Answer choices: (A) $2\pi/3$, (B) $2\pi$, (C) $5\pi/2$, (D) $8\pi/3$, (E) $3\pi$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #17 Visualize Spatial Relationships
The reachable region is an odd shape, so Tool #7 (Identify Subproblems) breaks it into pieces that are each a plain circular sector whose area is easy. Tool #1 (Draw a Diagram) pins down the angles: the doghouse blocks part of the swing at the tie-corner, and the rope can bend around the two neighboring corners. Tool #17 (Visualize Spatial Relationships) tracks how much rope is left after it wraps a corner and how far it can then swing. Add the sectors to get the total.
Execute — Answer: E
4.MD.C.7 Step 1 Swing the rope around the tie-corner
- Stand at the corner where the rope is tied.
- The doghouse fills the interior angle there, which for a regular hexagon is $120^\circ$.
- The rope can swing everywhere else around that corner, so it sweeps $360^\circ - 120^\circ = 240^\circ$ using its full $2$-yard length.
- That $240^\circ$ is $\tfrac{240}{360} = \tfrac{2}{3}$ of a full circle.
💡 The wall takes up the $120^\circ$ corner, so the rope is free to sweep the leftover $240^\circ$ around it.
7.G.B.4 Step 2 Area of the big sector
- A full circle of radius $2$ has area $\pi r^2 = \pi (2)^2 = 4\pi$.
- The dog only gets $\tfrac{2}{3}$ of that circle, so this main piece has area $\tfrac{2}{3}\times 4\pi = \tfrac{8\pi}{3}$.
💡 Taking a fraction of a circle's area is just that fraction times $\pi r^2$.
4.MD.C.7 Step 3 Rope wraps around each neighbor corner
- Pull the rope straight along one side to the next corner.
- That uses $1$ yard, leaving $2 - 1 = 1$ yard.
- The leftover rope can now bend around this corner and keep sweeping.
- Coming in straight along the side, it can turn until the next wall stops it; since the wall bends the outline by $180^\circ - 120^\circ = 60^\circ$, the leftover rope sweeps a $60^\circ$ sector of radius $1$.
- The same thing happens at the neighbor corner on the other side, giving a second $60^\circ$ sector.
💡 Past the corner the rope is shorter and only a small wedge opens up before the next wall blocks it.
7.G.B.4 Step 4 Area of the two small sectors
- Each small sector is $\tfrac{60}{360} = \tfrac{1}{6}$ of a circle of radius $1$, whose full area is $\pi(1)^2 = \pi$.
- So each small sector is $\tfrac{1}{6}\pi = \tfrac{\pi}{6}$, and there are two of them: $2\times \tfrac{\pi}{6} = \tfrac{\pi}{3}$.
💡 Two matching $60^\circ$ slivers of a unit circle add up to a $120^\circ$ slice, which is $\tfrac{1}{3}$ of the circle.
7.G.B.6 Step 5 Add every piece
- Put the big sector and the two small sectors together: $\tfrac{8\pi}{3} + \tfrac{\pi}{3} = \tfrac{9\pi}{3} = 3\pi$.
- That is the whole area the dog can reach outside the doghouse, so the answer is (E).
💡 The regions do not overlap, so the total reachable area is simply their areas added.
4.MD.C.7 Stand at the corner where the rope is tied. The doghouse fills the interior angl 7.G.B.4 A full circle of radius $2$ has area $\pi r^2 = \pi (2)^2 = 4\pi$. The dog only 4.MD.C.7 Pull the rope straight along one side to the next corner. That uses $1$ yard, le 7.G.B.4 Each small sector is $\tfrac{60}{360} = \tfrac{1}{6}$ of a circle of radius $1$, 7.G.B.6 Put the big sector and the two small sectors together: $\tfrac{8\pi}{3} + \tfrac Review
Reasonableness: A rough check: the main $240^\circ$ sweep alone is $\tfrac{8\pi}{3}\approx 8.4$, and the two little wraps add only about $1.0$ more, landing near $3\pi\approx 9.4$. The total must be more than $\tfrac{8\pi}{3}$ (choice D) because the wrap-around pieces genuinely add area, and among the choices only $3\pi$ exceeds $\tfrac{8\pi}{3}$. That pins the answer to (E).
Alternative: Combine the angles first instead of the areas. The full sweep is $240^\circ$ at radius $2$ plus two $60^\circ$ arcs at radius $1$. Written as fractions of a circle: $\tfrac{2}{3}(4\pi) + \tfrac{1}{3}(\pi) = \tfrac{8\pi}{3}+\tfrac{\pi}{3} = 3\pi$, the same result.
CCSS standards used (min grade 7)
7.G.B.4Know the formulas for area and circumference of a circle (Computing each swept region as a fraction of a circle's area $\pi r^2$: the $\tfrac{2}{3}$ sector of radius $2$ and the $\tfrac{1}{6}$ sectors of radius $1$.)4.MD.C.7Recognize angle measure as additive and solve addition and subtraction problems (Finding the sweep angles: $360^\circ - 120^\circ = 240^\circ$ at the tie-corner and $180^\circ - 120^\circ = 60^\circ$ at each neighbor corner.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Adding the non-overlapping sector areas $\tfrac{8\pi}{3} + \tfrac{\pi}{3} = 3\pi$ to get the total reachable region.)
⭐ When a rope longer than one wall bends around a corner, break the space it covers into circle slices — a big slice at the tie point plus small ones where the rope wraps — and add them up.
⭐ When a rope longer than one wall bends around a corner, break the space it covers into circle slices — a big slice at the tie point plus small ones where the rope wraps — and add them up.
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