AMC 8 · 2018 · #4
Grade 6 geometry-2dProblem
The twelve-sided figure shown has been drawn on 1 cm×1 cm graph paper. What is the area of the figure in cm2?
Pick an answer.
AMC 8 2018 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 twelve-sided polygon is drawn on $1 \text{ cm} \times 1 \text{ cm}$ graph paper with all twelve vertices on grid points. Find the area, in square centimeters, of the region enclosed by the polygon.
Givens: The graph paper has unit squares of side $1$ cm; The twelve vertices, read in order around the polygon, are $(1,3), (2,4), (2,5), (3,6), (4,5), (5,5), (6,4), (5,3), (5,2), (4,1), (3,2), (2,2)$; Every vertex lies on a lattice point; Answer choices: (A) $12$, (B) $12.5$, (C) $13$, (D) $13.5$, (E) $14$
Unknowns: The area of the twelve-sided polygon, in $\text{cm}^2$
Understand
Restated: A twelve-sided polygon is drawn on $1 \text{ cm} \times 1 \text{ cm}$ graph paper with all twelve vertices on grid points. Find the area, in square centimeters, of the region enclosed by the polygon.
Givens: The graph paper has unit squares of side $1$ cm; The twelve vertices, read in order around the polygon, are $(1,3), (2,4), (2,5), (3,6), (4,5), (5,5), (6,4), (5,3), (5,2), (4,1), (3,2), (2,2)$; Every vertex lies on a lattice point; Answer choices: (A) $12$, (B) $12.5$, (C) $13$, (D) $13.5$, (E) $14$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram
The figure has an awkward 12-sided boundary, but Tool #7 (Identify Subproblems) reveals a clean split: a central $3 \times 3$ square sitting on the grid from $(2,2)$ to $(5,5)$, plus four congruent right triangles poking outward (one above, one below, one left, one right). Each piece has an area that elementary geometry handles directly — a square as side $\times$ side, and a right triangle as $\tfrac{1}{2} \times$ base $\times$ height. Tool #1 (Draw a Diagram) is the supporting move: marking the four "point" triangles on the picture and the $3 \times 3$ square they hug makes the decomposition impossible to miss.
Execute — Answer: C
5.G.A.2 Step 1 - Sketch the polygon on the grid and trace the boundary in order.
- Mark the four outward-pointing tips at $(3,6)$ (top), $(6,4)$ (right), $(4,1)$ (bottom), $(1,3)$ (left).
- Notice that the eight remaining vertices — $(2,2), (5,2), (5,5), (2,5)$ together with the four "shoulders" $(3,2), (5,3), (4,5), (2,4)$ — all sit on the boundary of the $3 \times 3$ square with corners $(2,2)$ and $(5,5)$.
💡 Plotting points on a coordinate grid to see structure is exactly the Grade 5 coordinate-plane skill.
3.MD.C.7 Step 2 - Compute the area of the central $3 \times 3$ square.
- Its corners are $(2,2), (5,2), (5,5), (2,5)$, so each side is $5 - 2 = 3$ cm long.
💡 Area of a rectangle as side $\times$ side is a Grade 3 multiplication-as-area idea.
6.G.A.1 Step 3 - Compute the area of one outer triangle, say the top one with vertices $(2,5), (3,6), (4,5)$.
- Its base lies on the line $y = 5$ with length $4 - 2 = 2$ cm, and its apex $(3,6)$ sits $1$ cm above the base.
💡 Finding triangle area on a coordinate grid is a Grade 6 polygon-area standard.
4.OA.A.3 Step 4 - By symmetry, the other three outer triangles — right $(5,5),(6,4),(5,3)$; bottom $(4,1),(3,2),(5,2)$; left $(1,3),(2,4),(2,2)$ — are congruent to the top one.
- Each has base $2$ cm and height $1$ cm, so each has area $1$ cm$^2$, giving a combined outer-triangle area of $4 \times 1 = 4$ cm$^2$.
💡 Multiplying "how many congruent pieces" by "area per piece" is a Grade 4 multi-step word-problem move.
4.OA.A.3 Step 5 - Add the pieces.
- The total area is the central square plus the four outer triangles.
💡 Combining the part-areas back into the whole closes out the subproblem split — Grade 4 multi-step thinking.
5.G.A.2 Sketch the polygon on the grid and trace the boundary in order. Mark the four ou 3.MD.C.7 Compute the area of the central $3 \times 3$ square. Its corners are $(2,2), (5, 6.G.A.1 Compute the area of one outer triangle, say the top one with vertices $(2,5), (3 4.OA.A.3 By symmetry, the other three outer triangles — right $(5,5),(6,4),(5,3)$; bottom 4.OA.A.3 Add the pieces. The total area is the central square plus the four outer triangl Review
Reasonableness: The polygon fits inside a $5 \times 5$ bounding square from $(1,1)$ to $(6,6)$, so the area is at most $25$ and clearly more than the central $3 \times 3 = 9$. The four small triangular tips look like roughly $1$ cm$^2$ each by eye, so $9 + 4 = 13$ matches the picture. Choice (C) is the only option in that neighborhood.
Alternative: Tool #5 (Look for a Pattern) via Pick's Theorem on lattice polygons: count $B = 12$ boundary lattice points (the $12$ vertices, with no extra lattice points on any segment) and $I = 8$ interior lattice points (one at $(4,2)$, three at $y = 3$: $(2,3),(3,3),(4,3)$, three at $y = 4$: $(3,4),(4,4),(5,4)$, one at $(3,5)$). Then $\text{Area} = I + \tfrac{B}{2} - 1 = 8 + 6 - 1 = 13$ cm$^2$, confirming (C).
CCSS standards used (min grade 6)
5.G.A.2Represent real-world and mathematical problems by graphing points (Plotting the twelve vertices on the coordinate grid and visually identifying the central $3 \times 3$ square together with the four outward triangles.)3.MD.C.7Relate area to multiplication and addition operations (Computing the area of the central $3 \times 3$ square as $3 \times 3 = 9$ cm$^2$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area of one outward triangle as $\tfrac{1}{2} \times 2 \times 1 = 1$ cm$^2$ using the triangle area formula on a coordinate grid.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Multiplying $4 \times 1 = 4$ for the four congruent triangles, then adding $9 + 4 = 13$ to combine the central square with the outer triangles.)
⭐ This AMC 8 problem only needs the Grade 6 triangle-area rule (half of base times height) you already know, plugged into a simple square-plus-four-triangles split!
⭐ This AMC 8 problem only needs the Grade 6 triangle-area rule (half of base times height) you already know, plugged into a simple square-plus-four-triangles split!
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