AMC 8 · 2018 · #4

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-trianglesarea-rectanglescoordinate-geometryspatial-visualization area-differenceidentify-subproblems ↑ Prerequisites: area-rectanglesarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$ ?

$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$

Pick an answer.

(A)

(B)

12.5

(C)

(D)

13.5

(E)

View mode:

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

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

#1 Draw a Diagram 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 $(5,5)$.

$$\text{Decomposition: } \text{12-gon} = (3 \times 3 \text{ square}) + (\text{top tri}) + (\text{right tri}) + (\text{bottom tri}) + (\text{left tri})$$

💡 Plotting points on a coordinate grid to see structure is exactly the Grade 5 coordinate-plane skill.

#7 Identify Subproblems 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.

$$\text{Area}_{\text{square}} = 3 \times 3 = 9 \text{ cm}^2$$

💡 Area of a rectangle as side $\times$ side is a Grade 3 multiplication-as-area idea.

#7 Identify Subproblems 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.

$$\text{Area}_{\triangle} = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} \times 2 \times 1 = 1 \text{ cm}^2$$

💡 Finding triangle area on a coordinate grid is a Grade 6 polygon-area standard.

#7 Identify Subproblems 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$^2$.

$$4 \times 1 = 4 \text{ cm}^2$$

💡 Multiplying "how many congruent pieces" by "area per piece" is a Grade 4 multi-step word-problem move.

#7 Identify Subproblems 4.OA.A.3 Step 5

Add the pieces.
The total area is the central square plus the four outer triangles.

$$\text{Total area} = 9 + 4 = 13 \text{ cm}^2 \;\Rightarrow\; \textbf{(C)}$$

💡 Combining the part-areas back into the whole closes out the subproblem split — Grade 4 multi-step thinking.

[1] #1 5.G.A.2 Sketch the polygon on the grid and trace the boundary in order. Mark the four ou

[2] #7 3.MD.C.7 Compute the area of the central $3 \times 3$ square. Its corners are $(2,2), (5,

[3] #7 6.G.A.1 Compute the area of one outer triangle, say the top one with vertices $(2,5), (3

[4] #7 4.OA.A.3 By symmetry, the other three outer triangles — right $(5,5),(6,4),(5,3)$; bottom

[5] #7 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.2 Represent 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.7 Relate 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.1 Find 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.3 Solve 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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