Competition · AMC preparation · step 4 of 4

AMC 8 · 2018 · #4

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

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

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A twelve-sided polygon is drawn on 1 cm × 1 cm graph paper with all twelve vertices on grid points. Find the area, in square centimeters, of the region enclosed by the polygon.

Pick an answer.

(A)

(B)

12.5

(C)

(D)

13.5

(E)

AMC 8 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The figure has an awkward 12-sided boundary, but Tool #7 (Identify Subproblems) reveals a clean split: a central 3 × 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 × side, and a right triangle as 1/2 × base × height. Tool #1 (Draw a Diagram) is the supporting move: marking the four "point" triangles on the picture and the 3 × 3 square they hug makes the decomposition impossible to miss.

1STEP 1

Break the polygon apart

Plot the 12 vertices and trace the outline: it splits into a central 3 × 3 square from (2,2) to (5,5) with four triangular tips poking out.

Decomposition: 12-gon = (3 × 3 square) + (top tri) + (right tri) + (bottom tri) + (left tri)

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

The area of the twelve-sided figure equals the area of the central 3 × 3 square plus the areas of its four outward triangles.

▸ Why?

The central square and the four outward triangles fit together to form the whole figure with no gaps and no overlaps, so the figure's area is exactly the sum of those five pieces' areas.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

The central square runs from x=2 to x=5 and from y=2 to y=5, so it is 3 units wide and 3 units tall, holding 3 rows of 3 unit squares.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

▸ Why?

Each outward tip is a triangle with a base of 2 units and a peak sitting 1 unit away from the middle of that base, and the four tips are the same triangle repeated.

▸ Why?

Drawing the straight line from a tip's peak down to the middle of its base splits the tip into a left and a right right-triangle, and each of those is exactly half of a 1 × 1 unit square, so the tip covers 1/2 + 1/2 = 1 unit of area.

▸ Why?

That straight line from peak to base cuts the tip into two right triangles that overlap nowhere and leave no gap, so their two areas add up to the tip's area.

Whole is the sum of its partsBreak a thing into pieces with no gaps or overlaps and the pieces add back to the whole.

▸ Why?

A unit square's diagonal turns one triangular half exactly onto the other, so the two halves have equal area and each right triangle is half of the unit square.

Rigid motion preserves length and angleSliding, turning, or flipping a shape lays it exactly onto a copy, so every length and angle stays the same (folding is the flip case).

▸ Why?

The four tips are the same triangle placed at top, right, bottom, and left, so together they cover four equal groups of one tip's area.

Multiplication as equal groupsa times b means a equal groups of b — it is just repeated adding of the same amount.

5.G.A.2Draw A Diagram

2STEP 2

Find the square's area

The central square runs from (2,2) to (5,5), so each side is 5 - 2 = 3 cm and its area is 9 cm².

Area_square = 3 × 3 = 9 cm²

3STEP 3

Find one triangle's area

Take the top tip (2,5), (3,6), (4,5): base 4 - 2 = 2 cm and height 1 cm, so its area is 1 cm².

Area_△ = 1/2 × base × height = 1/2 × 2 × 1 = 1 cm²

4STEP 4

Total the four triangles

By symmetry the right, bottom, and left tips are congruent to the top one, so the four triangles total 4 cm².

4 × 1 = 4 cm²

5STEP 5

Add the pieces

Add the central square and the four tips: 9 + 4 gives the enclosed area of 13 cm².

Total area = 9 + 4 = 13 cm² → (C)

Answer

The polygon fits inside a 5 × 5 bounding square from (1,1) to (6,6), so the area is at most 25 and clearly more than the central 3 × 3 = 9. The four small triangular tips look like roughly 1 cm² each by eye, so 9 + 4 = 13 matches the picture. Choice (C) is the only option in that neighborhood.

💡Key takeaway

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!

Break the polygon apart
Find the square's area
Find one triangle's area
Total the four triangles
Add the pieces

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Break the polygon apart

Find the square's area

Find one triangle's area

Total the four triangles

Add the pieces

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