AMC 10 · 2014 · #13

Grade 8 geometry-2d

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

equilateral-trianglearea-trianglespythagorean-theorem identify-subproblemsspatial-visualization ↑ Prerequisites: area-trianglesequilateral-triangle

📏 Long solution 💡 3 insights 📊 Diagram

Problem

Equilateral $\triangle ABC$ has side length $1$ , and squares $ABDE$ , $BCHI$ , $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$ ?

Pick an answer.

(A)

$\dfrac{12+3\sqrt3}4$

(B)

$\dfrac92$

(C)

$3+\sqrt3$

(D)

$\dfrac{6+3\sqrt3}2$

(E)

AMC 10 2014 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.

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Toolkit + CCSS Solution

Understand

Restated: An equilateral triangle with side 1 has a square built outward on each of its three sides. Connecting the outer corners of the squares forms a six-sided figure. Find the area of that hexagon.

Givens: Triangle ABC is equilateral with side length 1; Squares ABDE, BCHI, CAFG each sit outside the triangle, one per side; Each square has side length 1, matching the triangle's side; Hexagon DEFGHI is formed by the six outer square-corners

Unknowns: The area of hexagon DEFGHI

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #17 Visualize Spatial Relationships

The hexagon is a compound shape. Instead of chasing its outer coordinates, cut it into pieces whose areas are easy: the central triangle, the three squares, and one small triangle tucked in at each vertex. Add the pieces back up.

Execute — Answer: C

#7 Identify Subproblems 6.G.A.1 Step 1

Cut the hexagon into pieces

The hexagon DEFGHI covers exactly three kinds of region: the central equilateral triangle ABC, the three squares glued to its sides, and three small triangles that fill the gaps at the corners (triangle AEF at vertex A, and the matching ones at B and C).
Nothing is counted twice and nothing is left out, so the hexagon's area is the sum of these seven pieces.

$$[DEFGHI] = [ABC] + 3\,[\text{square}] + 3\,[\text{corner triangle}]$$

💡 A messy shape becomes easy once you slice it into shapes you already know.

#1 Draw a Diagram 8.G.B.7 Step 2

Area of the central triangle

Triangle ABC is equilateral with side 1.
Dropping an altitude splits it into two right triangles; by the Pythagorean theorem the height is the square root of 1 minus one-quarter, which is root three over two.
So the area is one-half times base 1 times that height.

$$h=\sqrt{1^2-\left(\tfrac12\right)^2}=\frac{\sqrt3}{2},\quad [ABC]=\frac12\cdot 1\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{4}$$

💡 An equilateral triangle's height always comes from splitting it in half and using the Pythagorean theorem.

#7 Identify Subproblems 6.G.A.1 Step 3

Area of the three squares

Each square is built on a side of length 1, so each has area 1 times 1 = 1.
There are three of them.

$$3\times(1\cdot 1)=3$$

💡 A square on a unit side has area one, so three of them contribute exactly three.

#17 Visualize Spatial Relationships 7.G.B.5 Step 4

Find the corner angle

Look at vertex A.
Four regions meet there and their angles fill the full turn of 360 degrees: the triangle's own 60-degree angle, and a 90-degree angle from each of the two squares.
What is left over is the angle EAF of the small corner triangle, so it is 360 minus 60 minus 90 minus 90.

$$\angle EAF = 360^\circ-60^\circ-90^\circ-90^\circ = 120^\circ$$

💡 The angles crammed around one point must add to a full 360 degrees, so the leftover is forced.

#1 Draw a Diagram 8.G.B.7 Step 5

Area of one corner triangle

Triangle AEF has AE = AF = 1 (each is a square side) with the 120-degree angle between them.
Dropping an altitude from A splits it into two 30-60-90 right triangles with hypotenuse 1.
Each contributes half of the base root three over two and a height of one-half, giving area root three over four.
Three corner triangles give three times that.

$$[AEF]=\frac12\cdot\underbrace{\sqrt3}_{\text{base}}\cdot\underbrace{\tfrac12}_{\text{height}}=\frac{\sqrt3}{4},\quad 3\,[AEF]=\frac{3\sqrt3}{4}$$

💡 Splitting the 120-degree wedge in half makes two familiar 30-60-90 triangles you can measure.

#7 Identify Subproblems 7.G.B.6 Step 6

Add the pieces

Sum the seven regions: the central triangle root three over four, the three squares totaling 3, and the three corner triangles totaling three root three over four.
The two square-root parts combine into a single root three, so the total is 3 plus root three.
That matches choice (C).

$$\frac{\sqrt3}{4}+3+\frac{3\sqrt3}{4}=3+\sqrt3$$

💡 Once every piece is measured, the whole is just their sum.

[1] #7 6.G.A.1 The hexagon DEFGHI covers exactly three kinds of region: the central equilateral

[2] #1 8.G.B.7 Triangle ABC is equilateral with side 1. Dropping an altitude splits it into two

[3] #7 6.G.A.1 Each square is built on a side of length 1, so each has area 1 times 1 = 1. Ther

[4] #17 7.G.B.5 Look at vertex A. Four regions meet there and their angles fill the full turn of

[5] #1 8.G.B.7 Triangle AEF has AE = AF = 1 (each is a square side) with the 120-degree angle b

[6] #7 7.G.B.6 Sum the seven regions: the central triangle root three over four, the three squa

Review

Reasonableness: Rough size check: three unit squares alone give 3, and the leftover triangles are all thin slivers, so the total should sit a little above 3. The answer 3 + root three is about 4.73, which fits. Choices like 6 or 9/2 would need far more filler area than the small corner triangles provide.

Alternative: Extend the outer square edges until they meet, wrapping the figure in one big equilateral triangle of side 1 + root three. Its area minus the three obtuse corner triangles that stick out beyond the hexagon also gives 3 + root three, confirming the decomposition.

CCSS standards used (min grade 8)

6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing the hexagon into a triangle, three squares, and three corner triangles, and summing their areas)
7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles (Finding the 120-degree corner angle by subtracting the known angles that fill the 360 degrees around each vertex)
7.G.B.6 Solve real-world problems involving area, surface area, and volume (Combining the composite pieces into the final total area of the hexagon)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing the altitudes (root three over two, and the corner-triangle height) that give each triangle's area)

⭐ Slice a strange shape into a triangle, some squares, and the little triangles filling the corner gaps, then add every piece.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Cut the hexagon into pieces

Area of the central triangle

Area of the three squares

Find the corner angle

Area of one corner triangle

Add the pieces

Review

CCSS standards used (min grade 8)

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