AMC 8 · 2012 · #23
Grade 7 geometry-2dProblem
An equilateral triangle and a regular hexagon have equal perimeters. If the triangle's area is 4, what is the area of the hexagon?
Pick an answer.
AMC 8 2012 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: An equilateral triangle and a regular hexagon have the same perimeter. The triangle has area $4$. Find the area of the hexagon.
Givens: An equilateral triangle with area $4$; A regular hexagon; Perimeter of the triangle $=$ perimeter of the hexagon; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $4\sqrt{3}$, (E) $6\sqrt{3}$
Unknowns: The area of the regular hexagon
Understand
Restated: An equilateral triangle and a regular hexagon have the same perimeter. The triangle has area $4$. Find the area of the hexagon.
Givens: An equilateral triangle with area $4$; A regular hexagon; Perimeter of the triangle $=$ perimeter of the hexagon; Answer choices: (A) $4$, (B) $5$, (C) $6$, (D) $4\sqrt{3}$, (E) $6\sqrt{3}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
Tool #1 (Diagram) is the natural first move for a 2D geometry problem with no figure given — drawing the hexagon reveals a key fact: a regular hexagon is exactly $6$ small equilateral triangles joined at the center. That turns the hexagon's area into a sum we can control. Tool #7 (Subproblems) splits the work into two clean pieces: (a) compare the side of one small triangle to the side of the big triangle, and (b) use the side ratio to get the area of one small triangle, then multiply by $6$. Tool #9 (Easier Problem) covers the side-to-area scaling: equilateral triangles are all similar, so doubling the side multiplies the area by $2^2 = 4$ — no square-root-of-$3$ formula needed.
Execute — Answer: C
6.EE.B.7 Step 1 - Match perimeters to relate the sides.
- Let the triangle's side be $s_t$ and the hexagon's side be $s_h$.
- The triangle has $3$ sides, the hexagon has $6$, and the perimeters are equal: $3 s_t = 6 s_h$, so $s_t = 2 s_h$.
- The triangle's side is twice the hexagon's side.
💡 Writing one equation from the equal-perimeter condition and solving for $s_t$ is the Grade 6 one-step-equation move.
6.G.A.1 Step 2 - Draw the hexagon and split it into $6$ congruent equilateral triangles.
- A regular hexagon's center connects to all $6$ vertices, cutting it into $6$ identical equilateral triangles whose side length equals the hexagon's side $s_h$.
- So the hexagon's area is $6$ times the area of one small triangle.
💡 Decomposing a polygon into triangles to find its area is the Grade 6 "area by composition/decomposition" standard.
7.G.A.1 Step 3 - Compare the small triangle (side $s_h$) to the big triangle (side $s_t = 2 s_h$).
- Both are equilateral, so they are similar with side-length ratio $2$.
- For similar figures, the ratio of areas is the square of the ratio of sides, so the big triangle's area is $2^2 = 4$ times the small triangle's area.
💡 The "side scales by $k$ $\Rightarrow$ area scales by $k^2$" rule is the Grade 7 scale-drawing fact, used here without needing the $\tfrac{\sqrt{3}}{4}s^2$ formula.
6.EE.B.7 Step 4 - Plug in the given big-triangle area to find the small triangle's area.
- The big triangle has area $4$, and it is $4$ times the small one, so the small one has area $1$.
💡 Solving $4 \cdot A_{\text{small}} = 4$ for $A_{\text{small}}$ is a Grade 6 one-step equation.
6.G.A.1 Step 5 - Multiply by $6$ to get the hexagon's area.
- There are $6$ congruent small triangles inside the hexagon, each with area $1$.
💡 Adding up the $6$ congruent pieces of the decomposed hexagon gives its total area directly.
6.EE.B.7 Match perimeters to relate the sides. Let the triangle's side be $s_t$ and the h 6.G.A.1 Draw the hexagon and split it into $6$ congruent equilateral triangles. A regula 7.G.A.1 Compare the small triangle (side $s_h$) to the big triangle (side $s_t = 2 s_h$) 6.EE.B.7 Plug in the given big-triangle area to find the small triangle's area. The big t 6.G.A.1 Multiply by $6$ to get the hexagon's area. There are $6$ congruent small triangl Review
Reasonableness: The big triangle covers area $4$ with $3$ long sides; the hexagon has $6$ shorter sides ($s_h = \tfrac{1}{2} s_t$) but more of them, and it is much closer to a circle than a triangle is. With the same perimeter, the more circle-like shape should enclose more area — so the hexagon's area being larger than $4$ matches intuition. Answer $6$ also matches the "hexagon $=$ $6$ unit triangles" picture exactly. The other choices fail this picture: $4$ (no gain at all), $5$ (does not split into $6$ equal pieces of area $1$), and the $\sqrt{3}$ choices come from misusing the $\tfrac{\sqrt{3}}{4} s^2$ formula and forgetting that we never need a numeric side length here.
Alternative: Tool #13 (Convert to Algebra) with the explicit area formula also works. Let the triangle's side be $s_t$. Its area is $\tfrac{\sqrt{3}}{4} s_t^2 = 4$, so $s_t^2 = \tfrac{16}{\sqrt{3}}$. The hexagon's side is $s_h = \tfrac{s_t}{2}$, and a regular hexagon has area $\tfrac{3\sqrt{3}}{2} s_h^2 = \tfrac{3\sqrt{3}}{2} \cdot \tfrac{s_t^2}{4} = \tfrac{3\sqrt{3}}{8} \cdot \tfrac{16}{\sqrt{3}} = 6$. Same answer, but the toolkit path skips the $\sqrt{3}$ entirely — which is why we prefer it.
CCSS standards used (min grade 7)
6.EE.B.7Solve real-world and mathematical problems by writing and solving equations of the form $px = q$ (Setting $3 s_t = 6 s_h$ from the equal-perimeter condition and solving to get $s_t = 2 s_h$, and later solving $4 \cdot A_{\text{small}} = 4$ to get $A_{\text{small}} = 1$.)6.G.A.1Find the area of polygons by composing into rectangles or decomposing into triangles (Decomposing the regular hexagon into $6$ congruent equilateral triangles and adding their areas to get the hexagon's total area.)7.G.A.1Solve problems involving scale drawings of geometric figures, including computing actual lengths and areas (Using the rule that if the side scales by $2$, the area scales by $2^2 = 4$, so the small equilateral triangle's area is $\tfrac{1}{4}$ of the big triangle's area.)
⭐ Cut the hexagon into $6$ small equilateral triangles — the big triangle has the same shape but doubled side, so its area is $4$ times one small triangle. That makes each small triangle area $1$, and $6 \times 1 = 6$.
⭐ Cut the hexagon into $6$ small equilateral triangles — the big triangle has the same shape but doubled side, so its area is $4$ times one small triangle. That makes each small triangle area $1$, and $6 \times 1 = 6$.
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