AMC 10 · 2019 · #8
Grade 8 geometry-2dProblem
The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A square has four equilateral triangles drawn inward — one sitting on each side, all with side length $2$, and all four apexes meeting at the center of the square. Find the area inside the square but outside the triangles (the shaded region).
Givens: Each equilateral triangle has side $2$; One side of each triangle lies on one side of the square; All four apexes meet at the square's center; Choices: (A) $4$, (B) $12 - 4\sqrt{3}$, (C) $3\sqrt{3}$, (D) $4\sqrt{3}$, (E) $16 - 4\sqrt{3}$
Unknowns: Area of the shaded region (square minus the four triangles)
Understand
Restated: A square has four equilateral triangles drawn inward — one sitting on each side, all with side length $2$, and all four apexes meeting at the center of the square. Find the area inside the square but outside the triangles (the shaded region).
Givens: Each equilateral triangle has side $2$; One side of each triangle lies on one side of the square; All four apexes meet at the square's center; Choices: (A) $4$, (B) $12 - 4\sqrt{3}$, (C) $3\sqrt{3}$, (D) $4\sqrt{3}$, (E) $16 - 4\sqrt{3}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #16 Change Focus / Count the Complement, #3 Eliminate Possibilities
The shaded shape is irregular, but it equals (whole square) − (four triangles) — Tool #16 says count the complement. Split into (a) find the square's side from the triangle height (Tool #7 sub-question), (b) find the square's area, (c) find each triangle's area, (d) subtract. Tool #1 keeps the picture honest as we work.
Execute — Answer: B
8.G.B.7 Step 1 - Sub-question 1: How tall is one equilateral triangle of side $2$?
- Drop an altitude — it splits the equilateral triangle into two $30\text{-}60\text{-}90$ right triangles with hypotenuse $2$ and base $1$.
- By the Pythagorean theorem, the altitude $h$ satisfies $1^2 + h^2 = 2^2$.
💡 An equilateral triangle of side $s$ has altitude $\tfrac{s\sqrt{3}}{2}$ — here that's $\sqrt{3}$.
7.G.B.6 Step 2 - Sub-question 2: What's the side of the square?
- The apex of every triangle sits at the square's center.
- So the perpendicular distance from any side of the square to the center is $\sqrt{3}$, which is half the side.
- Therefore the side equals $2\sqrt{3}$.
💡 Center-to-side distance is half the square's width.
6.G.A.1 Step 3 - Sub-question 3: Square area.
- Side squared.
💡 Square area is just side times side.
6.G.A.1 Step 4 - Sub-question 4: Total triangle area.
- Each equilateral triangle has area $\tfrac{1}{2} \cdot 2 \cdot \sqrt{3} = \sqrt{3}$.
- There are four such triangles, and they don't overlap (each sits in a different ‘pie slice’ of the square meeting at the center).
💡 Four congruent non-overlapping triangles → multiply one area by four.
6.G.A.1 Step 5 Subtract triangle area from square area to get the shaded area.
💡 Whole minus the white triangles equals what's shaded.
7.NS.A.3 Step 6 - Sanity check against choices.
- $\sqrt{3} \approx 1.73$, so $4\sqrt{3} \approx 6.93$ and $12 - 4\sqrt{3} \approx 5.07$.
- That's roughly $42\%$ of the square's area of $12$ — plausible for the corner ‘pockets’ left over.
💡 Eyeballed area matches the picture.
8.G.B.7 Sub-question 1: How tall is one equilateral triangle of side $2$? Drop an altitu 7.G.B.6 Sub-question 2: What's the side of the square? The apex of every triangle sits a 6.G.A.1 Sub-question 3: Square area. Side squared. 6.G.A.1 Sub-question 4: Total triangle area. Each equilateral triangle has area $\tfrac{ 6.G.A.1 Subtract triangle area from square area to get the shaded area. 7.NS.A.3 Sanity check against choices. $\sqrt{3} \approx 1.73$, so $4\sqrt{3} \approx 6.9 Review
Reasonableness: The shaded area is positive (since $4\sqrt{3} \approx 6.93 < 12$) and clearly smaller than the square — good. The four triangles fill about $58\%$ of the square; what's left forms four small kite-shaped pockets at the corners. $12 - 4\sqrt{3}$ matches choice (B). Choices like $4$ or $4\sqrt{3}$ describe the triangles, not the leftover — those are trap answers.
Alternative: Tool #10 (Physical): cut the picture from paper. The square minus the four equilateral triangles leaves four congruent kite-shaped corners. Each kite's area is $\tfrac{1}{4}(12 - 4\sqrt{3}) = 3 - \sqrt{3}$; sum to $12 - 4\sqrt{3}$. Same answer (B).
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the altitude of the equilateral triangle ($h = \sqrt{3}$) from its half-base and side.)7.G.B.6Solve real-world problems involving area, surface area, and volume (Linking the triangle altitude to the square's side via the center-meeting condition.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the square's area $(2\sqrt{3})^2 = 12$ and each triangle's area $\tfrac{1}{2}\cdot 2\cdot\sqrt{3} = \sqrt{3}$.)7.NS.A.3Solve real-world problems involving the four operations with rational numbers (Subtracting $4\sqrt{3}$ from $12$ and estimating the result numerically for the sanity check.)
⭐ This AMC 10 problem only needs Grade 8 Pythagorean thinking you already know: each equilateral triangle has altitude $\sqrt{3}$, so the square has side $2\sqrt{3}$ and area $12$. The four triangles take up $4\sqrt{3}$, leaving $12 - 4\sqrt{3}$ for the shaded pockets.
⭐ This AMC 10 problem only needs Grade 8 Pythagorean thinking you already know: each equilateral triangle has altitude $\sqrt{3}$, so the square has side $2\sqrt{3}$ and area $12$. The four triangles take up $4\sqrt{3}$, leaving $12 - 4\sqrt{3}$ for the shaded pockets.
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