AMC 8 · 2012 · #21
Grade 6 geometry-3dProblem
Marla has a large white cube that has an edge of 10 feet. She also has enough green paint to cover 300 square feet. Marla uses all the paint to create a white square centered on each face, surrounded by a green border. What is the area of one of the white squares, in square feet?
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: A white cube has edge length $10$ feet. Marla paints a green border around a white square centered on every face, using up all of her green paint, which is enough to cover exactly $300$ square feet. What is the area of one white square, in square feet?
Givens: The cube has edge length $10$ feet, so each face is a $10 \times 10$ square; The cube has $6$ congruent faces; Total green paint available $= 300$ square feet, and Marla uses it all; Each face is split into a centered white square and a green border (which together cover the whole face); Answer choices: (A) $5\sqrt{2}$, (B) $10$, (C) $10\sqrt{2}$, (D) $50$, (E) $50\sqrt{2}$ (sq ft)
Unknowns: The area, in square feet, of one white square centered on a face
Understand
Restated: A white cube has edge length $10$ feet. Marla paints a green border around a white square centered on every face, using up all of her green paint, which is enough to cover exactly $300$ square feet. What is the area of one white square, in square feet?
Givens: The cube has edge length $10$ feet, so each face is a $10 \times 10$ square; The cube has $6$ congruent faces; Total green paint available $= 300$ square feet, and Marla uses it all; Each face is split into a centered white square and a green border (which together cover the whole face); Answer choices: (A) $5\sqrt{2}$, (B) $10$, (C) $10\sqrt{2}$, (D) $50$, (E) $50\sqrt{2}$ (sq ft)
Plan
Primary tool: #11 Use Symmetry
Secondary: #7 Identify Subproblems
Every face of the cube is treated the same way, so by symmetry the $300$ sq ft of green paint splits evenly into $6$ equal portions — one per face. That is Tool #11 (Use Symmetry): instead of tracking the whole cube, we only have to think about one face. Then Tool #7 (Identify Subproblems) handles that one face: its $100$ sq ft area is made of two pieces, white square $+$ green border, so the white area is just $100$ minus the green portion. No square roots or Pythagoras needed — the radical choices are distractors.
Execute — Answer: D
3.MD.C.7 Step 1 - Find the area of one face of the cube.
- Each face is a $10$ ft by $10$ ft square.
💡 Area of a rectangle as side $\times$ side is the Grade 3 area standard.
6.G.A.4 Step 2 - Use symmetry to share the green paint across the $6$ faces.
- The cube has $6$ congruent faces, each painted with the same pattern, so each face must receive the same share of the $300$ sq ft of green paint.
💡 Splitting the painted surface evenly across $6$ identical faces is exactly the "surface area as the sum of face areas" idea from Grade 6 nets.
4.MD.A.3 Step 3 - Break a single face into its two pieces.
- On any one face, the green border and the centered white square together cover the whole face, so their areas add to $100$ sq ft.
💡 Treating a face as the sum of its non-overlapping regions is the Grade 4 "area as additive" principle.
4.MD.A.3 Step 4 Plug in the green area on one face from Step 2 and solve for the white area.
💡 Subtracting the green piece from the whole face leaves the white piece — just additive area.
3.MD.C.7 Find the area of one face of the cube. Each face is a $10$ ft by $10$ ft square. 6.G.A.4 Use symmetry to share the green paint across the $6$ faces. The cube has $6$ con 4.MD.A.3 Break a single face into its two pieces. On any one face, the green border and t 4.MD.A.3 Plug in the green area on one face from Step 2 and solve for the white area. Review
Reasonableness: Total surface area of the cube is $6 \times 100 = 600$ sq ft, and Marla's $300$ sq ft of green paint is exactly half of that. So on every face, green covers half ($50$ sq ft) and white covers the other half ($50$ sq ft). That matches $50$ in answer (D), and the white square is smaller than the full face ($100$ sq ft) but takes up a meaningful chunk — a believable "centered square with border" picture. The radical answers $5\sqrt{2}$, $10\sqrt{2}$, $50\sqrt{2}$ are traps for students who guess the white square is tilted $45^\circ$ inside the face; nothing in the problem forces a tilt.
Alternative: Tool #6 (Guess and Check) on the choices. The white area on one face must be at most $100$ (the whole face) and the $6$ white areas plus $300$ sq ft of green must equal the total surface area $600$. So $6 \times (\text{white}) = 600 - 300 = 300$, giving white $= 50$ per face. Only choice (D) $= 50$ matches; the irrational choices cannot equal an exact half of an integer surface area.
CCSS standards used (min grade 6)
3.MD.C.7Relate area to multiplication; find the area of a rectangle by multiplying its side lengths (Computing the area of one face of the cube as $10 \times 10 = 100$ sq ft.)4.MD.A.3Apply area and perimeter formulas; treat area as additive over non-overlapping regions (Writing one face as white square $+$ green border $= 100$ sq ft and subtracting to get the white area.)6.G.A.4Represent three-dimensional figures using nets, and use the nets to find surface area (Recognizing the cube as $6$ congruent face-squares so the $300$ sq ft of green paint splits evenly into $\tfrac{300}{6} = 50$ sq ft per face.)
⭐ This AMC 8 problem only needs the Grade 6 idea that a cube has $6$ equal faces — then it's just $100 - 50 = 50$ on one face.
⭐ This AMC 8 problem only needs the Grade 6 idea that a cube has $6$ equal faces — then it's just $100 - 50 = 50$ on one face.
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