AMC 8 · 2024 · #3
Grade 3 geometry-2dProblem
Four squares of side length 4,7,9, and 10 are arranged in increasing size order so that their left edges and bottom edges align. The squares alternate in color white-gray-white-gray, respectively, as shown in the figure. What is the area of the visible gray region in square units?
Pick an answer.
AMC 8 2024 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: Four squares with side lengths $4, 7, 9,$ and $10$ are stacked so their left edges and bottom edges all meet at the same corner. From largest to smallest the colors alternate gray–white–gray–white. Find the total area (in square units) of the gray region that is still visible.
Givens: Four square side lengths: $4,\ 7,\ 9,\ 10$; All four squares share the same bottom-left corner (left edges aligned, bottom edges aligned); Colors alternate by size: side 10 = gray, side 9 = white, side 7 = gray, side 4 = white; Each smaller square covers part of the square beneath it; Answer choices: (A) 42, (B) 45, (C) 49, (D) 50, (E) 52
Unknowns: Total area of the visible gray region (in square units)
Understand
Restated: Four squares with side lengths $4, 7, 9,$ and $10$ are stacked so their left edges and bottom edges all meet at the same corner. From largest to smallest the colors alternate gray–white–gray–white. Find the total area (in square units) of the gray region that is still visible.
Givens: Four square side lengths: $4,\ 7,\ 9,\ 10$; All four squares share the same bottom-left corner (left edges aligned, bottom edges aligned); Colors alternate by size: side 10 = gray, side 9 = white, side 7 = gray, side 4 = white; Each smaller square covers part of the square beneath it; Answer choices: (A) 42, (B) 45, (C) 49, (D) 50, (E) 52
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #3 Eliminate Possibilities
The visible gray is not one blob — it is two separate L-shaped bands: an outer band between the side-10 gray and the side-9 white, plus an inner band between the side-7 gray and the side-4 white. Tool #7 (Identify Subproblems) lets us split the question into "find each gray band, then add them." Each band is just (big square area) − (small square area), a one-line calculation. Tool #1 (Draw a Diagram) confirms which gray is visible, and Tool #3 (Eliminate Possibilities) verifies against the multiple-choice list at the end.
Execute — Answer: E
2.G.A.1 Step 1 - Use the diagram to see exactly which gray is still visible.
- Since all four squares share the same bottom-left corner and grow outward, each smaller square sits inside a corner of the next larger one.
- The side-10 gray has a side-9 white covering one corner; inside that white sits the side-7 gray; inside that sits the side-4 white.
- So the visible gray is exactly two L-shaped bands: (1) the band between the side-10 gray and the side-9 white, and (2) the band between the side-7 gray and the side-4 white.
💡 Recognizing squares and seeing how they nest together is the kind of shape work introduced in Grade 2 geometry.
3.MD.C.7 Step 2 - Compute the outer band first.
- Take the area of the big gray square (side 10) and subtract the area of the white square (side 9) that covers part of it.
- The area of a square is side × side, so the side-10 square has area $10 \times 10 = 100$ and the side-9 square has area $9 \times 9 = 81$.
💡 Finding a square's area as side × side, then subtracting the inner shape's area to get the leftover band, is the heart of the Grade 3 area unit.
3.MD.C.7 Step 3 Do the same thing for the inner band: take the area of the side-7 gray square and subtract the area of the side-4 white square that sits in its corner.
💡 Exactly the same Grade 3 area idea as Step 2 — area of the outer square minus area of the inner square gives the band.
3.NBT.A.2 Step 4 The two gray bands do not overlap (each lives between a different pair of squares), so the total visible gray area is simply their sum.
💡 Adding two numbers under 1000 is fluent Grade 3 arithmetic.
3.NBT.A.2 Step 5 - Check against the answer choices.
- The candidates are 42, 45, 49, 50, 52.
- Our value 52 matches choice **(E)** exactly, and none of the others equal $19 + 33$.
- The answer also passes a sanity check: the largest gray square has total area 100 and the white covers 81, so the outer gray band must be smaller than 100; combined with the inner band of 33, a total of 52 is comfortably in range.
💡 Computing a small sum and matching it to a list of choices is direct use of Grade 3 add/subtract fluency.
2.G.A.1 Use the diagram to see exactly which gray is still visible. Since all four squar 3.MD.C.7 Compute the outer band first. Take the area of the big gray square (side 10) and 3.MD.C.7 Do the same thing for the inner band: take the area of the side-7 gray square an 3.NBT.A.2 The two gray bands do not overlap (each lives between a different pair of square 3.NBT.A.2 Check against the answer choices. The candidates are 42, 45, 49, 50, 52. Our val Review
Reasonableness: Intuition check: the side-10 gray has area 100 and the side-9 white covers 81 of it, so the outer gray band is only 19 — a thin strip, which makes sense because the two side lengths differ by just 1. The inner gray, between side 7 and side 4, has a much bigger side-length gap (3), so its band is fatter (33). Adding 19 + 33 = 52 sits well below the full 100 of the big square and is exactly the sum of two non-overlapping bands. Answer (E) 52 is reasonable.
Alternative: Tool #13 (Convert to Algebra) offers a shortcut via the difference-of-squares identity $a^2 - b^2 = (a-b)(a+b)$: $10^2 - 9^2 = (1)(19) = 19$ and $7^2 - 4^2 = (3)(11) = 33$. Same answer, but for an elementary student the straight "side × side, then subtract" route from Tool #7 is more natural and uses the same Grade 3 ideas.
CCSS standards used (min grade 3)
2.G.A.1Recognize and draw shapes having specified attributes (e.g., squares) (Identifying the four squares and seeing how they nest from a shared corner, so the visible gray splits cleanly into two L-shaped bands.)3.MD.C.7Relate area to the operations of multiplication and addition (area of a square = side × side; area of composite figures by adding/subtracting parts) (Computing each square's area as side × side and subtracting the inner square's area from the outer square's area to get each gray band.)3.NBT.A.2Fluently add and subtract within 1000 (Performing $100-81$, $49-16$, and $19+33$ and matching the result against the answer choices.)
⭐ This AMC 8 problem only needs Grade 3 square-area (side × side) and subtracting-the-inside-from-the-outside that you already know!
⭐ This AMC 8 problem only needs Grade 3 square-area (side × side) and subtracting-the-inside-from-the-outside that you already know!
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