Competition · AMC preparation · step 4 of 4
AMC 8 · 2024 · #3
Grade 3 geometry-2d
Pick an answer.
AMC 8 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Find which gray is visible
Shared corner means each gray square is covered by one white, leaving two visible gray bands: side-10 under side-9, side-7 under side-4.
Recognizing squares and seeing how they nest together is the kind of shape work introduced in Grade 2 geometry.
2.G.A.1Draw A DiagramMeasure the outer band
Outer band = big gray minus the white on it: 10 × 10 − 9 × 9 = 100 − 81 = 19.
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.
The outer visible gray region has area 10 × 10 - 9 × 9 = 19 square units, found by taking the whole side-10 gray square and removing the side-9 white square that covers part of it.
▸ Why?
Each square's area is its side counted rows-by-columns, so the side-10 gray square holds 10 × 10 = 100 unit squares and the side-9 white square holds 9 × 9 = 81.
▸ Why?
A square of side s is filled by s rows of unit squares with s in every row, so the number of unit squares — its area — is s added s times, that is s × s.
▸ Why?
The side-9 white square lies completely inside the side-10 gray square, so the gray you can still see is the big square's area with the covered part taken away.
▸ Why?
The gray you still see and the gray hidden under the white square together fill the whole side-10 square with no gap and no overlap, so those two parts add up to its full area of 100.
▸ Why?
Because the visible part plus the covered part equals the whole, taking the covered 81 away from the whole 100 leaves exactly the visible part.
Measure the inner band
Same idea for the inner band: side-7 gray minus side-4 white = 7 × 7 − 4 × 4 = 49 − 16 = 33.
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.MD.C.7Identify SubproblemsAdd the two bands
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.2Identify SubproblemsCheck against the choices
Match 19 + 33 = 52 against the choices — only (E) fits, and it sits well below the side-10 square's area of 100.
Computing a small sum and matching it to a list of choices is direct use of Grade 3 add/subtract fluency.
3.NBT.A.2Eliminate PossibilitiesThis AMC 8 problem only needs Grade 3 square-area (side × side) and subtracting-the-inside-from-the-outside that you already know!
- Find which gray is visible
- Measure the outer band
- Measure the inner band
- Add the two bands
- Check against the choices
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