Competition · AMC preparation · step 4 of 4
AMC 8 · 2001 · #9
Grade 7 geometry-2d
Pick an answer.
AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The kite is a picture problem, so start by sketching the large 18 × 21 rectangle with the kite inside (Tool #1). Once the picture is on paper, the question splits cleanly into three small subproblems (Tool #7): (a) area of the rectangle, (b) area of the kite, (c) subtract to get the waste. The picture also reveals a shortcut — the kite's diagonals are the full width and height of the rectangle, which forces the kite to fill exactly half. Sketching first, computing second, keeps the work on a Grade 6 area-of-polygons track instead of pushing toward heavier tools.
Find the foil area
Triple the 6 × 7 grid to an 18 × 21 foil rectangle; its area is 378 in².
Multiplying width by height for a rectangle is the Grade 6 area-of-polygons starting move.
6.G.A.1Draw A DiagramMark the kite's diagonals
Tripling the vertices puts the vertical diagonal at length 21 and the horizontal at 18, each spanning a full side.
Reading lengths from coordinates on a grid is the Grade 6 "polygons in the coordinate plane" standard.
6.G.A.3Draw A DiagramFind the kite's area
Subproblem — the kite's area is ½·d₁·d₂ = ½ × 21 × 18 = 189 in².
Half-the-product-of-diagonals is the Grade 7 area-of-quadrilateral move for a kite or rhombus.
The large kite covers 189 square inches, which is one half of 21 times 18 — the product of its two diagonals.
▸ Why?
The two diagonals cross at right angles, so slicing the kite along the vertical diagonal splits it into two triangles, and adding those two triangle areas gives the kite's whole area.
▸ Why?
The kite is one flat region cut into the two triangles with no gap and no overlap, so the pieces add back up to the whole kite.
▸ Why?
Each triangle takes the vertical diagonal, length 21, as its base, and its far tip sits 9 across from that base — half of the 18-long horizontal diagonal — so each triangle is half of 21 times 9, and the two together make half of 21 times 18.
▸ Why?
The area of each triangle is one half its base times its height.
Subtract to get the waste
Subproblem — the waste is what the kite leaves: 378 − 189 = 189 in², choice (D).
Composing/decomposing a region into kite + waste lets one subtraction finish the job.
6.G.A.1Identify SubproblemsSketch the 18 × 21 foil with the kite inside: the kite's diagonals are the full width and height, so the kite is exactly half the rectangle. The four corner scraps are the other half — 189 in{}², answer (D).
- Find the foil area
- Mark the kite's diagonals
- Find the kite's area
- Subtract to get the waste
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