AMC 8 · 2001 · #9

Grade 7 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-rectanglesarea-trianglessimilar-figurescoordinate-geometry area-differenceidentify-subproblems ↑ Prerequisites: area-rectanglessimilar-figures

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Genevieve's small kite sits on a 6 × 7 dot grid with vertices at (3,0),(0,5),(3,7),(6,5). For the large kite, every grid dimension is tripled. A rectangle of gold foil just covers the large grid, and the four corner pieces outside the kite are wasted. How many square inches of foil are wasted?

Pick an answer.

(A)

(B)

(C)

180

(D)

189

(E)

264

AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Triple the 6 × 7 grid to an 18 × 21 foil rectangle; its area is 378 in².

Foil area = 18 × 21 = 378 in²

2STEP 2

Tripling the vertices puts the vertical diagonal at length 21 and the horizontal at 18, each spanning a full side.

d₁ = 21 in, d₂ = 18 in

3STEP 3

Subproblem — the kite's area is ½·d₁·d₂ = ½ × 21 × 18 = 189 in².

Kite area =

\frac{1}{2}

× 21 × 18 =

\frac{378}{2}

= 189 in²

4STEP 4

Subproblem — the waste is what the kite leaves: 378 − 189 = 189 in², choice (D).

Waste = 378 - 189 = 189 in² → (D)

Answer

189

The kite's diagonals are the full 18 and 21 — exactly the sides of the rectangle. That means the kite occupies

\frac{1}{2}

d₁ d₂ =

\frac{1}{2}

· (rectangle area), so the kite is exactly half the foil and the four corner pieces are the other half. Half of 378 is 189, matching answer (D). A sanity check on scaling: tripling each length multiplies areas by 9. The small grid is 42 in{}² and the small kite has area

\frac{1}{2}

· 6 · 7 = 21 in{}², so the small waste is 42 - 21 = 21 in{}². Scaling waste by 9 gives 21 × 9 = 189 in{}². Same answer from two independent angles.

💡Key takeaway

Sketch 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).

More like this