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AMC 8 · 2001 · #9

Grade 7 geometry-2d
Archetype Compound Area by Decomposition / Difference
area-rectanglesarea-trianglessimilar-figurescoordinate-geometry area-differenceidentify-subproblems ↑ Prerequisites: area-rectanglessimilar-figures
📏 Long solution 💡 4 insights 📊 Diagram

Problem

To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.

The large kite is covered with gold foil. The foil is cut from a rectangular piece that just covers the entire grid. How many square inches of waste material are cut off from the four corners?

Pick an answer.

(A)
63
(B)
72
(C)
180
(D)
189
(E)
264
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Toolkit + CCSS Solution

Understand

Restated: Genevieve's small kite sits on a $6 \times 7$ 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?

Givens: Small grid is $6$ in wide and $7$ in tall (vertices span $x \in [0,6]$, $y \in [0,7]$); Small kite vertices: $(3,0),(0,5),(3,7),(6,5)$ — the diagonals of the kite are exactly the width and height of the grid; For the large kite both grid dimensions are tripled, so the large grid (and the foil) is $3 \times 6 = 18$ in by $3 \times 7 = 21$ in; Waste = area of foil rectangle $-$ area of large kite; Answer choices: (A) $63$, (B) $72$, (C) $180$, (D) $189$, (E) $264$

Unknowns: The area of the four corner pieces cut off from the rectangular foil, in square inches

Understand

Restated: Genevieve's small kite sits on a $6 \times 7$ 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?

Givens: Small grid is $6$ in wide and $7$ in tall (vertices span $x \in [0,6]$, $y \in [0,7]$); Small kite vertices: $(3,0),(0,5),(3,7),(6,5)$ — the diagonals of the kite are exactly the width and height of the grid; For the large kite both grid dimensions are tripled, so the large grid (and the foil) is $3 \times 6 = 18$ in by $3 \times 7 = 21$ in; Waste = area of foil rectangle $-$ area of large kite; Answer choices: (A) $63$, (B) $72$, (C) $180$, (D) $189$, (E) $264$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems

The kite is a picture problem, so start by sketching the large $18 \times 21$ 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.

Execute — Answer: D

#1 Draw a Diagram 6.G.A.1 Step 1
  • Sketch the large grid.
  • Tripling each side of the $6 \times 7$ small grid gives an $18 \times 21$ rectangle.
  • The foil rectangle has the same dimensions, so write down its area.
$$\text{Foil area} = 18 \times 21 = 378 \text{ in}^2$$

💡 Multiplying width by height for a rectangle is the Grade 6 area-of-polygons starting move.

#1 Draw a Diagram 6.G.A.3 Step 2
  • Locate the large kite on the sketch.
  • Tripling the small-kite vertices $(3,0),(0,5),(3,7),(6,5)$ gives $(9,0),(0,15),(9,21),(18,15)$.
  • The vertical diagonal runs from $(9,0)$ to $(9,21)$ — length $21$.
  • The horizontal diagonal runs from $(0,15)$ to $(18,15)$ — length $18$.
  • Both diagonals span the full sides of the rectangle.
$$d_1 = 21 \text{ in}, \quad d_2 = 18 \text{ in}$$

💡 Reading lengths from coordinates on a grid is the Grade 6 "polygons in the coordinate plane" standard.

#7 Identify Subproblems 7.G.B.6 Step 3
  • Subproblem: area of the kite.
  • With perpendicular diagonals $d_1, d_2$, the kite's area is $\tfrac{1}{2} d_1 d_2$.
  • Plug in the lengths from the sketch.
$$\text{Kite area} = \tfrac{1}{2} \times 21 \times 18 = \tfrac{378}{2} = 189 \text{ in}^2$$

💡 Half-the-product-of-diagonals is the Grade 7 area-of-quadrilateral move for a kite or rhombus.

#7 Identify Subproblems 6.G.A.1 Step 4
  • Subproblem: the waste is everything in the rectangle that the kite doesn't cover.
  • Subtract.
$$\text{Waste} = 378 - 189 = 189 \text{ in}^2 \;\Rightarrow\; \textbf{(D)}$$

💡 Composing/decomposing a region into kite $+$ waste lets one subtraction finish the job.

[1] #1 6.G.A.1 Sketch the large grid. Tripling each side of the $6 \times 7$ small grid gives a
[2] #1 6.G.A.3 Locate the large kite on the sketch. Tripling the small-kite vertices $(3,0),(0,
[3] #7 7.G.B.6 Subproblem: area of the kite. With perpendicular diagonals $d_1, d_2$, the kite'
[4] #7 6.G.A.1 Subproblem: the waste is everything in the rectangle that the kite doesn't cover

Review

Reasonableness: The kite's diagonals are the full $18$ and $21$ — exactly the sides of the rectangle. That means the kite occupies $\tfrac{1}{2} d_1 d_2 = \tfrac{1}{2} \cdot (\text{rectangle area})$, 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 $\tfrac{1}{2} \cdot 6 \cdot 7 = 21$ in${}^2$, so the small waste is $42 - 21 = 21$ in${}^2$. Scaling waste by $9$ gives $21 \times 9 = 189$ in${}^2$. Same answer from two independent angles.

Alternative: Tool #9 (Try a Simpler Case) on the small kite first. Small foil = $6 \times 7 = 42$ in${}^2$. Small kite = $\tfrac{1}{2} \cdot 6 \cdot 7 = 21$ in${}^2$. Small waste = $42 - 21 = 21$ in${}^2$. Then use the scaling rule: tripling lengths multiplies areas by $3^2 = 9$, so the large waste is $21 \times 9 = 189$ in${}^2$, again $(D)$. Solving the easier version first, then scaling, avoids ever multiplying $18 \times 21$.

CCSS standards used (min grade 7)

  • 6.G.A.1 Find the area of right triangles, other triangles, and special quadrilaterals by composing and decomposing into rectangles and triangles (Computing the foil rectangle's area as $18 \times 21 = 378$ and getting the waste by the decomposition rectangle $=$ kite $+$ waste.)
  • 6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side (Tripling the small-kite coordinates and reading the diagonal lengths $21$ and $18$ directly off the coordinates.)
  • 7.G.B.6 Solve real-world and mathematical problems involving area, volume, and surface area of two- and three-dimensional objects composed of triangles, quadrilaterals, polygons, cubes, and right prisms (Applying the kite-area formula $\tfrac{1}{2} d_1 d_2 = 189$ for the large kite.)

⭐ Sketch the $18 \times 21$ 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).

⭐ Sketch the $18 \times 21$ 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).

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