AMC 8 · 2001 · #8
Grade 5 geometry-2drate-ratioProblem
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.
Genevieve puts bracing on her large kite in the form of a cross connecting opposite corners of the kite. How many inches of bracing material does she need?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Genevieve draws a small kite on a one-inch dot grid with vertices at $(3,0)$, $(0,5)$, $(3,7)$, and $(6,5)$. For the large kite she triples every length on the grid. She then braces the large kite with a cross along its two diagonals (each diagonal connects opposite corners). How many inches of bracing material does she need?
Givens: Small-kite vertices on a one-inch grid: $(3,0)$, $(0,5)$, $(3,7)$, $(6,5)$; Large kite: every length on the grid is tripled (scale factor $3$); Bracing $=$ the two diagonals of the large kite, joined into a cross; Answer choices: (A) $30$, (B) $32$, (C) $35$, (D) $38$, (E) $39$
Unknowns: Total length, in inches, of the bracing material for the large kite
Understand
Restated: Genevieve draws a small kite on a one-inch dot grid with vertices at $(3,0)$, $(0,5)$, $(3,7)$, and $(6,5)$. For the large kite she triples every length on the grid. She then braces the large kite with a cross along its two diagonals (each diagonal connects opposite corners). How many inches of bracing material does she need?
Givens: Small-kite vertices on a one-inch grid: $(3,0)$, $(0,5)$, $(3,7)$, $(6,5)$; Large kite: every length on the grid is tripled (scale factor $3$); Bracing $=$ the two diagonals of the large kite, joined into a cross; Answer choices: (A) $30$, (B) $32$, (C) $35$, (D) $38$, (E) $39$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #9 Solve an Easier Related Problem
The figure is on a one-inch grid, so Tool #1 (Draw a Diagram) is built in — read each diagonal straight off the grid by counting units between opposite vertices. Tool #9 (Solve an Easier Related Problem) handles the scaling: get the answer for the easy small kite first, then multiply by $3$ for the large kite. Together these two tools avoid any need for algebra or distance formulas.
Execute — Answer: E
5.G.A.1 Step 1 - Use the grid to read the two diagonals of the small kite.
- The vertical diagonal joins $(3,0)$ to $(3,7)$, so its length is the difference in $y$-coordinates.
- The horizontal diagonal joins $(0,5)$ to $(6,5)$, so its length is the difference in $x$-coordinates.
💡 Both diagonals run along grid lines, so reading their length is just counting unit dots between the endpoints — exactly the Grade 5 coordinate-plane skill.
5.NF.B.5 Step 2 - Scale up to the large kite.
- Tripling the grid multiplies every length by $3$, so each diagonal of the large kite is $3$ times the corresponding small-kite diagonal.
💡 Solving the small (easier) kite first turns the large kite into a one-step scaling: multiply each known diagonal by the factor $3$.
4.OA.A.3 Step 3 The bracing is the two diagonals of the large kite, so the total length of bracing is the sum of those two diagonals.
💡 A cross made of two diagonals uses one piece for each diagonal, so adding the two diagonal lengths gives the total material — a one-step Grade 4 word-problem sum.
5.G.A.1 Use the grid to read the two diagonals of the small kite. The vertical diagonal 5.NF.B.5 Scale up to the large kite. Tripling the grid multiplies every length by $3$, so 4.OA.A.3 The bracing is the two diagonals of the large kite, so the total length of braci Review
Reasonableness: Check the scaling a different way. The whole small kite fits in a $6 \times 7$ bounding box, so tripling gives an $18 \times 21$ bounding box for the large kite. The horizontal diagonal spans the full width ($18$), and the vertical diagonal spans the full height ($21$); their sum is $18 + 21 = 39$. Also, the answer should be a multiple of $3$ since every length on the large kite is $3$ times an integer — and $39 = 3 \times 13$ is, while distractors $32$, $35$, and $38$ are not. Both checks point to (E).
Alternative: Tool #7 (Identify Subproblems): treat "diagonals of the large kite" as one subproblem (find each diagonal: $21$ and $18$) and "total bracing length" as a second subproblem (add them: $39$). Solving the two pieces in order and combining gives the same $39$ inches, confirming (E).
CCSS standards used (min grade 5)
5.G.A.1Use a pair of perpendicular number lines, called axes, to define a coordinate system; identify points in the plane by ordered pairs (Reading the small-kite vertices $(3,0)$, $(0,5)$, $(3,7)$, $(6,5)$ off the one-inch grid and computing each diagonal as a difference of coordinates ($7$ and $6$).)5.NF.B.5Interpret multiplication as scaling (resizing) (Recognizing that tripling the grid multiplies every length by $3$, so the small-kite diagonals $7$ and $6$ become $21$ and $18$ on the large kite.)4.OA.A.3Solve multistep word problems posed with whole numbers using the four operations (Combining the two large-kite diagonal lengths into the total bracing length $21 + 18 = 39$ inches.)
⭐ On a grid, a diagonal that runs straight across rows or columns is just a counting job — $7$ up, $6$ across. Triple the grid and you triple each length ($21$ and $18$), then add the two diagonals for the cross-shaped brace: $21 + 18 = 39$ inches, choice (E).
⭐ On a grid, a diagonal that runs straight across rows or columns is just a counting job — $7$ up, $6$ across. Triple the grid and you triple each length ($21$ and $18$), then add the two diagonals for the cross-shaped brace: $21 + 18 = 39$ inches, choice (E).
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