AMC 8 · 2001 · #7
Grade 6 geometry-2dProblem
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.
What is the number of square inches in the area of the small kite?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A kite is drawn on a one-inch grid. Reading the diagram, the four corners sit at $(3,0)$, $(0,5)$, $(3,7)$, and $(6,5)$. Find the area of this kite in square inches.
Givens: The kite is on a one-inch grid, so adjacent dots are $1$ inch apart; Vertices: bottom $(3,0)$, left $(0,5)$, top $(3,7)$, right $(6,5)$; Answer choices: (A) $21$, (B) $22$, (C) $23$, (D) $24$, (E) $25$
Unknowns: The area of the small kite, in square inches
Understand
Restated: A kite is drawn on a one-inch grid. Reading the diagram, the four corners sit at $(3,0)$, $(0,5)$, $(3,7)$, and $(6,5)$. Find the area of this kite in square inches.
Givens: The kite is on a one-inch grid, so adjacent dots are $1$ inch apart; Vertices: bottom $(3,0)$, left $(0,5)$, top $(3,7)$, right $(6,5)$; Answer choices: (A) $21$, (B) $22$, (C) $23$, (D) $24$, (E) $25$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The whole problem lives on the grid, so Tool #1 (Draw a Diagram) is the way in: place the kite on coordinates, read every length straight off the dots, and the horizontal segment from $(0,5)$ to $(6,5)$ jumps out as a natural cut. That cut splits the kite into a top triangle and a bottom triangle, which is Tool #7 (Identify Subproblems): one hard shape becomes two easy ones with a known base $\times$ height $\div 2$ formula.
Execute — Answer: A
5.G.A.2 Step 1 - Place the kite on the grid and read the corners.
- The bottom-left dot is $(0,0)$, and counting along the dots gives the four vertices: bottom $B=(3,0)$, left $L=(0,5)$, top $T=(3,7)$, right $R=(6,5)$.
- Notice that $L$ and $R$ share the same height $y=5$, so the segment $LR$ is horizontal.
💡 Grade 5 coordinate plane: each lattice point gives an exact $(x,y)$, and points with the same $y$ sit on a horizontal line.
6.G.A.1 Step 2 - Cut along the horizontal segment $LR$.
- The kite splits into a top triangle $\triangle TLR$ and a bottom triangle $\triangle BLR$.
- Both share the same base — the segment from $(0,5)$ to $(6,5)$, which has length $6 - 0 = 6$ inches.
💡 Grade 6: decomposing a polygon into triangles is the standard move for finding its area when no single formula applies.
5.G.A.2 Step 3 - Find the height of each triangle from its apex to the base $y=5$.
- The top apex $T=(3,7)$ is $7-5=2$ inches above the base; the bottom apex $B=(3,0)$ is $5-0=5$ inches below it.
💡 Vertical distance to a horizontal line is just the difference in $y$ — no Pythagorean work needed.
6.G.A.1 Step 4 - Apply $\text{area} = \tfrac{1}{2}\cdot\text{base}\cdot\text{height}$ to each triangle, then add.
- Both triangles use the same base $6$.
💡 Adding the two triangle areas is the Grade 6 "decompose, compute, recombine" recipe for area.
5.G.A.2 Place the kite on the grid and read the corners. The bottom-left dot is $(0,0)$, 6.G.A.1 Cut along the horizontal segment $LR$. The kite splits into a top triangle $\tri 5.G.A.2 Find the height of each triangle from its apex to the base $y=5$. The top apex $ 6.G.A.1 Apply $\text{area} = \tfrac{1}{2}\cdot\text{base}\cdot\text{height}$ to each tri Review
Reasonableness: Two checks. (1) Bounding box: the kite fits inside the $6 \times 7 = 42$ rectangle from $(0,0)$ to $(6,7)$, and a kite typically fills about half of its bounding rectangle — that predicts $\tfrac{42}{2} = 21$, matching the answer. (2) Kite-diagonal formula: the diagonals are $BT$ (length $7$) and $LR$ (length $6$), and $\tfrac{1}{2}d_1d_2 = \tfrac{1}{2}\cdot 7 \cdot 6 = 21$ — same answer from a totally different route.
Alternative: Tool #3 (Eliminate Possibilities) with the bounding-box estimate. The $6 \times 7$ rectangle has area $42$, so the kite's area must be under $42$ and well over half because the kite hugs most of the box. Choices $22$, $23$, $24$, $25$ would each require the kite to fill more than half plus a noticeable extra — but the four corner cutouts are clearly bigger than that. Only $21$ (just half) is consistent, pinning down (A).
CCSS standards used (min grade 6)
5.G.A.2Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane, and interpret coordinate values of points in the context of the situation (Reading the four kite vertices off the grid as ordered pairs and getting side lengths from differences of coordinates.)6.G.A.1Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes (Splitting the kite along its horizontal diagonal into two triangles and adding their areas to get $6 + 15 = 21$.)
⭐ Read the corners off the grid, slice the kite into two triangles along the horizontal $y=5$ line, and add: $\tfrac{1}{2}(6)(2) + \tfrac{1}{2}(6)(5) = 21$ — answer (A). A Grade 6 "decompose to find area" move handles an AMC 8 geometry problem.
⭐ Read the corners off the grid, slice the kite into two triangles along the horizontal $y=5$ line, and add: $\tfrac{1}{2}(6)(2) + \tfrac{1}{2}(6)(5) = 21$ — answer (A). A Grade 6 "decompose to find area" move handles an AMC 8 geometry problem.
More like this
Same archetype — closest grade level first.
