Competition · AMC preparation · step 4 of 4

AMC 8 · 2001 · #7

Grade 6 geometry-2d
coordinate-geometryarea-trianglesarea-rectanglesspatial-visualization coordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-triangles
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
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.

Pick an answer.

(A)
21
(B)
22
(C)
23
(D)
24
(E)
25

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

How to solve
Strategy Draw a Diagram

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 × height ÷ 2 formula.

1STEP 1

Read the kite's corners

Put the bottom-left dot at (0,0) and read the four corners; L=(0,5) and R=(6,5) share height y=5, so segment LR is horizontal.

B=(3,0), L=(0,5), T=(3,7), R=(6,5)
2STEP 2

Cut the kite in two

Cut along horizontal LR into a top triangle △TLR and a bottom triangle △BLR; both share base LR = 6 inches.

base = LR = 6 in
3STEP 3

Find each triangle's height

Height to base y=5: top apex T is 7-5 = 2 above, bottom apex B is 5-0 = 5 below.

h_top = 7-5 = 2, h_bot = 5-0 = 5
4STEP 4

Add the two triangle areas

Two triangle areas 12\frac{1}{2}(6)(2) = 6 and 12\frac{1}{2}(6)(5) = 15; add them: 6 + 15 = 21.

1/2(6)(2) + 1/2(6)(5) = 6 + 15 = 21 → (A)
Answer
21
Two checks. (1) Bounding box: the kite fits inside the 6 × 7 = 42 rectangle from (0,0) to (6,7), and a kite typically fills about half of its bounding rectangle — that predicts 422\frac{42}{2} = 21, matching the answer. (2) Kite-diagonal formula: the diagonals are BT (length 7) and LR (length 6), and 12\frac{1}{2}d₁d₂ = 12\frac{1}{2} · 7 · 6 = 21 — same answer from a totally different route.
💡Key takeaway

Read the corners off the grid, slice the kite into two triangles along the horizontal y=5 line, and add: 12\frac{1}{2}(6)(2) + 12\frac{1}{2}(6)(5) = 21 — answer (A). A Grade 6 "decompose to find area" move handles an AMC 8 geometry problem.

  • Read the kite's corners
  • Cut the kite in two
  • Find each triangle's height
  • Add the two triangle areas

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