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2001 AMC 8

Problem 11

Problem 11 · 2001 AMC 8 Medium
Geometry & Measurement coordinate-geometryarea

Points A, B, C, and D have these coordinates: A(3, 2), B(3, −2), C(−3, −2), and D(−3, 0). What is the area of quadrilateral ABCD?

ABCD
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Answer: C — 18.
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Hint 1 of 2
Always plot the points first — the picture tells you the shape. Notice A,B share x = 3 and C,D share x = −3, so two sides are vertical (parallel).
Still stuck? Show hint 2 →
Hint 2 of 2
Two parallel sides means trapezoid: use ½(b₁ + b₂)·h, treating the vertical sides as the bases and the horizontal distance between them as the height.
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Approach: trapezoid with two vertical sides
  1. Plotting reveals the parallel pair: AB is vertical (x = 3) and DC is vertical (x = −3). So ABCD is a trapezoid lying on its side.
  2. Base lengths: AB from y = 2 down to y = −2 has length 4; DC from y = 0 down to y = −2 has length 2. The "height" is the horizontal gap between the two verticals: 3 − (−3) = 6.
  3. Area = ½(b₁ + b₂)·h = ½(4 + 2)·6 = 18. The trapezoid formula works whichever way the parallel sides point — just measure the perpendicular distance between them.
Another way — Shoelace formula (works for any polygon):
  1. List the vertices in order A(3,2), B(3,−2), C(−3,−2), D(−3,0) and apply Shoelace: ½|Σ(xᵢyᵢ₊₁ − xᵢ₊₁yᵢ)|.
  2. Sum of x·(next y): 3·(−2)+3·(−2)+(−3)·0+(−3)·2 = −6−6+0−6 = −18; sum of y·(next x): 2·3+(−2)·(−3)+(−2)·(−3)+0·3 = 6+6+6+0 = 18.
  3. Area = ½|−18 − 18| = ½·36 = 18. Shoelace is your fallback when a coordinate shape isn't a tidy trapezoid.
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