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1999 AMC 8

Problem 14

Problem 14 · 1999 AMC 8 Medium
Geometry & Measurement pythagoreansymmetry
Figure for AMC 8 1999 Problem 14
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Answer: D — 34.
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Hint 1 of 2
The slant sides AB and CD are equal, so the trapezoid is symmetric β€” drop a vertical from B and from C and the bottom splits into the middle (equal to the top, 8) plus two equal overhangs. The two overhangs share the leftover 16 βˆ’ 8.
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Hint 2 of 2
Each slant side is now the hypotenuse of a right triangle with legs = height and overhang. Watch for a friendly Pythagorean triple.
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Approach: use symmetry to find the overhang, then a 3-4-5 triangle
  1. Because AB = CD, the figure is symmetric: dropping verticals from B and C carves a rectangle of width 8 in the middle, leaving 16 βˆ’ 8 = 8 split evenly into two overhangs of 4 each.
  2. Each slant side is the hypotenuse of a right triangle with legs 3 (height) and 4 (overhang): √(3Β² + 4Β²) = 5 β€” a 3-4-5 triangle, no messy square root.
  3. Perimeter = 16 + 8 + 5 + 5 = 34. Two transferable habits: exploit symmetry to find the overhang for free, and recognize 3-4-5 (and 5-12-13) so you spot the hypotenuse on sight.
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