Problem 20 · 2011 AMC 8
Medium
Geometry & Measurement
trapezoid-areadrop-altitudes
Quadrilateral ABCD is a trapezoid, AD = 15, AB = 50, BC = 20, and the altitude is 12. What is the area of the trapezoid?
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Answer: D — 750.
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Hint 1 of 2
Trapezoid area needs both parallel sides. You have the short one (50), the height (12), and the two slanted legs — so the only missing piece is how far each leg sticks out sideways. Drop a vertical from each top corner to split off two right triangles.
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Hint 2 of 2
Each leg is now the hypotenuse of a height-12 right triangle. Recognize the 3-4-5 family: 12-15 forces leg 9 (a 9-12-15 = 3-4-5×3), and 12-20 forces leg 16 (a 12-16-20 = 3-4-5×4) — no square roots needed.
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Approach: drop altitudes to extend the short base into the long one
- Drop verticals from the two top corners. They cut off a right triangle at each end, both of height 12. Left triangle: legs 12 and ?, hypotenuse 15 ⇒ ? = 9 (a 9-12-15 triangle). Right triangle: hypotenuse 20 ⇒ the other leg is 16 (a 12-16-20 triangle).
- The long base spans those two overhangs plus the rectangle in the middle: 9 + 50 + 16 = 75.
- Area = (1/2)(short + long)(height) = (1/2)(50 + 75)(12) = (1/2)(125)(12) = 750.
- Worth keeping: spotting 3-4-5 multiples (9-12-15, 12-16-20) lets you skip the Pythagorean square-root every time a known hypotenuse pairs with a leg of 12.
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