Problem 20 · 2002 AMC 8
Stretch
Geometry & Measurement
area-fractionsimilar-triangles
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Answer: D — 3 square inches.
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Hint 1 of 2
Don't chase exact lengths — work in *fractions of the whole 8*. The altitude XC bisects YZ, so it cuts the triangle into two equal halves; focus on the left half only.
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Hint 2 of 2
The little corner triangle cut off near X is a *half-scale* copy of that half (its sides are midpoint-halved). Half the lengths means a quarter of the area — so the shaded trapezoid is the half minus that quarter-of-a-half.
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Approach: halve the triangle, then remove the midpoint triangle
- Think in fractions, not measurements. Altitude XC bisects YZ, so triangle XYC is exactly half of XYZ: area 8 ÷ 2 = 4.
- Inside that half, the unshaded top triangle is built on midpoints, so it's a half-scale copy — and half the side lengths gives (½)² = ¼ the area, i.e. ¼ × 4 = 1.
- Shaded = 4 − 1 = 3 square inches.
- *The reusable idea:* you can find an area as a *fraction* of a known one without any side lengths — halving by a median, and the length²-to-area rule (half the sides → quarter the area) — then add and subtract the fractions.
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