Problem 18 · 2017 AMC 8
Medium
Geometry & Measurement
pythagorean-triplearea-decomposition
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Answer: B — Area 24.
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Hint 1 of 2
Draw the diagonal BD. The right angle at C makes ▵BCD a 3-4-5, handing you BD = 5 — and that 5 is the bridge to the rest of the figure.
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Hint 2 of 2
Watch the side lengths line up: with BD = 5, ▵ABD has sides 5, 12, 13 — a Pythagorean triple, so it's secretly a right triangle too. And since the quadrilateral is non-convex (C caves inward), its area is the big triangle − the notch: ▵ABD − ▵BCD.
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Approach: split on diagonal BD into two right triangles
- Add diagonal BD. ▵BCD has the right angle at C with legs BC = 4 and CD = 3, so BD = √(16+9) = 5. That hypotenuse is the key — it unlocks the second triangle.
- Now ▵ABD has sides BD = 5, AB = 12, AD = 13 — a 5-12-13 triple, so it's right-angled at B. (Recognizing the triple saves you a square-root computation.)
- Because C dents inward, the quadrilateral is the large ▵ABD with the small ▵BCD removed: area = (1/2)(12)(5) − (1/2)(4)(3) = 30 − 6 = 24.
- Why this transfers: for any odd quadrilateral, slice it on a diagonal into triangles; here the right angle hands you that diagonal for free, and spotting 3-4-5 / 5-12-13 lets you skip the Pythagorean arithmetic.
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