Problem 4 · 2019 AMC 8
Medium
Geometry & Measurement
pythagorean-triplearea
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Answer: D — 120 square meters.
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Hint 1 of 2
A rhombus only gives you a perimeter and one diagonal — so the hidden tool must be the special fact about a rhombus's diagonals: they meet at right angles and bisect each other. That instantly hands you a right triangle.
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Hint 2 of 2
The side (52÷4 = 13) is the hypotenuse and half of AC (24÷2 = 12) is a leg. Recognize 12-13-? — it's the 5-12-13 triple, so the missing leg is 5.
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Approach: perpendicular bisecting diagonals make a 5-12-13 right triangle
- Perimeter gives side = 52 ÷ 4 = 13; half of diagonal AC = 24 ÷ 2 = 12.
- The diagonals cut each other at right angles, so the center splits the rhombus into four identical right triangles with hypotenuse 13 (the side) and one leg 12 (half of AC). That's the 5-12-13 triple — no Pythagorean computation needed — so the other half-diagonal is 5 and the full diagonal BD = 10.
- Area of a rhombus = d1 × d22 = 24 × 102 = 120 sq m.
- You'll see it again as: whenever a problem hands you a rhombus (or a kite), reach first for "diagonals perpendicular" — it converts the figure into right triangles, and recognizing a Pythagorean triple (3-4-5, 5-12-13, 8-15-17) skips the square roots.
Another way — four little triangles:
- Each of the four right triangles has legs 12 and 5, so area 12 × 52 = 30.
- Four of them: 4 × 30 = 120 sq m — same answer, confirming the diagonal-product formula.
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