Problem 21 · 2015 AMC 8
Hard
Geometry & Measurement
angle-chaseequiangular-hexagonright-triangle-area
In the given figure hexagon ABCDEF is equiangular, ABJI and FEHG are squares with areas 18 and 32 respectively, ▵JBK is equilateral and FE = BC. What is the area of ▵KBC?
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Answer: C — Area 12.
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Hint 1 of 2
To get a triangle's area you need two sides and the angle between them — and B is exactly where everything meets. The squares hand you the two sides for free: a square's side is √(its area).
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Hint 2 of 2
For the missing angle, notice four angles fan out around point B (the square's corner, the equilateral triangle's corner, ∠KBC, and the hexagon's interior corner) and they must close up to 360°. An equiangular hexagon has every interior angle 120°, so the only unknown in that sum is ∠KBC.
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Approach: two sides from the squares + included angle from the 360° around B
- Sides from the squares: BK = JB = √18 = 3√2 (square side, copied by the equilateral ▵JBK), and BC = FE = √32 = 4√2.
- Angles fanning around B must total 360°: ∠ABJ (square) + ∠JBK (equilateral) + ∠KBC + ∠CBA (hexagon's 120° interior angle) = 90 + 60 + ∠KBC + 120 = 360, so ∠KBC = 90°.
- With a right angle at B, the two sides are the legs: area = ½(3√2)(4√2) = ½(12·2) = 12.
- Why this transfers: a vertex where several shapes meet lets you find a stubborn angle by 'going all the way around = 360°'; and the side of a square is always the square root of its area.
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