Question 9 • *CR CCSS Math 2 S2 [M2960OX]
Since 21 cm is one of the shorter sides, the longest side must be 2x, so 2x is at least 21. For an acute triangle, the square of the longest side must be less than the sum of the squares of the other sides: (2x)² < 21² + x². This gives 3x² < 441, so 2x < 2√147 ≈ 24.2 cm. Thus the greatest possible length, rounded to the nearest tenth, is 24.2 cm, choice B.
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