Problem 6 · 2014 AMC 8
Easy
Arithmetic & Operations
factoringsum-of-squares
Six rectangles each with a common base width of 2 have lengths of 1, 4, 9, 16, 25, and 36. What is the sum of the areas of the six rectangles?
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Answer: D — 182.
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Hint 1 of 2
Every rectangle shares the same width 2, so instead of computing six areas and adding, factor the 2 out: total = 2 × (sum of all the lengths). One multiply at the end.
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Hint 2 of 2
Bonus: the lengths 1, 4, 9, 16, 25, 36 are the first six perfect squares.
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Approach: factor out the common width (distributive property)
- All widths equal 2, so total area = 2×1 + 2×4 + … + 2×36 = 2 × (1 + 4 + 9 + 16 + 25 + 36).
- The lengths are the first six squares; their sum is 91.
- Total area: 2 × 91 = 182.
- Why this transfers: a common factor in every term can always be pulled outside the sum — that's the distributive property doing the heavy lifting, turning six multiplications into one.
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