1989 AMC 8 Problem 24

Attempt Problem 24 of the 1989 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 8 solutions, or check the answer key.

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24.

A square piece of paper is folded in half. The folded paper is then cut in half by a straight cut parallel to the fold. This forms three rectangles: one large rectangle and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?

12\dfrac{1}{2}

23\dfrac{2}{3}

34\dfrac{3}{4}

45\dfrac{4}{5}

56\dfrac{5}{6}

Answer: E
Concepts:paper foldingperimeter

Difficulty rating: 950

Solution:

Let the square have side 4.4. Folding in half makes a 2×42 \times 4 stack of two layers. Cutting parallel to the fold splits it into a strip containing the fold and a strip that does not.

The strip with the fold unfolds into the large rectangle, 2×4,2 \times 4, with perimeter 2(2+4)=12.2(2 + 4) = 12. The other strip is two separate small rectangles, each 1×4,1 \times 4, with perimeter 2(1+4)=10.2(1 + 4) = 10. The ratio is 1012=56.\frac{10}{12} = \frac{5}{6}.

Thus, the correct answer is E .

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