2001 AMC 8 Problem 16

Attempt Problem 16 of the 2001 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 2001 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

16.

A square piece of paper, 44 inches on a side, is folded in half vertically. Both layers are then cut in half parallel to the fold. Three new rectangles are formed, a large one and two small ones. What is the ratio of the perimeter of one of the small rectangles to the perimeter of the large rectangle?
13\dfrac{1}{3}
12\dfrac{1}{2}
34\dfrac{3}{4}
45\dfrac{4}{5}
56\dfrac{5}{6}
Answer: E
Concepts:paper foldingperimeter
Small hint:
After folding, the paper is a 4×24\times2 rectangle
Big hint:
The small rectangles are 4×14\times1
Solution:
The square is folded in half to create a 4×24\times2 rectangle.
Cutting the folded paper in half parallel to the fold gives two small 4×14\times1 rectangles and one large 4×24\times2 rectangle.
The ratio of the perimeter of a small rectangle to the perimeter of the large rectangle is 2(4+1)2(4+2)=1012=56.\frac{2(4+1)}{2(4+2)}=\frac{10}{12}=\frac56.
Thus, E is the correct answer.
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