2011 AMC 8 Problem 25

Below is the professionally curated solution for Problem 25 of the 2011 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 8 solutions, or check the answer key.

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Concepts:circle areasquare (geometry)Pythagorean Theorem

Difficulty rating: 1660

25.

A circle with radius 11 is inscribed in a square and circumscribed about another square as shown. Which fraction is closest to the ratio of the circle's shaded area to the shaded area between the two squares?

12\dfrac{1}{2}

11

32\dfrac{3}{2}

22

52\dfrac{5}{2}

Solution:

The circle's shaded area is equal to the area of the circle minus the area of the smaller square. The side length of the inner square can be calculated using the Pythagorean Theorem to get 12+12=2.\sqrt{1^2 + 1^2} = \sqrt{2}.

Therefore, the area of the inner square is 22=2.\sqrt{2}^2 = 2. The circle's shaded area is then 12π2=π2.1^2\pi - 2 = \pi - 2.

The area of the outside square is 22=4,2^2 = 4, so the shaded area between the two squares is 42=2.4 - 2 = 2.

The desired fraction is π223.142212.\dfrac{\pi - 2}{2} \approx \dfrac{3.14 - 2}{2} \approx \dfrac{1}{2}.

Thus, A is the correct answer.

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