2018 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 7 Easy

In the figure below, N congruent semicircles are drawn along a diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let A be the combined area of the small semicircles and B be the area of the region inside the large semicircle but outside the small semicircles. The ratio A:B is 1:18. What is N? (2018 AMC 10B Problem, Question#7)

  • A.

    16

  • B.

    17

  • C.

    18

  • D.

    19

  • E.

    36

Answer:D

Use the answer choices and calculate them. The one that works is (\rm D)~19.

Let the number of semicircles be n and let the radius of each semicircle to be r. To find the total area of all of the small semicircles, we have n\cdot\frac{{\pi}\cdot r^2}{2}.Next, we have to find the area of the larger semicircle. The radius of the large semicircle can be deduced to be n\cdot r. So, the area of the larger semicircle is \frac{\pi\cdot n^2\cdot r^2}{2} . Now that we have found the area of both A and B, we can find the ratio. \frac{A}{B}=\frac{1}{18}, so part-to-whole ratio is 1:19. When we divide the area of the small semicircles combined by the area of the larger semicircles, we get \frac{1}{n}. This is equal to \frac{1}{19}, By setting them equal, we find that n=19. This is our answer, which corresponds to choice (\rm D)~19.

Each small semicircle is \frac{1}{N^2} of the large semicircle. Since N small semicircles make \frac{1}{19} of the large one, \frac{N}{N^2}=\frac{1}{19}. Solving this, we get 19.

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