2018 AMC 10B Problems/Problem 7
Contents
Problem
In the figure below, congruent semicircles lie on the diameter of a large semicircle, with their diameters covering the diameter of the large semicircle with no overlap. Let be the combined area of the small semicircles and be the area of the region inside the large semicircle but outside the semicircles. The ratio is . What is ?
Solution 1 (Work using Answer Choices)
Use the answer choices and calculate them. The one that works is .
~Flame_Thrower
Note: This solution is time-consuming and practically useless.
Solution 2 (More Algebraic Approach)
Let the number of semicircles be and let the radius of each semicircle be . To find the total area of all of the small semicircles, we have .
Next, we have to find the area of the larger semicircle. The radius of the large semicircle can be deduced to be . So, the area of the larger semicircle is .
Now that we have found the area of both A and B, we can find the ratio. , so part-to-whole ratio is . When we divide the area of the small semicircles combined by the area of the larger semicircles, we get . This is equal to . By setting them equal, we find that . This is our answer, which corresponds to choice .
Solution by: Archimedes15
Note: you could have just set as in the beginning.
Solution 3
Each small semicircle is of the large semicircle. Since small semicircles make of the large one, . Solving this, we get . ...
Video Solution
~savannahsolver
Video Solution
https://youtu.be/NsQbhYfGh1Q?t=1120
~ pi_is_3.14
See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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All AMC 10 Problems and Solutions |
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