Difference between revisions of "2018 AMC 8 Problems/Problem 15"
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Let the radius of the large circle be <math>R</math>. Then the radii of the smaller circles are <math>\frac R2</math>. The areas the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is <math>\frac 14</math>. This means the combined area of the 2 smaller circles is <math>\frac 12</math> the larger circle, therefore the shaded region is equal to the combined area of the 2 smaller circles, which is <math>\boxed{\textbf{(D) } 1}</math> | Let the radius of the large circle be <math>R</math>. Then the radii of the smaller circles are <math>\frac R2</math>. The areas the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is <math>\frac 14</math>. This means the combined area of the 2 smaller circles is <math>\frac 12</math> the larger circle, therefore the shaded region is equal to the combined area of the 2 smaller circles, which is <math>\boxed{\textbf{(D) } 1}</math> | ||
+ | ==See Also== | ||
{{AMC8 box|year=2018|num-b=14|num-a=16}} | {{AMC8 box|year=2018|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:45, 21 November 2018
Problem 15
In the diagram below, a diameter of each of the two smaller circles is a radius of the larger circle. If the two smaller circles have a combined area of square unit, then what is the area of the shaded region, in square units?
Solution
Let the radius of the large circle be . Then the radii of the smaller circles are . The areas the circles are directly proportional to the square of the radii, so the ratio of the area of the small circle to the large one is . This means the combined area of the 2 smaller circles is the larger circle, therefore the shaded region is equal to the combined area of the 2 smaller circles, which is
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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