Difference between revisions of "2018 AMC 8 Problems/Problem 15"
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Therefore, the area of the shaded region is <math>\boxed{\textbf{(D) } 1}</math>. | Therefore, the area of the shaded region is <math>\boxed{\textbf{(D) } 1}</math>. | ||
− | ==Video Solution (CREATIVE | + | ==Video Solution (CREATIVE ANALYSIS!!!)== |
https://youtu.be/tYfMj2SSVJc | https://youtu.be/tYfMj2SSVJc | ||
Revision as of 03:17, 25 December 2023
Contents
Problem
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 1
Let the radius of the large circle be . Then, the radius of the smaller circles are . The areas of 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 half of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is .
Solution 2
Let the radius of the two smaller circles be . It follows that the area of one of the smaller circles is . Thus, the area of the two inner circles combined would evaluate to which is . Since the radius of the bigger circle is two times that of the smaller circles (the diameter), the radius of the larger circle in terms of would be . The area of the larger circle would come to .
Subtracting the area of the smaller circles from that of the larger circle (since that would be the shaded region), we have
Therefore, the area of the shaded region is .
Video Solution (CREATIVE ANALYSIS!!!)
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Video Solutions
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/51K3uCzntWs?t=1474
~ pi_is_3.14
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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