Difference between revisions of "2018 AMC 8 Problems/Problem 15"
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be the shaded region), we have <cmath>4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.</cmath> | be the shaded region), we have <cmath>4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.</cmath> | ||
− | Therefore, the area of the shaded region is <math>\boxed{(D) 1} | + | Therefore, the area of the shaded region is <math>\boxed{\textbf{(D) } 1}</math> |
==See Also== | ==See Also== |
Revision as of 22:16, 25 October 2020
Contents
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 1
Let the radius of the large circle be . Then the radii 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
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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