Difference between revisions of "2018 AMC 8 Problems/Problem 15"
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− | ==Problem | + | ==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 <math>1</math> square unit, then what is the area of the shaded region, in square units? | 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 <math>1</math> square unit, then what is the area of the shaded region, in square units? | ||
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filldraw(shift(1,0)*unitcircle,white,black); | filldraw(shift(1,0)*unitcircle,white,black); | ||
</asy> | </asy> | ||
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<math>\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}</math> | <math>\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}</math> | ||
− | ==Solution== | + | ==Solution 1== |
+ | |||
+ | 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 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 <math>\frac 14</math>. 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 <math>\boxed{\textbf{(D) } 1}</math> | ||
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+ | ==Solution 2== | ||
+ | |||
+ | Let the radius of the two smaller circles be <math>r</math>. It follows that the area of one of the smaller circles is <math>{\pi}r^2</math>. Thus, the area of the two inner circles combined would evaluate to <math>2{\pi}r^2</math> which is <math>1</math>. 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 <math>r</math> would be <math>2r</math>. The area of the larger circle would come to <math>(2r)^2{\pi} = 4{\pi}r^2</math>. | ||
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+ | Subtracting the area of the smaller circles from that of the larger circle(since that would be the shaded region), we have <cmath>4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.</cmath> | ||
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+ | Therefore, the area of the shaded region is <math>\boxed{\textbf{(D) } 1}</math> | ||
+ | |||
+ | ==Video Solutions == | ||
+ | https://youtu.be/-3WEf3EjGu0 | ||
+ | |||
+ | https://youtu.be/-JR7R0PyU-w | ||
− | + | ~savannahsolver | |
+ | ==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}} |
Latest revision as of 11:08, 18 February 2022
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 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
Video Solutions
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.