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 radius 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>. | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Therefore, the area of the shaded region is <math>\boxed{\textbf{(D) } 1}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | The area of a small circle is <math>\frac{1}{2}= \pi r^2</math>. Solving, we get <math>r = \sqrt{\frac{1}{2\pi}}</math>. | ||
+ | |||
+ | The radius of the large circle is <math>R=2r</math>. The area of the large circle is <math>{\pi}R^2={\pi}(2r)^2=4{\pi}r^2=4{\pi}\frac{1}{2\pi}=2</math>. | ||
+ | |||
+ | Subtract the area of the small circles from the area of the large circle to get the area of the shaded region: <math>2-1=</math> <math>\boxed{\textbf{(D) } 1}</math>. | ||
+ | |||
+ | ==Video Solution (CREATIVE ANALYSIS!!!)== | ||
+ | https://youtu.be/tYfMj2SSVJc | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solutions == | ||
+ | https://youtu.be/-3WEf3EjGu0 | ||
+ | |||
+ | https://youtu.be/-JR7R0PyU-w | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/51K3uCzntWs?t=1474 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
− | |||
==See Also== | ==See Also== |
Latest revision as of 10:15, 10 April 2024
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 .
Solution 3
The area of a small circle is . Solving, we get
.
The radius of the large circle is . The area of the large circle is
.
Subtract the area of the small circles from the area of the large circle to get the area of the shaded region:
.
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.