Difference between revisions of "2018 AMC 8 Problems/Problem 15"
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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 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> | 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> | ||
Revision as of 12:33, 20 August 2019
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?
![[asy] size(4cm); filldraw(scale(2)*unitcircle,gray,black); filldraw(shift(-1,0)*unitcircle,white,black); filldraw(shift(1,0)*unitcircle,white,black); [/asy]](http://latex.artofproblemsolving.com/1/8/1/1814b98003d8f9d763c70aa6a56a765673f27d59.png)
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 ![\[4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.\]](http://latex.artofproblemsolving.com/0/4/b/04b655be0be7f899ec8d33c6714293b59d27eb24.png)
Therefore, the area of the shaded region is or 
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. 
