2018 AMC 8 Problems/Problem 15: Difference between revisions
No edit summary |
No edit summary |
||
| Line 10: | Line 10: | ||
<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== | |||
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 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 <math>\frac 12</math> the larger circle, therefore the shaded region is equal to the combined area of the 2 smaller circles, which is <math>\boxed{\textbf{(D) } 1}</math> | |||
{{AMC8 box|year=2018|num-b=14|num-a=16}} | {{AMC8 box|year=2018|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:51, 21 November 2018
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
Let the radius of the large circle be 
. Then the radii of the smaller circles are 
. The areas 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 
the larger circle, therefore the shaded region is equal to the combined area of the 2 smaller circles, which is 
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
