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Difference between revisions of "2018 AMC 8 Problems/Problem 15"

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==Solution==
 
==Solution==
  
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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 <math>\frac 12</math> 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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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 <math>\frac 12 \$ 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}$
  
 
==See Also==
 
==See Also==

Revision as of 22:12, 1 December 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 $1$ 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]

$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } 1 \qquad \textbf{(E) } \frac{\pi}{2}$

Solution

Let the radius of the large circle be $R$. Then the radii of the smaller circles are $\frac R2$. 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 $\frac 14$. This means the combined area of the 2 smaller circles is $\frac 12 $ of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is$\boxed{\textbf{(D) } 1}$

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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