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Difference between revisions of "2018 AMC 8 Problems/Problem 15"
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==Solution 1==
 
==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 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==
 
==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>.  
+
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>
+
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
  
−
Therefore, the area of the shaded region is <math>\boxed{\textbf{(D) } 1}</math>
 
  
 
==See Also==
 
==See Also==

Latest revision as of 09:15, 10 April 2024

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 $1$ 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 1

Let the radius of the large circle be $R$. Then, the radius 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 half of the larger circle, and therefore the shaded region is equal to the combined area of the 2 smaller circles, which is $\boxed{\textbf{(D) } 1}$.

Solution 2

Let the radius of the two smaller circles be $r$. It follows that the area of one of the smaller circles is ${\pi}r^2$. Thus, the area of the two inner circles combined would evaluate to $2{\pi}r^2$ which is $1$. 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 $r$ would be $2r$. The area of the larger circle would come to $(2r)^2{\pi} = 4{\pi}r^2$.

Subtracting the area of the smaller circles from that of the larger circle (since that would be the shaded region), we have \[4{\pi}r^2 - 2{\pi}r^2 = 2{\pi}r^2 = 1.\]

Therefore, the area of the shaded region is $\boxed{\textbf{(D) } 1}$.

Solution 3

The area of a small circle is $\frac{1}{2}= \pi r^2$. Solving, we get $r = \sqrt{\frac{1}{2\pi}}$.

The radius of the large circle is $R=2r$. The area of the large circle is ${\pi}R^2={\pi}(2r)^2=4{\pi}r^2=4{\pi}\frac{1}{2\pi}=2$.

Subtract the area of the small circles from the area of the large circle to get the area of the shaded region: $2-1=$ $\boxed{\textbf{(D) } 1}$.

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

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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