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

(Solution 2)
(Solution 2)
 
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==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
+
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>.
  
smaller circles is <math>{\pi}r^2</math>. Thus, the area of the two inner circles combined would
+
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>
 
 
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>
 
Therefore, the area of the shaded region is <math>\boxed{\textbf{(D) } 1}</math>

Latest revision as of 15:48, 28 October 2020

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 1

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 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 $\boxed{\textbf{(D) } 1}$

See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
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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