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

(→‎Solution 2)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
−
==Problem 15==
+
==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 <math>1</math> square unit, then what is the area of the shaded region, in square units?
 
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 <math>1</math> square unit, then what is the area of the shaded region, in square units?
  
Line 19: Line 19:
 
==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 12:54, 18 January 2021

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 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 (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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_8_Problems/Problem_15&oldid=142629"
Invalid username
Login to AoPS