2018 AMC 8 Problems/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 $1$ 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 $\boxed{\textbf{(D) } 1}$ .

Video Solutions

https://youtu.be/-3WEf3EjGu0

https://youtu.be/-JR7R0PyU-w

~savannahsolver

2018 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2018 AMC 8 Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

Video Solutions

See Also