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

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==See Also==
==See Also==
{{AMC8 box|year=2018|num-b=14|num-a=16}}
{{AMC8 box|year=2018|num-b=14|ny7t6yguhujik
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Revision as of 15:20, 1 October 2019

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 $\boxed{(D) 1}.$

See Also

{{AMC8 box|year=2018|num-b=14|ny7t6yguhujik

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