Difference between revisions of "1960 AHSME Problems/Problem 7"

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[[Category:Introductory Geometry Problems]]

Latest revision as of 18:57, 17 May 2018

Problem

Circle $I$ passes through the center of, and is tangent to, circle $II$. The area of circle $I$ is $4$ square inches. Then the area of circle $II$, in square inches, is:

$\textbf{(A) }8\qquad \textbf{(B) }8\sqrt{2}\qquad \textbf{(C) }8\sqrt{\pi}\qquad \textbf{(D) }16\qquad \textbf{(E) }16\sqrt{2}$

Solutions

[asy] draw(circle((0,0),50)); draw(circle((-25,0),25)); [/asy]

Solution 1

Since Circle $I$ is tangent to circle $II$ and touches the center of circle $II$, the diameter of circle $I$ is the radius of circle $II$.

That means circle $II$ is twice as big as circle $I$, so the area of circle $II$ is four times as big as circle $I$.

The area of circle $II$ is $16$ square inches, so the answer is $\boxed{\textbf{(D)}}$.

Solution 2

Since Circle $I$ is tangent to circle $II$ and touches the center of circle $II$, the diameter of circle $I$ is the radius of circle $II$.

Applying the area formula $A = \pi r^2$, substitute $4$ for $A$ to solve for the radius of circle $1$. \[4 = \pi r^2\] \[\frac{4}{\pi} = r^2\] \[r = \frac{2}{\sqrt{\pi}}\]

That means the diameter of circle $I$ (or the radius of circle $II$) is $\frac{4}{\sqrt{\pi}}$. Apply the area formula again to find the area of circle $II$. \[A = \pi (\frac{2}{\sqrt{\pi}})^2 = \pi \cdot \frac{4}{\pi} = 16\]

The answer is $\boxed{\textbf{(D)}}$.

See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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All AHSME Problems and Solutions
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