1960 AHSME Problems/Problem 7

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

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1960 AHSME Problems/Problem 7

Contents

Problem

Solutions

Solution 1

Solution 2

See Also