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

Latest revision as of 17:57, 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 $\boxed{\textbf{(D)}}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

@@ Line 37: / Line 37: @@
 ==See Also==
-{{AHSME box|year=1960|num-b=6|num-a=8}}
+{{AHSME 40p box|year=1960|num-b=6|num-a=8}}
+[[Category:Introductory Geometry Problems]]

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