1955 AHSME Problems/Problem 15

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Problem 15

The ratio of the areas of two concentric circles is $1: 3$. If the radius of the smaller is $r$, then the difference between the radii is best approximated by:

$\textbf{(A)}\ 0.41r \qquad \textbf{(B)}\ 0.73 \qquad \textbf{(C)}\ 0.75 \qquad \textbf{(D)}\ 0.73r \qquad \textbf{(E)}\ 0.75r$

Solution

Observe that there are 3 feet in 1 yard, 9 square feet in 1 square yard, and 27 cubic feet in 1 cubic yard.

This means that if the ratio is $x$ in 1D, it is $x^2$ in 2D, and it is $x^3$ in 3D.

We can apply this thinking into our current question, which would have $1:3$ in the 2D-plane. Since the radius is 1D, we would have to square root both sides to get $1:\sqrt{3}$

If we plug in $r$ for the smaller radius, we get $r:\sqrt{3}r$ The difference between the two sides would be (approximately) $1.732r - r = \textbf{(D)} 0.73r$

See Also

1955 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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