1978 AHSME Problems/Problem 2

Problem 2

If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is

$\textbf{(A) }\frac{1}{\pi^2}\qquad \textbf{(B) }\frac{1}{\pi}\qquad \textbf{(C) }1\qquad \textbf{(D) }\pi\qquad \textbf{(E) }\pi^2$

Solution 1

Creating equations, we get $4\cdot\frac{1}{2\pi r} = 2r$. Simplifying, we get $\frac{1}{\pi r} = r$. Multiplying each side by $r$, we get $\frac{1}{\pi} = r^2$. Because the formula of the area of a circle is $\pi r^2$, we multiply each side by $\pi$ to get $1 = \pi r^2$. Therefore, our answer is $\boxed{\textbf{(C) }1}$

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See Also

 1978 AHSME (Problems • Answer Key • Resources) Preceded byProblem 1 Followed byProblem 3 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 All AHSME Problems and Solutions

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