1978 AHSME Problems/Problem 2

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