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 $4\cdot\frac{1}{2\pir} = 2r$ (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}$

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1978 AHSME Problems/Problem 2

Problem 2

Solution 1