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 $4\cdot\frac{1}{2\pi r} = 2r$. 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 $\boxed{\textbf{(C)  }1}$

~awin


See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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