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Difference between revisions of "1978 AHSME Problems/Problem 2"

(Created page with "== 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 <math>\textbf{(A) }\frac{1}{...")
 
(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
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Creating equations, we get <math>4\cdot\frac{1}{2\pir} = 2r</math>. Simplifying, we get <math>\frac{1}{\pir} = r</math>. Multiplying each side by r, we get <math>\frac{1}{\pi}</math> = r^2<math>. Because the formula of the area of a circle is </math>\pir^2<math>, we multiply each side by </math>\pi<math> to get </math>1 = \pir^2<math>.
+
Creating equations, we get <math>4\cdot\frac{1}{2\pir} = 2r</math>. Simplifying, we get <math>\frac{1}{\pir} = r</math>. Multiplying each side by r, we get <math>\frac{1}{\pi} = r^2</math>. Because the formula of the area of a circle is <math>\pir^2</math>, we multiply each side by <math>\pi</math> to get <math>1 = \pir^2</math>.
−
Therefore, our answer is </math>\boxed{\textbf{(C)  }1}$
+
Therefore, our answer is <math>\boxed{\textbf{(C)  }1}</math>F

Revision as of 14:32, 20 January 2020

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 $\pir^2$ (Error compiling LaTeX. Unknown error_msg), we multiply each side by $\pi$ to get $1 = \pir^2$ (Error compiling LaTeX. Unknown error_msg). Therefore, our answer is $\boxed{\textbf{(C) }1}$F

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