1978 AHSME Problems/Problem 2: Difference between revisions
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}{..." |
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==Solution 1== | ==Solution 1== | ||
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} | 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 < | 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
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 
. Because the formula of the area of a circle is $\pir^2$ (Error compiling LaTeX. Unknown error_msg), we multiply each side by 
to get $1 = \pir^2$ (Error compiling LaTeX. Unknown error_msg).
Therefore, our answer is 
F
