1978 AHSME Problems/Problem 2: Difference between revisions
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==Solution 1== | ==Solution 1== | ||
Creating equations, we get <math>4\cdot\frac{1}{2\pi r} = 2r</math>. Simplifying, we get <math>\frac{1}{\pi r} = r</math>. Multiplying each side by <math>r</math>, we get <math>\frac{1}{\pi} = r^2</math>. Because the formula of the area of a circle is <math>\pi r^2</math>, we multiply each side by <math>\pi</math> to get <math>1 = \pi r^2</math>. | Creating equations, we get <math>4\cdot\frac{1}{2\pi r} = 2r</math>. Simplifying, we get <math>\frac{1}{\pi r} = r</math>. Multiplying each side by <math>r</math>, we get <math>\frac{1}{\pi} = r^2</math>. Because the formula of the area of a circle is <math>\pi r^2</math>, we multiply each side by <math>\pi</math> to get <math>1 = \pi r^2</math>. | ||
Therefore, our answer is <math>\boxed{\textbf{(C) }1}</math> | Therefore, our answer is <math>\boxed{\textbf{(C) }1}</math> | ||
~awin | |||
Revision as of 14:33, 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 
. Simplifying, we get 
. Multiplying each side by 
, we get 
. Because the formula of the area of a circle is 
, we multiply each side by 
to get 
.
Therefore, our answer is 
~awin
