Difference between revisions of "1978 AHSME Problems/Problem 2"
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==Solution 1== | ==Solution 1== | ||
| − | Creating equations, we get <math>4\cdot\frac{1}{2\ | + | 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 | ||
| + | |||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME box|year=1978|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:59, 13 February 2021
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
See Also
| 1978 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
