Difference between revisions of "2010 AMC 8 Problems/Problem 16"
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== Solution == | == Solution == | ||
− | Let the side length of the square be <math>s</math>, and let the radius of the circle be <math>r</math>. Thus we have <math>s^2=r^2\pi</math>. Dividing each side by <math>r^2</math>, we get <math>\frac{s^2}{r^2}=\pi</math>. Since <math>\left(\frac{s}{r}\right)^2=\frac{s^2}{r^2}</math>, we have <math>\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}</math> | + | Let the side length of the square be <math>s</math>, and let the radius of the circle be <math>r</math>. Thus we have <math>s^2=r^2\pi</math>. Dividing each side by <math>r^2</math>, we get <math>\frac{s^2}{r^2}=\pi</math>. Since <math>\left(\frac{s}{r}\right)^2=\frac{s^2}{r^2}</math>, we have <math>\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}</math> |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=15|num-a=17}} | {{AMC8 box|year=2010|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:48, 25 December 2018
Problem 16
A square and a circle have the same area. What is the ratio of the side length of the square to the radius of the circle?
Solution
Let the side length of the square be , and let the radius of the circle be . Thus we have . Dividing each side by , we get . Since , we have
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 | ||
All AJHSME/AMC 8 Problems and Solutions |
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