Difference between revisions of "2010 AMC 8 Problems/Problem 16"
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− | == Problem == | + | == 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? | 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? | ||
<math> \textbf{(A)}\ \frac{\sqrt{\pi}}{2}\qquad\textbf{(B)}\ \sqrt{\pi}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ 2\pi\qquad\textbf{(E)}\ \pi^{2} </math> | <math> \textbf{(A)}\ \frac{\sqrt{\pi}}{2}\qquad\textbf{(B)}\ \sqrt{\pi}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ 2\pi\qquad\textbf{(E)}\ \pi^{2} </math> |
Revision as of 21:04, 22 October 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
- Timmy Deng ~ North Carolina
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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