Difference between revisions of "2010 AMC 8 Problems/Problem 16"
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==Solution 2== | ==Solution 2== | ||
| − | Let the area of both the circle and square be <math>\pi</math>. We know that the radius of a circle with area <math>\pi</math> is 1 and the side length of the square is <math>\sqrt{\pi}</math>. Now we can calculate <math>\frac{\sqrt{\pi}}{1} \Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}</math> | + | Let the area of both the circle and square be <math>\pi</math>. We know that the radius of a circle with area <math>\pi</math> is 1 and the side length of the square is <math>\sqrt{\pi}</math>. Now we can calculate the ratio <math>\frac{\sqrt{\pi}}{1} \Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}</math> |
==Video by MathTalks== | ==Video by MathTalks== | ||
Revision as of 19:07, 19 January 2024
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 
Solution 2
Let the area of both the circle and square be 
. We know that the radius of a circle with area is 1 and the side length of the square is 
. Now we can calculate the ratio 
Video by MathTalks
https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
