Difference between revisions of "2010 AMC 8 Problems/Problem 16"
(→Solution 2) |
|||
(2 intermediate revisions by one other user not shown) | |||
Line 6: | Line 6: | ||
== 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> | ||
+ | |||
+ | niggers a niggers fr fr | ||
==Video by MathTalks== | ==Video by MathTalks== | ||
https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be | https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be | ||
− | |||
− | |||
− | |||
Latest revision as of 22:40, 21 April 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
niggers a niggers fr fr
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.