Difference between revisions of "2010 AMC 8 Problems/Problem 16"

Revision as of 17:26, 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?

$\textbf{(A)}\ \frac{\sqrt{\pi}}{2}\qquad\textbf{(B)}\ \sqrt{\pi}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ 2\pi\qquad\textbf{(E)}\ \pi^{2}$

Solution

Let the side length of the square be $s$ , and let the radius of the circle be $r$ . Thus we have $s^2=r^2\pi$ . Dividing each side by $r^2$ , we get $\frac{s^2}{r^2}=\pi$ . Since $\left(\frac{s}{r}\right)^2=\frac{s^2}{r^2}$ , we have $\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}$

Video by MathTalks

https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be

Solution 2

Let the area of both the circle and square be $\pi$ . We know that the radius of a circle with area $\pi$ is 1 and the side length of the square is $\sqrt{\pi}$ . Now we can calculate $\frac{\sqrt{\pi}}{1} \Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}$

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.

@@ Line 11: / Line 11: @@
 https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
+==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>

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