1. ## Distance

A boxing ring is in the shape of a square, 33 feet on each side. How far apart are the fighters when one is in the center and the other is in the corner.

2. Originally Posted by pashah
A boxing ring is in the shape of a square, 33 feet on each side. How far apart are the fighters when one is in the center and the other is in the corner.
The diagnol of a square is the line from one corner to the opposite corner (it's the hypotenuse of a right triangle with the same sides as the square)

therefore we find the distance from one corner to the opposite corner:

$\displaystyle \sqrt{a^2+b^2}=c$

$\displaystyle \sqrt{33^2+33^2}=c$

$\displaystyle \sqrt{1089+1089}=c$

$\displaystyle \sqrt{2178}=c$

$\displaystyle \sqrt{2178}=c$

but we only want to go half that distance, so we divide that answer by 2:

$\displaystyle \frac{\sqrt{2178}}{2}=\frac{c}{2}$

$\displaystyle \frac{\sqrt{2178}}{\sqrt4}=\frac{c}{2}$

$\displaystyle \sqrt{\frac{2178}{4}}=\frac{c}{2}$

$\displaystyle \sqrt{544.5}=\frac{c}{2}$

$\displaystyle 23.3345\approx\frac{c}{2}$

so the distance from a corner to the center is about 23 feet

3. Originally Posted by pashah
A boxing ring is in the shape of a square, 33 feet on each side. How far apart are the fighters when one is in the center and the other is in the corner.
The length of the square diagnol is,
$\displaystyle s\sqrt{2}$
Thus,
$\displaystyle 33\sqrt{2}\approx 46.53$

4. here's a diagram of what I'm talking about (note: x and y are boxers)(also note: it's supposed to be a square):

Code:

-------x
|    / |
|   /  | a
|  y   |
| /    |
|/     |
-------
b
To Hacker: one of the boxers is in the CENTER of the square

5. $\displaystyle \sqrt{\bigg({\frac{33}{2}\bigg)^2}+{\bigg(\frac{33 }{2}\bigg)^2}}} \approx 23.33$