# Distance

• Aug 2nd 2006, 05:35 PM
pashah
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.
• Aug 2nd 2006, 05:42 PM
Quick
Quote:

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:

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

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

$\sqrt{1089+1089}=c$

$\sqrt{2178}=c$

$\sqrt{2178}=c$

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

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

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

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

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

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

so the distance from a corner to the center is about 23 feet
• Aug 2nd 2006, 05:51 PM
ThePerfectHacker
Quote:

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,
$s\sqrt{2}$
Thus,
$33\sqrt{2}\approx 46.53$
• Aug 2nd 2006, 05:53 PM
Quick
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
• Aug 2nd 2006, 07:58 PM
c_323_h
$\sqrt{\bigg({\frac{33}{2}\bigg)^2}+{\bigg(\frac{33 }{2}\bigg)^2}}} \approx 23.33$