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.

Printable View

- Aug 2nd 2006, 05:35 PMpashahDistance
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 PMQuickQuote:

Originally Posted by**pashah**

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 - Aug 2nd 2006, 05:51 PMThePerfectHackerQuote:

Originally Posted by**pashah**

$\displaystyle s\sqrt{2}$

Thus,

$\displaystyle 33\sqrt{2}\approx 46.53$ - Aug 2nd 2006, 05:53 PMQuick
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

- Aug 2nd 2006, 07:58 PMc_323_h
$\displaystyle \sqrt{\bigg({\frac{33}{2}\bigg)^2}+{\bigg(\frac{33 }{2}\bigg)^2}}} \approx 23.33$