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)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