A boxing ring is in the shape of a square, 20 ft on each side. How far apart are the fighters when they are in opposite corners of the ring?
2007-05-31 16:23:29 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The diagonal of the square is 20*sqrt(2).
Pythagoras theorem.
2007-05-31 16:27:33 · answer #1 · answered by Dr D 7 · 1⤊ 1⤋
Using pythagorean (sp) theorem, a2+b2 = c2 where the 2's are squares you get
400+400=c2
the boxers are the square root of 800 ft away from each other, so just over 28.3 feet apart
2007-05-31 16:29:21 · answer #2 · answered by J O 2 · 0⤊ 0⤋
The diagonal of a 20x20 square is 20rad2 exactly.
This equals 28.284271247461900976033774484194
to 32 significant figures. ;-)
About 28 feet.
But then you can't assume that the two fighters are right at the corners and also there's the whole uncertainty principle thing too!
2007-05-31 16:27:54 · answer #3 · answered by gebobs 6 · 0⤊ 0⤋
Pythagoras' Theorem
a^2 + b^2 = c^2
a=b=20ft
c=sqrt(20^2+20^2)
=sqrt(800)
= 20sqrt(2) ft --------> exact answer
= 28.284 ft -------> approximate answer
2007-05-31 16:27:56 · answer #4 · answered by gudspeling 7 · 0⤊ 0⤋