SOLUTION: I asked a question about a boxing ring yesterday and have studied up and down on the pythoragean theorum and have gotten nowhere with it. My question is this: A boxing ring in th

Algebra ->  Triangles -> SOLUTION: I asked a question about a boxing ring yesterday and have studied up and down on the pythoragean theorum and have gotten nowhere with it. My question is this: A boxing ring in th      Log On


   

Question 1930: I asked a question about a boxing ring yesterday and have studied up and down on the pythoragean theorum and have gotten nowhere with it. My question is this:
A boxing ring in the shape of a square, 20 ft on each side. How far apart would the boxers be when they are in opposite corners of the ring?
I think it would be 20 ft but I cannot figure out how to explain it.
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
draw a square, label the 4 corners clockwise A, B, C and D, starting from any of the corners.
Assume boxers are in corners A and C, opposite each other. Draw straight line AC
Now have a right-angled triangle, ABC where AC is the hypotenuse and AB and BC are sides of the ring, each 20ft, so
AC%5E2+=+AB%5E2+%2B+BC%5E2
AC%5E2+=+20%5E2+%2B+20%5E2
AC%5E2+=+400+%2B+400
AC%5E2+=+800
AC+=+sqrt%28800%29
AC = 28.28ft approx.
Jon.