SOLUTION: A baseball diamond is actually a square 90 feet on a side. A catcher fields a bunt along the third baseline 10 feet from home plate. How far would the catcher have to throw the bal

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: A baseball diamond is actually a square 90 feet on a side. A catcher fields a bunt along the third baseline 10 feet from home plate. How far would the catcher have to throw the bal      Log On


   

Question 440835: A baseball diamond is actually a square 90 feet on a side. A catcher fields a bunt along the third baseline 10 feet from home plate. How far would the catcher have to throw the ball to second base. This is a homework problem and I need to show all work.
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
(80)2+(90)2=x2
6400+8100=x2
14500=x2
√14500=x=120.416 ft..