In a softball game, how far must the catcher throw to second base?

The distance to be thrown can be seen in the diagram below:
Line AB is the required distance.
The distance can be calculated using the Pythagorean Theorem for solving right triangles given to be:
[tex]\begin{gathered} c^2=a^2+b^2 \\ where \\ c=hypotenuse \\ a,b=legs \end{gathered}[/tex]From the image, we have that:
[tex]\begin{gathered} c=AB \\ a=60 \\ b=60 \end{gathered}[/tex]Therefore, we can calculate the distance to be:
[tex]\begin{gathered} AB^2=60^2+60^2 \\ AB^2=3600+3600=7200 \\ AB=\sqrt{7200} \\ AB=84.85\text{ feet} \end{gathered}[/tex]The distance is 84.85 feet.