The catcher is trying to throw out a base runner trying to steal second base. if the distance between bases is 90 feet , approximately how far is the catcher's throw? (round to the nearest whole number)

see attachment below.

I have a slight problem going into this question: I don't know much
about baseball.

I know it's called a baseball "diamond", but is it really a square ?
I mean is the distance from Home to 2nd base the same as the
distance between 1st and 3rd base ?

You know what ?  If it's not a square, then we don't have enough
information to solve this problem.  But since they gave it to you to
solve, with just that much information, I guess it must be square.
OK.  The baseball diamond is actually a square ... at least THIS
one is.

That means there's a right angle at each base.

Great !
Get your scissors and carefully cut the baseball diamond in half,
along the line from Home to 2nd base.  Then throw away one half.

Now what do you hold in your hand ?

It's a right triangle.  The legs are the two 90-ft base lines, the right angle
is at that base, and the hypotenuse is the line from Home to 2nd base ...
exactly the distance the catcher has to throw to Second.

Can you find the length of the hypotenuse when the legs are both 90 ft ?

Remember the hypotenuse is

     the square root of (one leg squared plus the other leg squared).

You can finish it off from here.  Take it, Braydon !

What’s the improper fraction?