Question

The distance between consecutive bases is 90 feet. An outfielder catches the ball on the third base line about 40 feet behind third base. How far would the outfielder have to throw the ball to first base?

Answer 1

If we assume 3rd base is at (0,0), the ball is caught at a spot along the line y = -x such that (x,y) is 45 from (0,0). In other words, it's a right angle triangle with two equal angles, and the hypotenuse is 45. Find the other two sides.

Next, the distance across the diamond from 3rd base to 1st base is the hypotenuse of another right triangle with two equal angles, except this time we know the two sides and need to find the hypotenuse.

Having found the two sides of the little triangle behind 3rd base, and the hypotenuse of the triangle formed by home plate, 1st base and 3rd base, we can set up yet another right triangle and use Pythagorean theorem to find the hypotenuse, which will be the distance the ball must be thrown. Alternatively, we can use the distance formula (which is just the Pythagorean theorem in disguise) and the coordinates of the spot the ball was caught and the coordinates of 1st base. 3rd base is at (0,0), so 1st base will be at (h,0) where h is the value we found for the hypotenuse of the triangle formed by home, 1st, 3rd.