Question

A baseball diamond is a square that is 90feet on each side. What is the distance a catcher has to throw the ball from home to second base?

Answer 1

Answer:

127.3 feet.

Step-by-step explanation:

We know that a baseball diamond has square form, which means all sides are congruent.

So, we already know that each side of the square is 90 feet.

Notice that the distance from home to second base is a diagonal of the square, which can be found using pythagorean theorem where the diagonal is the hypotenuse and both legst are 90 feet long.

$d^{2}=90^{2} +90^{2} \\d=\sqrt{8100+8100}\\ d=\sqrt{16200} \\d \approx 127.3$

Therefore, the distance a catcher has to throw the ball from home to second base is around 127.3 feet.

Answer 2

Answer:

well, since a real baseball diamond is 90 feet on each side then the distance from homebase to second is 127 feet  3 3/8 inches

hope this helps :D