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A good question and let me see if I can remember some basic geometry here. We know that it is 60.5 feet from the pitcher's rubber to home plate. We also know that it is 90 feet from home plate to the pitcher's mound. So using the Pythagoreum Theory, which states that the sum of the squares of the sides of a right triangle (distance between home and rubber, distance between first base and rubber) are equal to the square of the hypotenuse (distance between home and first base). This equasion is

a2 + b2 = c2 where a is the distance between home plate and the pitching rubber, b is the distance between the pitching rubber and first base, and c is the distance between home plate and first base. So using the information we already know, we can say that:

60.52 + b2 = 902 next, multiply to find the squares

3660.25 + b2 = 8100 next, subtract the 3660.25 from both sides of the equasion

b2 = 4439.75 next, find the square root of 4439.75

b = 66.631449031219484653750691943936

The distance between the pitching rubber and first base is 66.631449031219484653750691943936 feet.

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The above answer is incorrect, because it assumes that the pithcer's mound is at the exact center of the diamond, when it is, in fact, closer to home plate than 2nd base. The triangle described above is not in fact right, so the Pythagorean Theorem cannot be used. The below answer is correct.

PM to 1st = PM to 3rd = 63.717 feet, Law of Cosines:c^2=(60.5)^2+90^2-2(60.5)(90)(Cos 45 degrees)

PM to 2nd = 66.779 Feet, Pythagoreum Theorem and Subtraction: c^2 = 90^2 + 90^2, c minus 60.5

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Q: How far is the pitchers mound from all of the other bases?

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