We can use the Pythagorean theorem to solve this problem. Let's draw a diagram of the baseball diamond:
A ---- B
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D ---- C
Let A be home plate, B be first base, C be second base, and D be third base. The pitching rubber is located on a line connecting home plate and second base, 58.5 feet from home plate. Let's call the point where this line intersects the side AD "E". We want to find the distance from E to B.
Since the baseball diamond is a square, we know that AB = BC = CD = DA = 92 feet. We also know that AE + EC = AC = 92 feet. Using the Pythagorean theorem, we can find AE:
AE^2 + EC^2 = AC^2
AE^2 + (92 - AE)^2 = 92^2
Simplifying and solving for AE:
2AE^2 - 2(92)AE + 92^2 - 58.5^2 = 0
AE = (92 ± sqrt(4016.25)) / 2
AE ≈ 44.84 feet (we discard the negative solution since AE is a distance)
Finally, we can use the Pythagorean theorem again to find the distance from E to B:
EB^2 = AB^2 - AE^2
EB^2 = 92^2 - 44.84^2
EB ≈ 80.78 feet
Therefore, it is about 80.78 feet from the pitching rubber to first base.