Question

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

Explanation:

Step1: Recognize the problem as a right - triangle problem

A baseball diamond is a square. The distance from home to second base is the length of the diagonal of the square. If the side of the square \(a = 90\) feet, and we consider the right - triangle formed by two adjacent sides of the square (e.g., from home to first and first to second) and the diagonal (home to second). By the Pythagorean theorem \(c^{2}=a^{2}+b^{2}\), and since \(a = b=90\) (sides of the square) in this case.

Step2: Apply the Pythagorean theorem

Substitute \(a = 90\) and \(b = 90\) into the formula \(c=\sqrt{a^{2}+b^{2}}\). So \(c=\sqrt{90^{2}+90^{2}}=\sqrt{8100 + 8100}=\sqrt{16200}\).
We can simplify \(\sqrt{16200}=\sqrt{8100\times2}=90\sqrt{2}\approx90\times1.414 = 127.26\) feet.

Answer:

The distance from home to second base is \(90\sqrt{2}\approx127.26\) feet.