Question

a baseball diamond is a square, 89.0 ft on a side, with home plate and the three bases as vertices. the pitchers position is 60.5 ft from home plate. find the distance from the pitchers position to each of the bases. the distance from the pitchers position to 1st base, 2nd base, and 3rd base is □ ft, □ ft, and □ ft, respectively. (round to the nearest tenth as needed.)

Explanation:

Step1: Find the length of the diagonal of the square

The formula for the diagonal \(d\) of a square with side length \(s\) is \(d = s\sqrt{2}\). Given \(s=89.0\) ft, so \(d = 89.0\times\sqrt{2}\approx89.0\times1.414 = 125.846\) ft.

Step2: Use the distance from pitcher to home plate to find the distance from pitcher to each base

Let the distance from pitcher to base be \(x\). The distance from home - plate to base is \(d/2\) (half of the diagonal of the square). We can use the law of cosines. In a square, the angle between the line from home - plate to pitcher and home - plate to base is \(45^{\circ}\).
By the law of cosines \(x^{2}=60.5^{2}+(d/2)^{2}-2\times60.5\times(d/2)\times\cos45^{\circ}\)
Substitute \(d = 89.0\sqrt{2}\), then \(d/2=89.0\times\frac{\sqrt{2}}{2}\approx63.0\)
\(x^{2}=60.5^{2}+63.0^{2}-2\times60.5\times63.0\times\frac{\sqrt{2}}{2}\)
\(x^{2}=3660.25 + 3969-2\times60.5\times63.0\times0.707\)
\(x^{2}=3660.25+3969-(7623\times0.707)\)
\(x^{2}=3660.25 + 3969 - 5399.4\)
\(x^{2}=2229.85\)
\(x=\sqrt{2229.85}\approx47.2\)

Answer:

The distance from the pitcher's position to 1st base, 2nd base, and 3rd base is \(47.2\) ft, \(47.2\) ft, and \(47.2\) ft respectively.