Questions: A baseball diamond is a square, 94.0 ft on a side, with home plate and the three bases as vertices. The pitcher's position is 62.5 ft from home plate. Find the distance from the pitcher's position to each of the bases. The distance from the pitcher's position to 1st base, 2nd base, and 3rd base is (Round to the nearest tenth as needed.) ft, ft, and ft.

Solution Steps

The problem involves finding the distance from the pitcher's position to each of the bases on a baseball diamond, which is a square with each side measuring 94.0 ft. The pitcher's position is 62.5 ft from home plate.

Assume the baseball diamond is a coordinate plane with home plate at (0,0), 1st base at (94,0), 2nd base at (94,94), and 3rd base at (0,94). The pitcher's position is 62.5 ft from home plate along the line connecting home plate and 2nd base.

Since the pitcher's position is along the diagonal from home plate to 2nd base, we use the ratio of distances to find the coordinates. The diagonal distance of the square is \( 94\sqrt{2} \) ft. The pitcher's position is \( \frac{62.5}{94\sqrt{2}} \) of the way along the diagonal.

\[ \text{Pitcher's coordinates} = \left( \frac{62.5}{94\sqrt{2}} \times 94, \frac{62.5}{94\sqrt{2}} \times 94 \right) = \left( \frac{62.5 \times 94}{94\sqrt{2}}, \frac{62.5 \times 94}{94\sqrt{2}} \right) = \left( \frac{62.5}{\sqrt{2}}, \frac{62.5}{\sqrt{2}} \right) \]

\[ \text{Pitcher's coordinates} = \left( \frac{62.5\sqrt{2}}{2}, \frac{62.5\sqrt{2}}{2} \right) = \left( 31.25\sqrt{2}, 31.25\sqrt{2} \right) \]

Use the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \) to find the distance from the pitcher's position to each base.

\[ d = \sqrt{(31.25\sqrt{2} - 0)^2 + (31.25\sqrt{2} - 0)^2} = \sqrt{(31.25\sqrt{2})^2 + (31.25\sqrt{2})^2} = \sqrt{2 \times (31.25\sqrt{2})^2} = \sqrt{2 \times 1953.125} = \sqrt{3906.25} = 62.5 \text{ ft} \]

\[ d = \sqrt{(94 - 31.25\sqrt{2})^2 + (0 - 31.25\sqrt{2})^2} \]

\[ d = \sqrt{(94 - 31.25\sqrt{2})^2 + (94 - 31.25\sqrt{2})^2} \]

Final Answer