Suppose a certain baseball diamond is a square 95 feet on a side. The pitching rubber is located 61.5 feet from home plate on a line joining home plate and second base. How far is it from the pitching rubber to first base? The distance from the pitching rubber to first base is about \( \square \) feet. (Round to the nearest tenth as needed.)

To find the distance from the pitching rubber to first base, we can visualize the baseball diamond as a square with home plate (H), first base (1B), second base (2B), and third base (3B) at the corners. The pitching rubber is located 61.5 feet from home plate along the diagonal towards second base. The distance from home plate to first base forms a right triangle where: - One leg (the distance from home to first base) is the side of the square, which is 95 feet. - The other leg is the vertical distance from the pitching rubber (61.5 feet from home plate) straight towards the line connecting home to first base. Using the Pythagorean theorem, we can calculate the distance to first base (let's call this distance d): 1. Identify the base of the triangle: - The distance from home plate to first base is 95 feet. - The vertical leg from the pitching rubber to home is the distance remaining from the pitching rubber down to the baseline, which is \( 61.5^2 + x^2 = 95^2 \). Therefore: \[ d = \sqrt{95^2 + (61.5)^2} = \sqrt{9025 + 3782.25} = \sqrt{12807.25} \approx 113.2 \text{ feet} \] The distance from the pitching rubber to first base is about \( 113.2 \) feet.