Problem 49 - Algebra and Trigonometry

Question

A Major League baseball diamond has four bases forming a square whose sides measure 90 feet each. The pitcher's mound is 60.5 feet from home plate on a line joining home plate and second base. Find the distance from the pitcher's mound to first base. Round to the nearest tenth of a foot.

Step-by-Step Solution

Answer

The distance from the pitcher's mound to first base, when rounded to the nearest tenth of a foot, is approximately 49 feet.

Step 1: Identify the triangle

Visualize the baseball diamond as a square and note the pitcher's mound, home plate, and first base forming a right triangle. The diagonal of a square divides it into two congruent right triangles. The distance from home plate to second base forms the hypotenuse of the right triangle.

Step 2: Calculate hypotenuse

Calculate the distance from home plate to second base, which is the hypotenuse of the right triangle. Using the Pythagorean theorem, the hypotenuse \(c\) of a right triangle can be calculated as the square root of the sum of the squares of the other two sides. Given that both sides of the square are 90 feet, the hypotenuse is calculated as \(\sqrt{90^2 + 90^2}\).

Step 3: Calculate the distance from the pitcher's mound to first base

Our main aim is to find the third side of the smaller right triangle formed by the distance from home plate to the pitcher's mound and from the pitcher's mound to first base. This can be done using the Pythagorean theorem, where the hypotenuse is the diagonal of the square. However, we know the hypotenuse of the smaller right triangle (from home plate to the pitcher's mound) is 60.5 feet, and the hypotenuse of the large right triangle (the square's diagonal) is found in step 2. The distance from the pitcher's mound to first base \(d\) can be calculated as \(\sqrt{\text{Diagonal}^2 - 60.5^2}\).