A baseball diamond is a square, 87.0 ft on a side, with
home plate and the three bases as vertices.
The pitcher's position is 60.5 ft from home plate. Find
the distance from the pitcher's position to each of the
bases.
3rd base
***
2nd base
60.5 ft
The distance from the pitcher's position to 1st base, 2nd base, and 3rd base is
(Round to the nearest tenth as needed.)
87.0 ft
ft,
1st base
ft, and
ft, respectively.

A baseball diamond is a square 870 ft on a side with home plate and the three bases as vertices The pitchers position is 605 ft from home plate Find the distanc class=

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Answer:

So, the answers are:

- Distance to first base: 104.3 ft

- Distance to second base: 87 ft

- Distance to third base:104.3 ft

Step-by-step explanation:

The distance from the pitcher's position to first base is the hypotenuse of a right triangle with legs of length 60.5 ft and 87 ft. Using the Pythagorean theorem:

\[ \sqrt{60.5^2 + 87^2} \approx 104.3 \ \text{ft} \ (rounded to the nearest tenth) \]

The distance from the pitcher's position to second base is the same as the distance from first base to home plate, which is simply the side length of the square, or 87 ft.

The distance from the pitcher's position to third base can be found using the Pythagorean theorem again, with legs of length 60.5 ft and 87 ft:

\[ \sqrt{60.5^2 + 87^2} \approx 104.3 \ \text{ft} \ (rounded to the nearest tenth) \]

So, the answers are:

- Distance to first base: 104.3 ft

- Distance to second base: 87 ft

- Distance to third base:104.3 ft