Question

A baseball diamond is a square that is 90 feet on each side. A player threw the ball from 2nd base to home.
How far did the ball travel? Round to the nearest tenth.

Answer 1

To find the distance the ball travels from 2nd base to home plate on a baseball diamond, we can visualize the baseball diamond as a square with vertices at the four bases.

We can set up the square as follows:
- 1st base (A) is located at (90, 0)
- 2nd base (B) is at (90, 90)
- 3rd base (C) is at (0, 90)
- Home plate (D) is at (0, 0)

The coordinates for 2nd base (B) are (90, 90) and for home plate (D) are (0, 0).

To find the distance between these two points, we can use the distance formula:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

Substituting in the coordinates of 2nd base and home plate:

\[
x_1 = 90, \quad y_1 = 90, \quad x_2 = 0, \quad y_2 = 0
\]

Now we can calculate:

\[
d = \sqrt{(0 - 90)^2 + (0 - 90)^2}
\]

Calculating the squares:

\[
d = \sqrt{(-90)^2 + (-90)^2} = \sqrt{8100 + 8100} = \sqrt{16200}
\]

Now we can simplify this:

\[
d = \sqrt{16200} \approx 127.279
\]

Rounding this to the nearest tenth gives:

\[
d \approx 127.3 \text{ feet}
\]

Therefore, the distance the ball traveled from 2nd base to home plate is approximately **127.3 feet**.