Question

In a softball game, how far must the catcher throw to second base?

Answer 1

The distance to be thrown can be seen in the diagram below:

Line AB is the required distance.

The distance can be calculated using the Pythagorean Theorem for solving right triangles given to be:

`\begin{gathered} c^2=a^2+b^2 \\ where \\ c=hypotenuse \\ a,b=legs \end{gathered}`

From the image, we have that:

`\begin{gathered} c=AB \\ a=60 \\ b=60 \end{gathered}`

Therefore, we can calculate the distance to be:

`\begin{gathered} AB^2=60^2+60^2 \\ AB^2=3600+3600=7200 \\ AB=√(7200) \\ AB=84.85\text{ feet} \end{gathered}`

The distance is 84.85 feet.