Question

a baseball \diamond\ is actually a square with side lengths of 90 feet. in a game, a runner tries to steal second base. how far must the catcher throw from home to second base in order to get the runner out? 90\sqrt{2} feet 90\sqrt{3} feet 90 feet 180 feet

Explanation:

Step1: Identify the geometric shape

The baseball - diamond is a square with side length \(a = 90\) feet. The distance from home - plate to second - base is the length of the diagonal of the square.

Step2: Apply the Pythagorean theorem

For a square of side length \(a\), if the diagonal is \(d\), in a right - triangle formed by two adjacent sides of the square, by the Pythagorean theorem \(d^{2}=a^{2}+a^{2}\) (since the two legs of the right - triangle have length \(a\)). Given \(a = 90\) feet, then \(d^{2}=90^{2}+90^{2}=2\times90^{2}\).

Step3: Solve for \(d\)

Taking the square root of both sides of the equation \(d^{2}=2\times90^{2}\), we get \(d=\sqrt{2\times90^{2}} = 90\sqrt{2}\) feet.

Answer:

A. \(90\sqrt{2}\) feet