Question

a baseball field is in the shape of a square that is 90 feet on a side as shown. A player catches a ball at point x, which is three-quarters of the way between 2nd base & 3rd base, and throws the ball to 1st base.

What is the distance the player throws the call?

Answer 1

Answer:

The distance the player throws the ball is $112.5\ ft$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

In the right triangle ABX

$AB=90\ ft$

$BX=(3/4)90=67.5\ ft$

XA --------> is the distance the player throws the ball (hypotenuse of the right triangle)

Applying the Pythagoras Theorem

$XA^{2}=AB^{2}+BX^{2}$

substitute the values

$XA^{2}=90^{2}+67.5^{2}$

$XA^{2}=12,656.25$

$XA=112.5\ ft$

Answer 2

The distance player throws the call was calculated using Pythagoras theorem and it is 112.5 feet.

It is given that

Side of the square = 90 feet

Distance of point x from the second base = 3/4 * 90 = 67.5

The distance the player throws the call let us say D will be the shortest distance between x and 1st base.

Distance D can be calculated by Pythagoras theorem

What is Pythagoras theorem?

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the perpendicular and the base.

$D = \sqrt{90^{2} +67.5^{2} }$

$D = \sqrt{8100+4556.25} \\\\D = \sqrt{12656.25\\}$

$D=\sqrt{12656.25} = 112.5$

Therefore, the distance the player throws the call is 112.5 feet.

To get more about Pythagoras theorem, visit:

https://brainly.com/question/343682