# In a baseball game, baseball dimond is a square with sides 27m. how far does the first player have to throw the ball to get a runner out a third base.

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### 2 Answers

The answer to this is that the first baseman must throw 38.18 meters, if he is on first base, to a person standing on the third base bag. This can be found by using the Pythagorean Theorem.

If you draw out a diagram of the baseball diamond, you will see that the baselines from home to third and home to first make a right angle at home plate. This means that a line from third to first is the hypotenuse of a right triangle.

The two legs of the triangle are each 27 meters. Therefore:

27^2 + 27^2 = c^2 (the distance from 1st to 3rd).

1458 = c^2

c = 38.18

We know that the throw is the horizontal distance to be covered alon the diagonal length of the square.

The diagonal length of suare of side is given by:

d^2 = s^2+s^2 , by Pythagoras theorem. s is the length of the side of the square.

d^2 = 27^2 +27^2

d ^2 = 2*27^2.

Taking the square root, we get:

d = 27* sqrt 2

d = 27*1.414214 =38.1838 meter.