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.

2 Answers

pohnpei397's profile pic

pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted on

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

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.