The base-path of a baseball diamond form a square. if it is 90 ft from home to first, how far does the catcher have to throw to catch someone stealing second base?

Question

Genevieve

Mathematics

12 March, 14:03

The base-path of a baseball diamond form a square. if it is 90 ft from home to first, how far does the catcher have to throw to catch someone stealing second base?

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Answers (1)

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Jeffery Pearson · Answer 1

Draw a line from home plate to second base. The square will be divided into two congruent right triangles.

Focusing on one of the right triangles, we have two legs of 90 feet each and an unknown hypotenuse. Call the hypotenuse x for now

a = 90

b = 90

c = x

Pythagorean theorem

a^2+b^2 = c^2

90^2+90^2 = x^2

8100+8100 = x^2

16200 = x^2

x^2 = 16200

x = sqrt (16200)

x = 127.279

The last value is approximate

So the distance from home plate to second base is approximately 127.279 feet