SOLUTION: A baseball diamond is shown. The four corners of the diamond are "Home plate," "1st base," "2nd base," and "3rd base." There is a mound near the center of the diamond. The followin

Algebra ->  Pythagorean-theorem -> SOLUTION: A baseball diamond is shown. The four corners of the diamond are "Home plate," "1st base," "2nd base," and "3rd base." There is a mound near the center of the diamond. The followin      Log On


   

Question 1167707: A baseball diamond is shown. The four corners of the diamond are "Home plate," "1st base," "2nd base," and "3rd base." There is a mound near the center of the diamond. The following measurements are given.
Each side of the diamond is labeled 90 ft.
The distance between "Home plate" and the mound near the center of the diamond is labeled 60 ft, 6 inches.
In baseball, the pitcher's mound is 60 feet, 6 inches from home plate. How far (in ft) from the mound is second base?
The answer is not 127ft
Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
Of course it is not 127ft, that is the distance from home plate to 2nd base
( sqrt%2890%5E2+%2B+90%5E2%29+=+90%2Asqrt%282%29+=+127.28ft+ )


Subtract 60ft 6in (60.5ft) from 127.28ft to get the mound-to-2nd base distance:
66.78ft or approx. 66ft 9in.