Question

A baseball diamond is a square with sides 22.4m. The pitcher's mound is 16.8m from home plate on the line joining home plate and second base. How far is the pitcher's mound from first base
?

Answer 1

c^2 = a^2 + b^2

22.1^2= 16.8^2 + b^2

solve for 'b' using pythag. theorem

Answer 2

Thank you Charrice

Answer 3

I think there's a slight error here. The solution given works only if the angle at the pitcher's mound is 90 degrees. However, the center of the square is 22.4/√2 = 15.8m from home plate.

If you make the first-base line the x-axis, with home plate at (0,0), then the pitcher's mound is at (16.8/√2,16.8/√2) = (11.88,11.88)

The distance from, (11.88,11.88) to (22.4,0) is

√(22.4-11.88)^2 + 11.88^2 = 15.87m

The solution above gives 14.36m