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
?
c^2 = a^2 + b^2
22.1^2= 16.8^2 + b^2
solve for 'b' using pythag. theorem
Thank you Charrice
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
To find the distance from the pitcher's mound to first base, we can use the Pythagorean theorem.
The Pythagorean theorem states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In this case, the distance from the pitcher's mound to first base forms the hypotenuse of a right triangle, with the distance from the pitcher's mound to home plate as one side and the distance from home plate to first base as the other side.
Let's call the distance from the pitcher's mound to first base "x".
Using the Pythagorean theorem, we have:
x^2 = (16.8m)^2 + (22.4m)^2
Simplifying:
x^2 = 282.24m^2 + 501.76m^2
x^2 = 784m^2
Taking the square root of both sides, we find:
x = √(784m^2)
x = 28m
Therefore, the distance from the pitcher's mound to first base is 28 meters.