A baseball diamond is a square with sides 27.4m. The pitchers mound is 18.4m from home plate on the line joining home plate and second base. How far is the pitcher's mound from first base?
How we got that the angle is 45
It’s a square so each angle would be 90 degrees divide that by 2 since there would be a diagonal line and 90 divided by 2 is 45
your answer is right but the work is wrong - 812.99 should be 712.99
To find the distance between the pitcher's mound and 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, we can consider the distance from home plate to first base as one side of the triangle, the distance from home plate to the pitcher's mound as the other side, and the distance from first base to the pitcher's mound as the hypotenuse.
Let's label the distance from home plate to first base as "x" (which is what we want to find), and the distance from home plate to the pitcher's mound as "18.4m".
Using the Pythagorean theorem, we have:
x^2 = (27.4m)^2 - (18.4m)^2
x^2 = 750.76m^2 - 338.56m^2
x^2 = 412.2m^2
Now, we can take the square root of both sides of the equation to solve for x:
x = √412.2m^2
x ≈ 20.3m
Therefore, the distance between the pitcher's mound and first base is approximately 20.3 meters.
At 19.39 meters
18.4^2+27.4^2-2*18.4*27.4Cos 45
D^2= 1089.32-812.99
D Sq root of 376.33
D= 19.39