A baseball diamond is actually a square with each side approximately 27.4 m. The pitcher's mound is 18.4 m from home plate on the diagonal of the square - note that the pitcher's mound is NOT located at the centre of the diamond!

a) How far (to 3 decimal places) from the centre of the diamond is the pitcher's mound? Is it closer to home plate or second base?

I don't know if it's that my diagram is awry or what, but I can't seem to get the answer. I don't know how easily one might be able to understand this without the diagram I drew, but here's my math:

(I made x the distance from one corner of the square/diamond (H for home plate) to the centre of the diamond. The 45° was not exactly given, but I assumed it since it's a square...)

Find x:
cos45° = 27.4 m / x
x = 27.4 m / cos45°
x = 38.7

38.7 m - 18.4 m = 20.3 m therefore, the pitcher's mound is 20.3 m from the centre of the diamond.

The answer should be 0.975 m, way off.

should have been

cos 45 = x/27.4, you have it upside down, cos ? = adjacent/hypotenuse

x = 27.4(cos45) = 19.3747

so the centre is 19.3747 ft from home
making the pitcher mount 19.3747 - 18.4 or
.9747 ft from the centre, and being closer to home than 2nd base.

BTW, you should have know that your answer could not be correct, since entire distance from home to 2nd is ...

d^2 = 27.4^2 + 27.4^2
d = √1501.52 = 38.75 ft , which would have put your pitching mount beyond 2nd base in the outfield.

I think I may need to rethink this question because in the diagram I drew, 27.4 m is adjacent to the 45° angle.

And the proof you showed me at the end cannot be visualised in my diagram.. Would it be possible for you to explain in words (I know it's really hard..) what your diagram looks like?

Oh, never mind about not being able to visualise your proof at the end, I completely misread that, wow. But my diagram is still incorrect somehow because 27.4 is the adjacent side..

Oh! Wait, sorry, again. Never mind. That was really stupid of me. I don't know how I didn't see it, but I realise what was wrong now, thank you

the baseball diamond is a square with sides 27.4

If you draw the two diagonals, they will meet at the centre, giving you 4 identical right-angled triangles.
in each of these triangles 27.4 is the hypotenuse and the line going to the centre is the adjacent.

To solve this problem, you need to consider the geometry of the baseball diamond. Start by drawing a diagram of the diamond and labeling the relevant distances.

Let's start by finding the distance from one corner of the square (home plate) to the center of the diamond. To do this, we can use the concept of right-angled triangles. The diagonal of the square connecting home plate to second base is the hypotenuse of a right-angled triangle. The two legs of this triangle are each 27.4 m / √2, which comes from dividing the side length of the square by the square root of 2.

Using the Pythagorean theorem, we can find the length of the hypotenuse (x) which represents the distance from one corner of the square to the center of the diamond:

x² = (27.4 m / √2)² + (27.4 m / √2)²

x² = 27.4 m² / 2 + 27.4 m² / 2

x² = 27.4 m²

x = √(27.4 m²)

x ≈ 27.4 m

So, the distance from one corner of the square to the center of the diamond is approximately 27.4 meters.

Next, we need to find the distance from the center of the diamond to the pitcher's mound. To do this, we can subtract the length of the diagonal from the center to home plate (x) by the given distance from home plate to the pitcher's mound (18.4 m).

Distance from center to pitcher's mound = x - 18.4 m

Distance from center to pitcher's mound ≈ 27.4 m - 18.4 m

Distance from center to pitcher's mound ≈ 9 m

Therefore, the pitcher's mound is approximately 9 meters from the center of the diamond. Since the pitcher's mound is closer to home plate than second base, it is closer to home plate.