Question by olla: Distance..?
A baseball diamond has four bases (forming a squre) that are 90 feet apart; the pitcher’s mound is 60.5 feet from home plate Approximate the distance, to the nearest thousandth, from the pitcher’s mound to each of the other three bases. (Hint: the pitcher’s monud is NOT in the center of the baseball diamond.)

1.Distance to first base:

2.Distance to second base:

3.Distance to third base:

Best answer:

Answer by Steven X
The coordinates of the pitchers mound (taking home base to be the origin) are 60.5x(0.707…, 0.707…).

The coordinates of first base are (90, 0), so the distance to first base is |(90-60.5×0.707, 60.5×0.707)|

The coordinates of the second base are (90, 90), so the distance is |(90-60.5×0.707, 90-60.5×0.707)|

The distance to third base is the same as the distance to first base.

These distances (magnitudes of vectors) can be calculated as the square roots of the sum of the squares of the x coordinate and y coordinate.

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