Question

A baseball diamond is a square with 90 foot sides. How far must the third baseman throw the ball to make it to first base?

Answer 1

`127ft` (3s.f.)

Explanation:

there are two adjacent sides between third base and first base.

since the shape in question is a square, the two points are opposite.

the line between two opposite sides of a square is the diagonal.

the third baseman must throw along the length of the diagonal to make it to first base.

the length of one side of the square (and therefore each side) can be used to find the length of a diagonal.

pythagoras' theorem: `a^2+b^2 = c^2`

the diagonal is the hypotenuse, `c`.

`a` and `b` are the two adjacent sides of the square in between the two bases,

since all sides of a square are equal, `a` and `b` must be equal.

the two can therefore be put under one variable.

`a^2 + a^2 = c^2`.

it is given that `a`, the side length, is `90`.

`a^2 + a^2 = c^2`

`2a^2 = c^2`

`2 * 90^2 = 2 * 8100 = 16200`

`c^2 = 16200`

`c = sqrt(16200)`

`= 127.279...`

`127.279.. = 127`, to `3` significant figures.

the length of the diagonal is `127` feet, to `3` significant figures.

the third baseman must throw `~~127` feet, to reach first base.