Suppose a certain baseball diamond is a square 95 feet on a side. The pitching rubber is located 61.5 feet from home plate on a line joining home plate and second base. How far is it from the pitching rubber to first base? The distance from the pitching rubber to first base is about $$ \square $$ feet. (Round to the nearest tenth as needed.)

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UpStudy AI · Accepted Answer

We begin by placing the baseball diamond on a coordinate plane. Let

- Home plate be at $$ (0, 0) $$,
- First base at $$ (95, 0) $$,
- Second base at $$ (95, 95) $$.

The pitching rubber is located along the line from home plate to second base. This line has the direction vector $$ \langle 1,1 \rangle $$, so a unit vector in that direction is
$$
\left\langle \frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right\rangle.
$$
Since the pitching rubber is $$61.5$$ feet from home plate along this line, its coordinates are
$$
\left( \frac{61.5}{\sqrt{2}}, \frac{61.5}{\sqrt{2}} \right).
$$

The distance from the pitching rubber to first base (located at $$ (95, 0) $$) is given by the distance formula:
$$
\text{Distance} = \sqrt{\left(95-\frac{61.5}{\sqrt{2}}\right)^2+\left(0-\frac{61.5}{\sqrt{2}}\right)^2}.
$$

To approximate, first compute:
$$
\frac{61.5}{\sqrt{2}} \approx \frac{61.5}{1.414} \approx 43.5.
$$
So the pitching rubber is approximately at $$ (43.5, 43.5) $$.

Then the differences in coordinates are:
$$
\Delta x \approx 95 - 43.5 = 51.5 \quad \text{and} \quad \Delta y \approx  0 - 43.5 = -43.5.
$$
Thus, the distance is approximately:
$$
\sqrt{(51.5)^2 + (43.5)^2} \approx \sqrt{2652.25 + 1892.25} \approx \sqrt{4544.5} \approx 67.4.
$$

The distance from the pitching rubber to first base is about $$ \boxed{67.4} $$ feet.
The distance from the pitching rubber to first base is approximately 67.4 feet.

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