The problem asks to find the volume of a pentagonal pyramid. The base is a regular pentagon with side length 10 mm. The height of the pyramid is 15 mm.

1. Problem Description

The problem asks to find the volume of a pentagonal pyramid. The base is a regular pentagon with side length 10 mm. The height of the pyramid is 15 mm.

2. Solution Steps

First, we need to find the area of the pentagonal base. The area of a regular pentagon with side length $s$ is given by

$A = \frac{5s^2}{4 \tan(\frac{\pi}{5})}$.

Alternatively, we can express the area as $A = \frac{1}{4}\sqrt{25 + 10\sqrt{5}} s^2 \approx 1.72048s^2$.

In this case, $s = 10$ mm.

$A = \frac{5(10^2)}{4 \tan(\frac{\pi}{5})} = \frac{500}{4 \tan(\frac{\pi}{5})} = \frac{125}{\tan(\frac{\pi}{5})} \approx \frac{125}{0.7265} \approx 172.048$

Also $A = \frac{1}{4} \sqrt{25 + 10\sqrt{5}} (10)^2 = \frac{100}{4} \sqrt{25+10\sqrt{5}} = 25\sqrt{25+10\sqrt{5}} \approx 25\sqrt{25+22.36} = 25\sqrt{47.36} = 25(6.88) \approx 172$.

The volume of a pyramid is given by the formula

$V = \frac{1}{3} \times A \times h$,

where $A$ is the area of the base and $h$ is the height of the pyramid.

In our case, $A = 172.048 \text{ mm}^2$ and $h = 15$ mm.

So, $V = \frac{1}{3} \times 172.048 \times 15 = 172.048 \times 5 = 860.24 \text{ mm}^3$.

3. Final Answer

The volume of the pentagonal pyramid is approximately 860.24 $mm^3$.