We are given a right prism whose bases are congruent regular pentagons. The side length of each pentagon is approximately $10.2$ mm, and the apothem measures $7$ mm. The height of the prism is $4$ mm. We need to find the volume of the prism, rounding to the nearest tenth.

1. Problem Description

We are given a right prism whose bases are congruent regular pentagons. The side length of each pentagon is approximately $10.2$ mm, and the apothem measures $7$ mm. The height of the prism is $4$ mm. We need to find the volume of the prism, rounding to the nearest tenth.

2. Solution Steps

The volume $V$ of a prism is given by the formula

$V = A_{base} \cdot h$,

where $A_{base}$ is the area of the base and $h$ is the height of the prism.

The area of a regular polygon with $n$ sides is given by

$A = \frac{1}{2} \cdot a \cdot P$,

where $a$ is the apothem and $P$ is the perimeter.

In our case, the base is a regular pentagon, so $n=5$.

The side length is $10.2$ mm, so the perimeter is $P = 5 \cdot 10.2 = 51$ mm.

The apothem is $a = 7$ mm.

So the area of the pentagon base is

$A_{base} = \frac{1}{2} \cdot 7 \cdot 51 = \frac{357}{2} = 178.5$ mm$^2$.

The height of the prism is $h = 4$ mm.

So the volume of the prism is

$V = A_{base} \cdot h = 178.5 \cdot 4 = 714$ mm$^3$.

3. Final Answer

The volume of the prism is $714.0$ mm$^3$.