Notice that s stands for the slant height of the pyramid, and this height is the height of each one of the lateral triangular faces.
We will need to use the next formulas
[tex]\begin{gathered} A_{\text{hex}}=\frac{3\sqrt[]{3}}{2}a^2\to\text{Area of a regular hexagon (a is one of its sides)} \\ A_{\text{tri}}=\frac{bh}{2}\to\text{Area of a triangle} \end{gathered}[/tex]
Therefore, the surface area is
[tex]\begin{gathered} A_{\text{figure}}=A_{\text{hex}}+6\cdot A_{\text{tri}}=\frac{3\sqrt[]{3}(3)^2}{2}+6\cdot\frac{(3\cdot10.3)}{2} \\ =\frac{27\sqrt[]{3}}{2}+92.7 \\ \Rightarrow A_{\text{figure}}=\frac{27\sqrt[]{3}}{2}+92.7 \\ \Rightarrow A_{\text{figure}}\approx23.4+92.7=116.1 \\ \Rightarrow A_{\text{figure}}=116.1 \end{gathered}[/tex]
Lateral Area=116.1 m^2
As for the volume, the volume of a regular pyramid is
[tex]V=\frac{A_{\text{base}}\cdot h}{3}[/tex]
In our case,
[tex]V=\frac{3\sqrt[]{3}a^2}{2}\cdot\frac{h}{3}=\frac{\sqrt[]{3}}{2}a^2h[/tex]
Therefore,
[tex]\Rightarrow V=\frac{\sqrt[]{3}}{2}(3)^2\cdot10=45\sqrt[]{3}[/tex]
Volume= 77.9 m^3
