Question

Find the lateral surface area and volume of the solid object in picture

Answer 1

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

`\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}}=(bh)/(2)\to\text{Area of a triangle} \end{gathered}`

Therefore, the surface area is

`\begin{gathered} A_{\text{figure}}=A_{\text{hex}}+6\cdot A_{\text{tri}}=\frac{3\sqrt[]{3}(3)^2}{2}+6\cdot((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}`

Lateral Area=116.1 m^2

As for the volume, the volume of a regular pyramid is

`V=\frac{A_{\text{base}}\cdot h}{3}`

In our case,

`V=\frac{3\sqrt[]{3}a^2}{2}\cdot(h)/(3)=\frac{\sqrt[]{3}}{2}a^2h`

Therefore,

`\Rightarrow V=\frac{\sqrt[]{3}}{2}(3)^2\cdot10=45\sqrt[]{3}`

Volume= 77.9 m^3