![\begin{gathered} \text{LSA}\approx312m^2 \\ V\approx794m^3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jyle82po7jtzfsjwk5ybfpi5bs3gdjzh7q.png)
1) Note that this solid, is a cross-sectioned cone. We can find the Lateral Surface Area, using this formula:
![LSA=\pi\cdot s\cdot(R+r)_{}](https://img.qammunity.org/2023/formulas/mathematics/college/nzcqjkomgklnbef129tv3er4qegyk3ex5r.png)
Where R, is the biggest radius, s for the slant height, r for the smallest radius.
2) So, plugging the measures into that we can write out:
![\text{LSA}=\pi\cdot9.63(6.20+4.10)=311.61m^2\approx312m^2](https://img.qammunity.org/2023/formulas/mathematics/college/7evm4qdwb3w6dss1aro2bd5yd88oj4ek0b.png)
Note that we have rounded it off to the nearest whole. So now, let's find the volume of that solid:
![V=(\pi h)/(3)(R^2+Rr+r^2)](https://img.qammunity.org/2023/formulas/mathematics/college/nh2pbkrsha616alzrxq6c8b6bc98ig2x0q.png)
Plugging into that the given dimensions:
![\begin{gathered} V=(\pi h)/(3)(R^2+Rr+r^2) \\ V=(\pi\cdot9.4)/(3)(6.2^2+6.2\cdot4.1+4.1^2) \\ V\approx794m^3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/t06tm89aqiqzi94virk4y69lci6m3neznx.png)
Note the volume in cubic meters.
And that is the answer