Given
[tex]\begin{gathered} r=8m \\ l=25 \end{gathered}[/tex]The given formula
[tex]\begin{gathered} Surface\text{ area of a cone =}\pi rl+B \\ where\text{ l is slant height} \\ and\text{ B is the area base} \end{gathered}[/tex]The base is circle
[tex]\begin{gathered} Area\text{ of a circle=}\pi r^2 \\ Area\text{ of a circle=}\pi8^2 \\ Area\text{ of a circle =64}\pi \end{gathered}[/tex]Now back to the given formula
[tex]\begin{gathered} Surface\text{ of area of cone=}\pi rl+B \\ B=64\pi \\ l=25 \\ r=8 \\ \end{gathered}[/tex][tex]\begin{gathered} Surface\text{ of area of cone =}\pi\times8\times25+64\pi \\ Surface\text{ofareaofcone=200}\pi\text{+64\pi} \\ Surface\text{ofareaofcone=264\pi} \end{gathered}[/tex]The final answer
[tex]264\pi m^2[/tex]