Check the picture below.
if the diameter of its base is 12, then its radius must be half that, or 6.
[tex]\textit{surface area of a cone}\\\\ SA=\pi r\stackrel{\stackrel{slant~height}{sh}}{\sqrt{r^2+h^2}} + \pi r^2~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=6\\ sh=18 \end{cases}\implies SA=\pi (6)(\stackrel{sh}{18})+\pi (6)^2 \\\\\\ SA=108\pi +36\pi \implies SA=144\pi[/tex]
