check the picture below.
[tex]\bf \textit{lateral area of a cone}\\\\
LA=\pi r\stackrel{slant-height}{\sqrt{r^2+h^2}}~~
\begin{cases}
LA=539\pi \\
r=49
\end{cases}\implies 539\pi =\pi 49\sqrt{r^2+h^2}
\\\\\\
\cfrac{539\pi }{49\pi }=\sqrt{r^2+h^2}\implies 11=\sqrt{r^2+h^2}[/tex]
