So we are given in the question the surface area of the curved part:
![SA= \pi rl](https://img.qammunity.org/2019/formulas/mathematics/middle-school/gvaw7xdsaewls5m9xwcbn7ro5cf9xr3rcl.png)
We have those values already, so we can readily solve for that:
![SA= \pi (5)(9)=45 \pi](https://img.qammunity.org/2019/formulas/mathematics/middle-school/3dg4jj4hx0e94qwqsh85ylynd14i1s8oxj.png)
And the question said to leave it in terms of
![\pi](https://img.qammunity.org/2019/formulas/mathematics/middle-school/w2mw2141nk3383oncfx010frgx49xcj0c8.png)
, so we do not need to simplify further.
Let us solve for the base now. The base is a circle, and we know that the area of a circle is found using the equation:
![A= \pi r^(2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/g7edsuqma4l537ot5s0b6gm9jmckeeg5w2.png)
We can also readily solve for this using 5 for r:
![A= \pi (5)^(2) =25 \pi](https://img.qammunity.org/2019/formulas/mathematics/middle-school/912j6llvv90opdukk6twofowyjjvle4xzd.png)
And now that we have both surfaces, we can add them together to find the total surface area:
![SA=45 \pi +25 \pi =70 \pi](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ew1c1eui6uhtk0jiq04plu388pmzz41zam.png)
So now we know that
the total surface area of this cone is
![70 \pi](https://img.qammunity.org/2019/formulas/mathematics/middle-school/a8cbr2aawsg3c018v73etzkudrhh5rfihn.png)
.