cones

If you could envision rotating $\triangle \text{ABC}$ around $\overline{\text{AB}}$, then the diagram would look like the following diagram. I realize the diagram is not drawn to scale. As of now, I have not yet worked out how to draw 3-dimensional figures efficiently.

The surface area of right cones are $\text{SA}_\text{right cone} = \pi r^2 + \pi rL$ where r is the radius of the right cone and the L is the slant height of the right cone. We already know the radius of this cone, but we can find L with the use of Pythagorean's Theorem.

$L^2 = 1^2 + 3^2 \\ L^2 = 1 + 9 \\ L^2 = 10 \\ L = \sqrt{10} \text{ or } L = -\sqrt{10}$

Since we are dealing with lengths, reject $L = -\sqrt{10}$. Now that we have found all the necessary lengths, we can find the surface area of this right cone.

$\begin{align*} \text{SA}_\text{right cone} &= \pi r^2 + \pi rL \\ &= \pi * 3^2 + \pi * 3 * \sqrt{10} \\ &= 9\pi + 3\sqrt{10}\pi \\ &= (9 + 3\sqrt{10}) \pi \end{align*}$

The question asks what the surface area is pi times what number? Well, that number would be $9 + 3\sqrt{10}$.