A cone is generated by rotating triangle $ABC$ around side $\overline{AB}$. Its total surface area is $\pi$ times what number?
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 \(9 + 3\sqrt{10}\).