An equilateral triangle of side $12$ centimeters is rotated about a side to form a solid. What is the number of square centimeters in the surface area of the solid?

Answer 1

We can start by finding the sidelength of the solid. When the triangle is rotated about a side, the solid formed is a cone whose base is an equilateral triangle. Since the side length of the triangle is $12$ centimeters, the altitude of the triangle is $12\sqrt{3}/2=6\sqrt{3}$. The radius of the cone formed by rotating the triangle is equal to the height of the triangle.

Next, we find the slant height of the cone. Let $r$ be the radius of the cone (which is also the height of the triangle). Then, the altitude of the triangle is $\sqrt{12^2-6^2}=6\sqrt{3}$ by the Pythagorean Theorem. We apply the Pythagorean Theorem again to find the slant height $s$ of the cone:\[s=\sqrt{r^2+\left(\frac{12\sqrt{3}}{2}\right)^2}=\sqrt{r^2+108}.\]Because the solid is formed by rotating an equilateral triangle, the slant height of the cone is equal to the side length of the triangle. We have $s=r$ (the slant height is equal to the radius by $\triangle ABC$ below), so we can substitute this to find the radius of the cone:\[\sqrt{r^2+108}=r\]Squaring both sides yields $r^2+108=r^2$, so $108=0$. This is a contradiction, so our original assumption must be false. Therefore, the solid formed by rotating the triangle about a side is not a cone. Because we cannot model the solid in any of the above forms, we must be missing part of the solid. Thus, we have to read the problem again and find that the solid is actually a cone with another cone removed from it to leave a cylindrical hole. Since the altitude of the cone is $6\sqrt{3}$ and the base is equilateral, we know the base of the triangle is $6$ centimeters. The height of the cylindrical hole is the same as the slant height of the solid cone, $6\sqrt{3}$, so the diameter of the hole is equal to the altitude of the original triangle times 2. Therefore, the radius of the hole is $\left(\frac{6\sqrt{3}}{2}\right)\cdot 2=6\sqrt{3}$ centimeters. The total surface area of the solid is the curved surface area of the original cone minus the curved surface area of the hole. The curved surface area of the original cone is $\pi r s=\pi r^2=\pi (6\sqrt{3})^2=108\pi$. The hole is a cylinder of radius $6\sqrt{3}$ and height $6\sqrt{3}$, so the curved surface area of the hole is $2\pi r h=72\pi$. Thus, the total surface area of the solid is $108\pi-72\pi=\boxed{36\pi\text{ sq cm}}$.