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?
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 $108\pi-72\pi=\boxed{36\pi\text{ sq cm}}$.
To find the surface area of the solid formed by rotating an equilateral triangle, we need to determine the lateral area and add it to the area of the two circular bases.
The lateral area can be found by calculating the circumference of the circle formed by rotating the triangle and multiplying it by the length of the side of the triangle. Since the triangle is equilateral, each side is of length 12 cm.
The circumference of a circle can be found using the formula:
Circumference = 2 * π * radius.
In this case, the triangle is rotated about one of its sides, which acts as the radius of the circle. Therefore, the radius of the circle is equal to half the length of one side of the triangle.
Radius = 12 cm / 2 = 6 cm.
Using the formula for circumference, we can calculate:
Circumference = 2 * π * 6 cm = 12π cm.
Since the triangle is rotated about one of its sides, the lateral area is equal to the circumference multiplied by the length of the side of the triangle:
Lateral Area = 12π cm * 12 cm = 144π cm^2.
To find the area of the circular bases, we need to determine the radius of the circle formed by rotating the triangle. The radius is equal to the length of one side of the triangle, which is 12 cm.
The area of a circle can be found using the formula:
Area = π * radius^2.
Therefore, the area of each circular base is:
Area = π * (12 cm)^2 = 144π cm^2.
Since we have two circular bases, we need to multiply this area by 2:
Total Area of the Circular Bases = 2 * 144π cm^2 = 288π cm^2.
To find the surface area, we add the lateral area to the total area of the circular bases:
Surface Area = Lateral Area + Total Area of the Circular Bases
Surface Area = 144π cm^2 + 288π cm^2
Surface Area = 432π cm^2.
Therefore, the surface area of the solid formed by rotating the equilateral triangle is 432π square centimeters.