2001 AMC 10 Problem 17

1. Picture rolling up the sector like a paper party hat. The center point of the sector becomes the tip of the cone.
2. Each straight edge of the sector runs from the tip down to the rim of the cone, along the outside. That slanted distance is the slant height, so the slant height is $10$. This rules out (B) and (D), which say the height (straight up through the middle) is $10$.
3. The curved edge of the sector wraps all the way around to form the circle at the bottom of the cone. So its length equals the distance around the base.
4. A full circle of radius $10$ has circumference $2 \times \pi \times 10 = 20\pi$. The sector is $252^\circ$ out of $360^\circ$, which is $\frac{252}{360} = \frac{7}{10}$ of the circle. Its curved edge is
$$\frac{7}{10} \times 20\pi = 14\pi.$$
5. The base circle has circumference $14\pi$. Since circumference is $2\pi r$, we get $2\pi r = 14\pi$, so $r = 7$.

The answer is $\boxed{\textbf{(C)}\ \text{slant height } 10 \text{ and radius } 7}$.

Rolling paper does not stretch it, so lengths carry over: the sector's radius becomes the slanted side, and its curved edge becomes the base circle. A handy check: the fraction of the full circle ($\frac{7}{10}$) equals the base radius divided by the slant height ($\frac{7}{10}$).

Compare areas. The sector's area is $\frac{7}{10}$ of a circle of radius $10$: $\frac{7}{10} \times 100\pi = 70\pi$. The curved surface of a cone with base radius $r$ and slant height $10$ has area $\pi \times r \times 10$. Setting $10\pi r = 70\pi$ gives $r = 7$.