2001 AMC 10 Problem 17

When the two straight edges are glued together, every radius of the sector runs from the apex of the cone down to its rim. So the slant height of the cone is $10$; this rules out (B) and (D), which describe the height.

The curved edge of the sector wraps around to become the circle at the base of the cone. Its length is the fraction $\frac{252}{360} = \frac{7}{10}$ of the full circumference $2\pi \cdot 10 = 20\pi$:
$$\frac{7}{10} \cdot 20\pi = 14\pi.$$
A circle of circumference $14\pi$ has radius $\frac{14\pi}{2\pi} = 7$.

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

Rolling a sector into a cone preserves lengths: radius to slant height, arc to base circumference. Both conversions are one-line computations once you identify which piece becomes which. The ratio "central angle over $360^\circ$" equals "base radius over slant height," a fact worth remembering for any cone-net problem.

Match areas: the sector has area $\frac{7}{10}\pi \cdot 10^2 = 70\pi$, and a cone's lateral area is $\pi r \ell$ with $\ell = 10$. Then $10\pi r = 70\pi$ gives $r = 7$.

• Treating the sector radius 10 as the cone's height rather than its slant height, which points to choices (B) or (D).
• Using the arc length $14\pi$ as the base diameter or as the radius, giving $r = 14$ or forcing a guess between (A) and (E).