2020 AMC 10B Problem 10

Picture rolling the paper sector into a cone. Two lengths carry over:

• the straight edges (the radius $4$) run from the tip of the cone down to the rim, so they become the slant height, $\ell = 4$;
• the curved edge wraps once around the bottom, so it becomes the circumference of the base.

1. Length of the curved edge. A full circle of radius $4$ has circumference $2\pi \times 4 = 8\pi$. The sector is three quarters of the circle, so its curved edge is $\tfrac34 \times 8\pi = 6\pi$.
2. Base radius. The base circle has circumference $6\pi$, so $2\pi r = 6\pi$ and $r = 3$.
3. Height. The height, the base radius and the slant height form a right triangle, with the slant height as the longest side:
$$h = \sqrt{4^2 - 3^2} = \sqrt{16 - 9} = \sqrt7.$$
4. Volume. A cone has volume $\tfrac13 \times (\text{base area}) \times (\text{height})$:
$$V = \frac13 \times \pi \times 3^2 \times \sqrt7 = 3\pi\sqrt7.$$

The answer is $\boxed{\textbf{(C)}\ 3 \pi \sqrt7}$.

Rolling paper does not stretch it, so lengths are kept. The sector's radius becomes the slant height and its arc becomes the base circumference. Those two facts give the base radius, and the Pythagorean theorem gives the height. Every "roll a sector into a cone" problem follows these same three steps.

When the sector is rolled up, its radius $4$ becomes the slant height $\ell$ of the cone, and its arc becomes the circumference of the base.

Arc length of a three-quarter sector of radius $4$: $\tfrac34 \cdot 2\pi \cdot 4 = 6\pi$. Setting $2\pi r = 6\pi$ gives base radius $r = 3$.

Height from the right triangle with hypotenuse $\ell = 4$ and leg $r = 3$:
$$h = \sqrt{4^2 - 3^2} = \sqrt{7}.$$

Volume:
$$V = \frac13 \pi r^2 h = \frac13 \pi \cdot 9 \cdot \sqrt7 = 3\pi\sqrt7.$$

The answer is $\boxed{\textbf{(C)}\ 3 \pi \sqrt7}$.