2020 AMC 12B Problem 9

Below is the professionally curated solution for Problem 9 of the 2020 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 12B solutions, or check the answer key.

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Concepts:cone volume Pythagorean Theorem

Difficulty rating: 1470

A three-quarter sector of a circle of radius $4$ inches together with its interior can be rolled up to form the lateral surface of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?

$3\pi \sqrt{5}$

$4\pi \sqrt{3}$

$3\pi \sqrt{7}$

$6\pi \sqrt{3}$

$6\pi \sqrt{7}$

Solution:

The sector's arc length is $\tfrac34 \cdot 2\pi \cdot 4 = 6\pi,$ which becomes the base circumference: $2\pi r = 6\pi,$ so $r = 3.$

The slant height is the sector radius $4,$ so the height is $h = \sqrt{4^2 - 3^2} = \sqrt{7}.$ The volume is $\frac13 \pi r^2 h = \frac13 \pi \cdot 9 \cdot \sqrt{7} = 3\pi\sqrt{7}.$

Thus, the correct answer is C.

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