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2020 AMC 10B Problem 10

2020 AMC 10B Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Solid Geometry · difficulty 2 of 5 · about 3 min · position P1-10

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

A)
$3\pi \sqrt5$
B)
$4\pi \sqrt3$
C)
$3 \pi \sqrt7$
D)
$6\pi \sqrt3$
E)
$6\pi \sqrt7$

Answer

Key insight

The sector's arc (three quarters of 8 pi) becomes the base circumference, so the base radius is 3 and the slant height is 4.

Solution

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

Why this works

Rolling a sector into a cone preserves two lengths: the sector radius (now the slant height) and the arc (now the base circumference). Those two facts determine $r$ and $\ell$ ; the Pythagorean theorem supplies $h$ . Every sector-to-cone problem follows this same three-step template.

The trap

Using 4 as the cone's radius or height instead of its slant height, or forgetting the 1/3 in the cone volume formula.

Common mistakes

Using 4 as the cone's radius or height instead of its slant height, or forgetting the 1/3 in the cone volume formula.
Taking the base radius to be $\tfrac34 \cdot 4 = 3$ by luck but with wrong reasoning, then using $h = 4$ to get $12\pi$ , which is not a choice; recheck via the Pythagorean theorem.

Techniques

Set up the equation/formula and compute; no special trick needed