AMC 10 · 2020 · #9

Grade 8 geometry-3d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

volume-conepythagorean-theoremarea-circlesperimeter identify-subproblemsphysical-representation ↑ Prerequisites: area-circlespythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A three-quarter sector of a circle of radius 4 is rolled and taped along its two radii to form the lateral surface of a right circular cone. Find the cone's volume in cubic inches.

Pick an answer.

(A)

$3\pi \sqrt5$

(B)

$4\pi \sqrt3$

(C)

$3 \pi \sqrt7$

(D)

$6\pi \sqrt3$

(E)

$6\pi \sqrt7$

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram) — sketch the flat sector and the resulting cone side by side, labeling what maps to what. Tool #10 (Physical) — for a younger reader, cutting a paper sector and rolling it makes the slant/arc/circumference correspondences obvious. Tool #7 (Subproblems) then breaks the volume calculation into three small pieces: (a) base radius from the arc, (b) cone height from Pythagorean theorem, (c) plug into V = 1/3π r² h. Tool #3 verifies against the answer choices.

1STEP 1

Identify the slant height

The sector's radius becomes the slant height.

ℓ = 4 inches

2STEP 2

Find the arc length

Compute the sector's arc.

arc = 3/4 · 8π = 6π inches

3STEP 3

Find the base radius

The arc becomes the base circumference.

2π r = 6π → r = 3

4STEP 4

Find the height

Pythagoras gives the height.

h² + r² = ℓ² → h² + 9 = 16 → h = √(7)

5STEP 5

Compute the volume

Plug into the formula.

V = 1/3 π r² h = 1/3 π · 9 · √(7) = 3π√(7)

6STEP 6

Match the choice

The volume is three pi root seven.

3π√(7) → (C)

Answer

3 π √7

Sanity-check the geometry. A full circle of radius 4 rolled into a cone would give r = 4 (degenerate — flat disk). A half-circle would give r = 2 (a steep cone). Three-quarters lies between, so r = 3 is in the right ballpark. ✓ With r = 3 and slant 4, h = √(16 - 9) = √(7) ≈ 2.65 — shorter than the slant, as required. ✓ Numerically V = 3π√(7) ≈ 3 · 3.14 · 2.65 ≈ 24.9 cubic inches, a sensible volume for a small paper cone. ✓

💡Key takeaway

This AMC 12 problem only needs Grade 8 cone formulas you already know — the 3/4 sector's arc 6π becomes the base circumference, giving r = 3; the slant 4 and base 3 give height √(7) by Pythagorean; then V = 1/3π(9)(√(7)) = 3π√(7).