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AMC 10 · 2020 · #10

Grade 8 geometry-3d
Archetype Compound Area by Decomposition / Difference
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 44 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?

Pick an answer.

(A)
$3\pi \sqrt5$
(B)
$4\pi \sqrt3$
(C)
$3 \pi \sqrt7$
(D)
$6\pi \sqrt3$
(E)
$6\pi \sqrt7$
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Toolkit + CCSS Solution

Understand

Restated: A three-quarter sector of a circle with 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.

Givens: The sector has radius $4$ inches; The sector spans three-quarters of the full circle ($270^\circ$); The two straight radii of the sector get taped together to form the cone's slant edge; The arc of the sector becomes the circular rim of the cone's base; Choices: (A) $3\pi\sqrt{5}$, (B) $4\pi\sqrt{3}$, (C) $3\pi\sqrt{7}$, (D) $6\pi\sqrt{3}$, (E) $6\pi\sqrt{7}$

Unknowns: The cone's volume in cubic inches

Understand

Restated: A three-quarter sector of a circle with 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.

Givens: The sector has radius $4$ inches; The sector spans three-quarters of the full circle ($270^\circ$); The two straight radii of the sector get taped together to form the cone's slant edge; The arc of the sector becomes the circular rim of the cone's base; Choices: (A) $3\pi\sqrt{5}$, (B) $4\pi\sqrt{3}$, (C) $3\pi\sqrt{7}$, (D) $6\pi\sqrt{3}$, (E) $6\pi\sqrt{7}$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #10 Create a Physical Representation, #3 Eliminate Possibilities

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 $V = \tfrac13\pi r^2 h$. Tool #3 verifies against the answer choices.

Execute — Answer: C

#1 Draw a Diagram 7.G.A.3 Step 1
  • Draw the flat sector: a pie-slice of radius $4$ covering $\tfrac34$ of a full circle.
  • When rolled, the two straight edges of length $4$ tape together to form the slant side of the cone, and the curved arc becomes the circular rim of the cone's base.
  • So the cone's slant height is $\ell = 4$.
$$\ell = 4 \text{ inches}$$

💡 The two flat edges become the cone's slant; the arc becomes the base rim.

#7 Identify Subproblems 7.G.B.4 Step 2
  • Compute the arc length of the $\tfrac34$ sector.
  • The full circle of radius $4$ has circumference $2\pi(4) = 8\pi$.
  • Three-quarters of that is the arc.
$$\text{arc} = \tfrac{3}{4} \cdot 8\pi = 6\pi \text{ inches}$$

💡 Three-quarters of the full circumference.

#7 Identify Subproblems 7.G.B.4 Step 3
  • This arc becomes the cone's base circumference.
  • Set $2\pi r = 6\pi$ to recover the base radius.
$$2\pi r = 6\pi \;\Rightarrow\; r = 3$$

💡 Base circumference equals the arc length.

#7 Identify Subproblems 8.G.B.7 Step 4
  • The cone's height $h$, base radius $r$, and slant $\ell$ form a right triangle ($h$ is vertical, $r$ is the horizontal leg from the cone's tip-projection to the base rim, and $\ell$ is the hypotenuse).
  • Apply Pythagorean theorem.
$$h^2 + r^2 = \ell^2 \;\Rightarrow\; h^2 + 9 = 16 \;\Rightarrow\; h = \sqrt{7}$$

💡 Slant, height, and base radius make a right triangle.

#7 Identify Subproblems 8.G.C.9 Step 5

Apply the cone volume formula.

$$V = \tfrac{1}{3} \pi r^2 h = \tfrac{1}{3} \pi \cdot 9 \cdot \sqrt{7} = 3\pi\sqrt{7}$$

💡 Plug $r = 3, h = \sqrt{7}$ into $\tfrac13\pi r^2 h$.

#3 Eliminate Possibilities 4.NBT.A.2 Step 6

$3\pi\sqrt{7}$ matches choice (C).

$$3\pi\sqrt{7} \Rightarrow \textbf{(C)}$$

💡 Read the matching answer choice.

[1] #1 7.G.A.3 Draw the flat sector: a pie-slice of radius $4$ covering $\tfrac34$ of a full ci
[2] #7 7.G.B.4 Compute the arc length of the $\tfrac34$ sector. The full circle of radius $4$ h
[3] #7 7.G.B.4 This arc becomes the cone's base circumference. Set $2\pi r = 6\pi$ to recover t
[4] #7 8.G.B.7 The cone's height $h$, base radius $r$, and slant $\ell$ form a right triangle (
[5] #7 8.G.C.9 Apply the cone volume formula.
[6] #3 4.NBT.A.2 $3\pi\sqrt{7}$ matches choice (C).

Review

Reasonableness: 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 $4$, $h = \sqrt{16 - 9} = \sqrt{7} \approx 2.65$ — shorter than the slant, as required. ✓ Numerically $V = 3\pi\sqrt{7} \approx 3 \cdot 3.14 \cdot 2.65 \approx 24.9$ cubic inches, a sensible volume for a small paper cone. ✓

Alternative: Tool #10 (Physical): cut a $\tfrac34$-circle of radius $4$ inches from paper, roll it, and tape the two radii. Measure the base diameter ($\approx 6$ inches, so $r = 3$) and the height ($\approx 2.6$ inches, matching $\sqrt{7}$). Same volume calculation.

CCSS standards used (min grade 8)

  • 7.G.A.3 Describe the two-dimensional figures that result from slicing three-dimensional figures (Tracking the unrolling/rolling correspondence: sector radius becomes cone slant, sector arc becomes base circumference.)
  • 7.G.B.4 Know the formulas for area and circumference of a circle (Computing the sector arc length $\tfrac34 \cdot 2\pi(4) = 6\pi$ and reading $r = 3$ from $2\pi r = 6\pi$.)
  • 8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Recovering the cone height from $h^2 + r^2 = \ell^2$, i.e. $h = \sqrt{16 - 9} = \sqrt{7}$.)
  • 8.G.C.9 Know the formulas for volumes of cones, cylinders, and spheres (Computing $V = \tfrac13 \pi r^2 h = \tfrac13 \pi \cdot 9 \cdot \sqrt{7} = 3\pi\sqrt{7}$.)
  • 4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Matching the volume $3\pi\sqrt{7}$ to choice (C).)

⭐ This AMC 10 problem only needs Grade 8 cone formulas you already know — the $\tfrac34$ sector's arc $6\pi$ becomes the base circumference, giving $r = 3$; the slant $4$ and base $3$ give height $\sqrt{7}$ by Pythagorean; then $V = \tfrac13\pi(9)(\sqrt{7}) = 3\pi\sqrt{7}$.

⭐ This AMC 10 problem only needs Grade 8 cone formulas you already know — the $\tfrac34$ sector's arc $6\pi$ becomes the base circumference, giving $r = 3$; the slant $4$ and base $3$ give height $\sqrt{7}$ by Pythagorean; then $V = \tfrac13\pi(9)(\sqrt{7}) = 3\pi\sqrt{7}$.

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