AMC 10 · 2008 · #19

Grade 8 geometry-3d
Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid
circular-sectorvolume-cylinderthirty-sixty-ninety-triangle identify-subproblems ↑ Prerequisites: area-circles
📏 Long solution 💡 3 insights
Problem
A cylinder of radius 4 feet and length 9 feet lies on its side and holds water to a depth of 2 feet, measured from the lowest point. Find the volume of the water, in cubic feet.

Pick an answer.

(A)
$\ 24\pi - 36 \sqrt {2}$
(B)
$\ 24\pi - 24 \sqrt {3}$
(C)
$\ 36\pi - 36 \sqrt {3}$
(D)
$\ 36\pi - 24 \sqrt {2}$
(E)
$\ 48\pi - 36 \sqrt {3}$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The tank is 3D, but tool #1 (Draw a Diagram) collapses it: a cylinder on its side has the same circular cross-section all along, so the water is one flat shape stretched over the 9-ft length, and the volume is that wet area times 9. That turns the problem into finding the area of the wet part of a circle. Tool #7 (Identify Subproblems) breaks the wet region — a circular segment cut off by the flat water surface — into a sector minus a triangle, two shapes with known area formulas. Tool #4 (Introduce a Variable) handles the missing lengths and the central angle: dropping a perpendicular from the center to the water chord makes a right triangle, and the Pythagorean theorem plus a 30-60-90 shape give the half-chord and the 120° angle. Tool #8 (Analyze the Units) closes it out: square feet of area times feet of length gives cubic feet of volume.

1STEP 1

Slice the tank crosswise

Every crosswise slice is the same circle, so the volume is one circle's wet area times 9.

V = A_wet × 9
2STEP 2

Locate the water line

The center sits 4 ft above the bottom, so the water surface is a chord 2 ft below the center, cutting off a circular segment.

distance from center to water chord = 4 - 2 = 2
3STEP 3

Find the central angle

A perpendicular from the center meets the chord 2 ft down, and half the chord is 232\sqrt{3} — a 30-60-90 triangle, so the central angle is 120°.

MA=√(16-4)=2√(3), ∠ AOB = 2 × 60° = 120°
4STEP 4

Area of the sector

A 120° wedge is one third of the circle, and the whole circle has area 16π16\pi, so the sector is 16π3\frac{16\pi}{3}.

sector = 120/360 · π · 4² = 16π/3
5STEP 5

Area of the triangle

The triangle on the chord has base 434\sqrt{3} and height 2, so its area is 434\sqrt{3}; the wet segment is 16π3−43\frac{16\pi}{3} - 4\sqrt{3}.

△ AOB = 1/2 · 4√(3) · 2 = 4√(3); A_wet=16π/3-4√(3)
6STEP 6

Multiply by the length

Multiply the wet area by the length 9: the volume is 48π−36348\pi - 36\sqrt{3} cubic feet, choice (E).

9(16π/3-4√(3)) = 48π - 36√(3)
Answer
48π - 36 √(3)
The water is only 2 ft deep in a circle of radius 4, so it should fill less than half the cross-section. The wet area 16π/3-4√(3)≈ 16.76-6.93≈ 9.8 square feet is well under half of the full 16π≈ 50.3, as expected. The volume 48π-36√(3)≈ 150.8-62.4≈ 88.4 cubic feet is far below the full-tank volume π · 4² · 9 = 144π≈ 452 cubic feet, consistent with a tank filled well below halfway. This matches choice (E).
💡Key takeaway

For a tank on its side, the water volume is just the wet slice of the circle — a sector minus a triangle — stretched along the tank's length.

  • Slice the tank crosswise
  • Locate the water line
  • Find the central angle
  • Area of the sector
  • Area of the triangle
  • Multiply by the length

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