AMC 10 · 2008 · #19
Grade 8 geometry-3dA cylindrical tank with radius 4 feet and height 9 feet is lying on its side. The tank is filled with water to a depth of 2 feet. What is the volume of water, in cubic feet?
Pick an answer.
AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: The tank is a cylinder with radius $4$ ft, resting on its side.; The length of the cylinder (its horizontal axis) is $9$ ft.; The water is $2$ ft deep, measured up from the lowest point of the tank.
Unknowns: The volume of water inside the tank, in cubic feet.
Understand
Restated: 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.
Givens: The tank is a cylinder with radius $4$ ft, resting on its side.; The length of the cylinder (its horizontal axis) is $9$ ft.; The water is $2$ ft deep, measured up from the lowest point of the tank.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable, #8 Analyze the Units
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 $120^\circ$ angle. Tool #8 (Analyze the Units) closes it out: square feet of area times feet of length gives cubic feet of volume.
Execute — Answer: E
7.G.B.6 Step 1 Slice the tank crosswise
- The tank is a cylinder lying on its side, so every vertical slice across it is the same circle, and the water forms the same shape along the whole $9$-foot length.
- That means the volume of water equals the area of the wet part of one circular cross-section times the length $9$.
- The messy 3D question becomes a flat 2D area question.
💡 A cylinder is one slice repeated, so its volume is a single slice's area stretched along the length.
7.G.B.4 Step 2 Locate the water line
- Draw the circular cross-section with radius $4$.
- Its center sits $4$ feet above the lowest point of the tank.
- The water is $2$ feet deep, so the water surface is a horizontal chord sitting $4-2=2$ feet below the center.
- The wet region is the small slice of the circle below that chord.
- A slice cut off by a chord is a circular segment, and its area is the pie-slice (sector) from the center minus the triangle between the center and the chord.
💡 The flat water surface is a chord, and the piece under it is a sector with the top triangle taken away.
8.G.B.7 Step 3 Find the central angle
- Let $O$ be the center and $M$ the midpoint of the water chord; the segment $OM$ has length $2$ (the distance just found) and meets the chord at a right angle.
- Let $A$ and $B$ be the two endpoints of the chord, so $OA=OB=4$ are radii.
- By the Pythagorean theorem the half-chord is $MA=\sqrt{4^2-2^2}=\sqrt{12}=2\sqrt{3}$.
- The right triangle $OMA$ has sides $2$, $2\sqrt{3}$, $4$ — the $1:\sqrt{3}:2$ ratio of a 30-60-90 triangle — so the angle at $O$ is $60^\circ$.
- By symmetry the full central angle across the chord is $\angle AOB = 2\times 60^\circ = 120^\circ$.
💡 The perpendicular from the center splits the chord into two equal right triangles whose side ratio reveals a 30-60-90 shape.
7.G.B.4 Step 4 Area of the sector
- The sector is the pie slice from the center $O$ out to the bottom arc, spanning the $120^\circ$ central angle.
- Since $120^\circ$ is $\frac{120}{360}=\frac{1}{3}$ of a full turn, the sector is one third of the whole circle.
- The whole circle has area $\pi r^2 = \pi\cdot 4^2 = 16\pi$, so the sector area is $\frac{1}{3}\cdot 16\pi = \frac{16\pi}{3}$.
💡 A $120^\circ$ wedge is exactly one third of the circle, so take one third of the circle's area.
6.G.A.1 Step 5 Area of the triangle
- The triangle $AOB$ has the whole water chord as its base and the center $O$ as its apex.
- The base is the full chord $AB = 2\times MA = 4\sqrt{3}$, and the height from $O$ straight down to the chord is $OM = 2$.
- So its area is $\frac{1}{2}\times \text{base}\times \text{height} = \frac{1}{2}\times 4\sqrt{3}\times 2 = 4\sqrt{3}$.
- The wet segment is the sector minus this triangle: $\frac{16\pi}{3} - 4\sqrt{3}$.
💡 Base times height over two gives the triangle, and removing it from the sector leaves just the water slice.
7.G.B.6 Step 6 Multiply by the length
- The wet cross-section has area $\frac{16\pi}{3}-4\sqrt{3}$ square feet.
- Multiply by the tank's length $9$ feet to get the volume: $9\left(\frac{16\pi}{3}-4\sqrt{3}\right) = \frac{9\cdot 16\pi}{3} - 9\cdot 4\sqrt{3} = 48\pi - 36\sqrt{3}$ cubic feet.
- This matches choice $(E)$.
💡 Square feet of area times feet of length gives cubic feet of volume.
7.G.B.6 The tank is a cylinder lying on its side, so every vertical slice across it is t 7.G.B.4 Draw the circular cross-section with radius $4$. Its center sits $4$ feet above 8.G.B.7 Let $O$ be the center and $M$ the midpoint of the water chord; the segment $OM$ 7.G.B.4 The sector is the pie slice from the center $O$ out to the bottom arc, spanning 6.G.A.1 The triangle $AOB$ has the whole water chord as its base and the center $O$ as i 7.G.B.6 The wet cross-section has area $\frac{16\pi}{3}-4\sqrt{3}$ square feet. Multiply Review
Reasonableness: 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 $\frac{16\pi}{3}-4\sqrt{3}\approx 16.76-6.93\approx 9.8$ square feet is well under half of the full $16\pi\approx 50.3$, as expected. The volume $48\pi-36\sqrt{3}\approx 150.8-62.4\approx 88.4$ cubic feet is far below the full-tank volume $\pi\cdot 4^2\cdot 9 = 144\pi\approx 452$ cubic feet, consistent with a tank filled well below halfway. This matches choice $(E)$.
Alternative: Use the standard circular-segment formula directly. With radius $r=4$ and the chord a distance $d=2$ from the center, the segment area is $r^2\cos^{-1}\!\left(\frac{d}{r}\right) - d\sqrt{r^2-d^2} = 16\cos^{-1}\left(\frac{1}{2}\right) - 2\sqrt{12} = 16\cdot\frac{\pi}{3} - 4\sqrt{3} = \frac{16\pi}{3}-4\sqrt{3}$, the same wet area. Multiplying by the length $9$ again gives $48\pi-36\sqrt{3}$.
CCSS standards used (min grade 8)
7.G.B.6Solve real-world problems involving area, surface area, and volume (Turning the cylinder into (wet cross-section area) times length, and computing the final volume $48\pi-36\sqrt{3}$.)7.G.B.4Know the formulas for area and circumference of a circle (Computing the full circle area $16\pi$ and taking the $120^\circ$ sector as one third of it.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the half-chord $2\sqrt{3}$ and recognizing the 30-60-90 triangle that gives the $120^\circ$ central angle.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area $4\sqrt{3}$ of the center-to-chord triangle that is subtracted from the sector.)
⭐ 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.
⭐ 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.
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