AMC 8 · 2008 · #21
Grade 8 geometry-3dProblem
Jerry cuts a wedge from a 6-cm cylinder of bologna as shown by the dashed curve. Which answer choice is closest to the volume of his wedge in cubic centimeters?
Pick an answer.
AMC 8 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 bologna is $6$ cm long and $8$ cm across (so radius $4$ cm). Jerry's dashed cut is a slanted oval that passes through the cylinder's central axis, dividing it into two matching wedges. Estimate the volume of one wedge in cubic centimeters.
Givens: The cylinder's diameter is $8$ cm, so its radius is $r = 4$ cm; The cylinder's length (height) is $h = 6$ cm; The dashed cut is a plane that contains the cylinder's axis; Answer choices: (A) $48$, (B) $75$, (C) $151$, (D) $192$, (E) $603$
Unknowns: The volume of the wedge, rounded to the nearest choice
Understand
Restated: A cylinder of bologna is $6$ cm long and $8$ cm across (so radius $4$ cm). Jerry's dashed cut is a slanted oval that passes through the cylinder's central axis, dividing it into two matching wedges. Estimate the volume of one wedge in cubic centimeters.
Givens: The cylinder's diameter is $8$ cm, so its radius is $r = 4$ cm; The cylinder's length (height) is $h = 6$ cm; The dashed cut is a plane that contains the cylinder's axis; Answer choices: (A) $48$, (B) $75$, (C) $151$, (D) $192$, (E) $603$
Plan
Primary tool: #15 Visualize / Use Symmetry
Secondary: #9 Solve an Easier Related Problem
The dashed oval looks fancy, but it is just the slanted view of a flat cut that goes straight through the cylinder's middle line (its axis). Tool #15 (Visualize / Symmetry) lets us see that this single slice splits the cylinder into two mirror-image halves of equal size. Once we believe that, the hard 3-D shape question shrinks to Tool #9 (Easier Related Problem): find the whole cylinder's volume and take half. No calculus, no fancy slicing.
Execute — Answer: C
7.G.B.6 Step 1 - Read off the dimensions from the figure.
- The label "$8$ cm" marks the diameter of the round end, and "$6$ cm" marks the length along the cylinder.
- So the radius is $r = 8 \div 2 = 4$ cm and the height (length) is $h = 6$ cm.
💡 Grade 7 geometry: pull the radius and the height straight off a labeled solid before plugging into a formula.
8.G.A.1 Step 2 - See the symmetry of the cut.
- The dashed oval is the slanted picture of a flat plane that goes through the cylinder's center axis.
- Any plane through the axis splits the cylinder into two congruent (mirror-image) wedges, so each wedge has exactly half the cylinder's volume.
💡 Grade 8 rigid-motion thinking: a reflection across the cutting plane sends one wedge onto the other, so their volumes match.
8.G.C.9 Step 3 - Compute the cylinder's volume with the standard formula $V = \pi r^2 h$.
- Plug in $r = 4$ and $h = 6$.
💡 Grade 8 "volume of a cylinder" is just area of the circle times the length.
7.G.B.4 Step 4 - Take half of the cylinder's volume to get the wedge, then approximate using $\pi \approx 3.14$.
- Compare with the choices.
💡 Grade 7 "know the formulas for area and circumference of a circle" includes using $\pi \approx 3.14$ for numerical estimates.
7.G.B.6 Read off the dimensions from the figure. The label "$8$ cm" marks the diameter o 8.G.A.1 See the symmetry of the cut. The dashed oval is the slanted picture of a flat pl 8.G.C.9 Compute the cylinder's volume with the standard formula $V = \pi r^2 h$. Plug in 7.G.B.4 Take half of the cylinder's volume to get the wedge, then approximate using $\pi Review
Reasonableness: Sanity check the size. The whole cylinder is $96\pi \approx 301.6$ cm$^3$, so half is about $150.8$ cm$^3$ — right between (B) $75$ and (D) $192$, and matching (C) $151$ very closely. Choice (A) $48$ forgets the $\pi$, (D) $192$ doubles the wedge by mistake, and (E) $603$ is the whole cylinder times $2$. Only (C) is consistent with "half of $96\pi$".
Alternative: Tool #9 (Easier Related Problem) with a familiar shape: imagine slicing a round cake of radius $4$ cm and height $6$ cm straight down through the middle. You get two equal half-cakes. Each half has half the cake's volume, which is $\tfrac{1}{2}\pi r^2 h = \tfrac{1}{2}\pi (16)(6) = 48\pi \approx 151$ cm$^3$. Same answer, same picture.
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Using $\pi \approx 3.14$ to turn $48\pi$ into the numerical estimate $\approx 150.8$ so it can be matched with the answer choices.)7.G.B.6Solve real-world and mathematical problems involving area, volume and surface area of two- and three-dimensional objects (Reading the radius and height off the labeled cylinder and identifying which dimensions go into the volume formula.)8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Recognizing that a reflection across the cutting plane maps one wedge onto the other, so the two wedges are congruent and share equal volume.)8.G.C.9Know the formulas for the volumes of cones, cylinders, and spheres and use them to solve real-world and mathematical problems (Computing $V_{\text{cylinder}} = \pi r^2 h = 96\pi$ cm$^3$ as the parent volume before halving.)
⭐ When a cut goes straight through the center of a round shape, both pieces are equal — so this AMC 8 problem is really just "half of $\pi r^2 h$".
⭐ When a cut goes straight through the center of a round shape, both pieces are equal — so this AMC 8 problem is really just "half of $\pi r^2 h$".
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