HSG.GMD.A.3: Volume of Cylinders, Pyramids, Cones and Spheres
In plain English: HSG.GMD.A.3 is the Common Core geometry standard that asks students to use the volume formulas for cylinders, pyramids, cones and spheres to solve problems. Students find volumes, work backward from a volume to a missing radius or height, and combine solids in real objects such as tanks, silos and cups. It is usually taught in high school Geometry.
Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.
Common Core State Standards for Mathematics · Domain: Geometric Measurement and Dimension (GMD) · Cluster: Explain volume formulas and use them to solve problems Also written as HSG-GMD.A.3 or G-GMD.3 · Official standard
Students already met the volume formulas for cylinders, cones and spheres in Grade 8. This lesson adds pyramids and moves from plugging in numbers to solving problems: choosing the right formula for a real object, finding a missing radius or height from a known volume, getting the height of a pyramid from its slant height, and adding or subtracting solids to find the volume of a composite object.
Throughout, students keep track of what each letter means (radius, not diameter; perpendicular height, not slant height), leave exact answers in terms of π before rounding, and convert cubic units to liters when the context calls for it.
Learning Objectives
By the end of this lesson, students will be able to:
Choose and apply the volume formula for a cylinder (V = πr²h), a pyramid (V = ⅓Bh), a cone (V = ⅓πr²h) and a sphere (V = ⁴⁄₃πr³)
Solve for a missing radius, height or base edge when the volume is known
Find the volume of a composite solid by adding or subtracting cylinders, cones, pyramids, spheres and hemispheres
Give answers in exact form and as rounded decimals with correct cubic units, and convert to liters when needed
Prior Knowledge Required
Students should already be comfortable with:
Area of a circle 7.G.B.4
Volume of right prisms, V = Bh 7.G.B.6
The Pythagorean Theorem in two and three dimensions 8.G.B.7
Volume formulas for cones, cylinders and spheres 8.G.C.9
Show a cone, a sphere and a cylinder that all have radius 6 cm, where the cone and the cylinder are 12 cm tall (so the height equals the diameter of the sphere). Students first predict the order of the volumes, then compute them.
Warm-Up Prompt
"A cone, a sphere and a cylinder each have radius 6 cm. The cone and the cylinder are 12 cm tall. Rank them from smallest to largest volume. Then compute each volume in terms of π. What do you notice about the three answers?"
The volumes are 144π cm³ (cone), 288π cm³ (sphere) and 432π cm³ (cylinder), in the ratio 1 : 2 : 3 (Diagram 2). Many students guess that the sphere is the largest. Use the ratio to preview the three formulas: the cone is one third of its cylinder, and a sphere is two thirds of the cylinder that just contains it.
Direct Instruction20 minutes
Use Diagram 1 to name the parts of each solid, then state the formulas. Where the formulas come from is the subject of HSG.GMD.A.1; here the goal is to use them well.
Cylinder: V = Bh = πr²h. The base is a circle, so B = πr².
Pyramid: V = ⅓Bh, where B is the area of the base polygon (a square, rectangle or triangle) and h is the perpendicular height from the apex to the base.
Cone: V = ⅓Bh = ⅓πr²h. A cone is to a cylinder what a pyramid is to a prism: one third of the solid with the same base and height.
Sphere: V = ⁴⁄₃πr³. A hemisphere is half of that, ⅔πr³.
Problem-solving routine: sketch and label the solid, write the formula, substitute (radius, not diameter), keep π exact, round only at the end, and write cubic units. Useful conversion: 1 L = 1000 cm³ and 1 m³ = 1000 L.
Cylinder in context
A cylindrical water tank has an inside diameter of 1.2 m and a height of 1.5 m. How many liters does it hold?
Equation: r = 0.6 m; V = π(0.6)²(1.5) = 0.54π ≈ 1.696 m³ ≈ 1696 L
Pyramid from the slant height
A square pyramid has base edge 10 m and slant height 13 m (measured along a face from the apex to the midpoint of a base edge). Find its volume.
Equation: h = √(13² - 5²) = 12 m; V = ⅓(10²)(12) = 400 m³
Cone in context
A waffle cone has a rim diameter of 5 cm and a height of 12 cm. How much ice cream fits inside if it is filled level with the rim?
Equation: V = ⅓π(2.5)²(12) = 25π ≈ 78.5 cm³
Sphere, working backward
A spherical balloon holds 4500π cm³ of air. What is its radius?
Equation: ⁴⁄₃πr³ = 4500π, so r³ = 3375 and r = 15 cm
Composite solid
A grain silo is a cylinder with radius 3 m and height 10 m, topped by a hemisphere with the same radius. Find its volume.
Pairs solve three problems on whiteboards and hold them up after each one. (a) A cylindrical vase with inside radius 5 cm must hold 1 liter of water. How tall must it be at least? (h = 1000 ÷ (25π) ≈ 12.7 cm.) (b) A conical pile of sand is 8 m across at the bottom and 3 m tall. Find its volume. (16π ≈ 50.3 m³.) (c) A hemispherical mixing bowl has inside radius 8 cm. How much does it hold? (1024π/3 ≈ 1072.3 cm³, about 1.07 L.) Listen for these errors: using the diameter as the radius, dropping the ⅓ for the cone, squaring instead of cubing the radius, and rounding π too early.
Independent Practice15 minutes
Students work alone on four problems and check with a partner at the end. (1) A cylinder has radius 2 in and height 7 in. Find its volume (28π ≈ 88.0 in³). (2) A rectangular pyramid has a base 8 ft by 6 ft and height 9 ft. Find its volume (144 ft³). (3) A cone has volume 96π cm³ and radius 6 cm. Find its height (8 cm). (4) A sphere has diameter 10 cm. Find its volume (500π/3 ≈ 523.6 cm³). For problem 3, students write one sentence explaining how they undid the ⅓.
Closure5-10 minutes
Exit ticket: (1) A cone and a cylinder both have radius 3 cm and height 7 cm. Find both volumes and explain how they compare (63π and 21π cm³: the cone is one third of the cylinder). (2) Find the volume of a sphere with radius 1.5 cm (4.5π ≈ 14.1 cm³). (3) Name one mistake you will watch for next time.
Differentiation Strategies
For Struggling Students
Give a formula card with a labeled sketch of each solid and a blank row for "r = ___, h = ___, B = ___" to fill in before substituting
Start with problems that give the radius directly, then move to problems that give the diameter
Let students keep answers in terms of π first and use the calculator only for the final rounding step
For Advanced Students
Ask for the height of a cone that has the same volume as a sphere of radius r and the same radius r (h = 4r)
Ask how the volume changes when every length of a solid is multiplied by k, and why the factor is the same for all four solids
Design a cylindrical can that holds 355 cm³ with a height between 10 cm and 13 cm, and report the range of possible radii
Assessment Guidance
What to Look For
Check that students choose the right formula from the description of the object, not from a list of numbers. Strong work labels r and h on a sketch, uses the radius rather than the diameter, uses the perpendicular height of a pyramid or cone, and shows the step that undoes the ⅓ or the ⁴⁄₃ when solving backward. For composite solids, look for a plan (which pieces, added or subtracted) before any arithmetic. Final answers should carry cubic units, and liters when the context asks for capacity.
02
Classroom Activities
3 Activities
1
Pour and Compare
20 minGroups of 3
Groups fill hollow solids with rice to see the cone-cylinder and sphere-cylinder relationships, then check them against the formulas using their own measurements. The pouring connects to the informal arguments of HSG.GMD.A.1; the calculations practice this standard.
Procedure
Measure the radius and height of the hollow cylinder, cone and sphere with a ruler. A typical set has radius 3.5 cm, and the cone and cylinder are 7 cm tall
Fill the cone with rice, level it, and pour it into the cylinder. Count how many cones fill the cylinder (3)
Fill the sphere and pour it into the cylinder. Estimate what fraction of the cylinder it fills (about ⅔)
Compute all three volumes from your measurements. With r = 3.5 cm and h = 7 cm: cylinder 85.75π ≈ 269.4 cm³, cone ≈ 89.8 cm³, sphere ≈ 179.6 cm³
Discussion Questions
Your pouring gave a result close to, but not exactly, 3 cones. Is the difference a measuring issue or a mathematical one?
Why does the sphere fill two thirds of the cylinder only when the cylinder's height equals the sphere's diameter?
Which of the four formulas could you now rebuild from the cylinder formula alone?
Modification for Distance Learning
Students use kitchen objects instead: a drinking glass as the cylinder and a paper cone rolled to the same rim and height. They fill the cone with water and pour it into the glass, then compute both volumes from ruler measurements.
2
Half-Liter Package Design
20 minPairs
Pairs design four containers that each hold 500 cm³ (half a liter). Each design fixes one dimension, and students solve the volume formula for the other one.
The 4 Designs
Design A, cylinder can: the radius is 4 cm. Find the height (500 ÷ 16π ≈ 9.9 cm)
Design B, cone cup: the rim radius is 5 cm. Find the height (1500 ÷ 25π ≈ 19.1 cm)
Design C, square pyramid bottle: the base edge is 10 cm. Find the height (15 cm)
Design D, spherical bottle: find the radius (r³ = 375 ÷ π, so r ≈ 4.9 cm)
Procedure
Solve each formula for the unknown before substituting any numbers
Sketch each design to scale on grid paper, 1 square = 1 cm
Choose the design you would sell and write two sentences that justify the choice (easy to hold, easy to stack, fits in a lunch bag)
Challenge Variation
Pairs redesign the cone cup so that it is no taller than 12 cm and still holds 500 cm³, and find the smallest rim radius that works (r = √(1500 ÷ 12π) ≈ 6.3 cm).
3
Volume Stations with Real Objects
20 minGroups of 3-4
Groups rotate through four stations, one for each solid. At each station they measure a real object (or read its dimensions from the card), choose the formula and compute the volume.
The 4 Stations
Station 1, cylinder: a round cake pan 9 in across and 2 in deep. Volume: 40.5π ≈ 127.2 in³
Station 2, pyramid: a camping tent shaped like a square pyramid, base 2.4 m by 2.4 m and height 1.8 m. Volume: ⅓(5.76)(1.8) ≈ 3.46 m³
Station 3, cone: a paper water cup, rim diameter 7 cm and height 9 cm. Volume: 36.75π ≈ 115.5 cm³
Station 4, sphere: a tennis ball, diameter about 6.7 cm. Volume: ⁴⁄₃π(3.35)³ ≈ 157.5 cm³
Procedure
Spend 5 minutes at each station. Record the formula, the measurements and the volume on the group sheet
At each station, write which measurement was hardest to take and how it could change the answer
Challenge Variation
At Station 3, find the height of a cylindrical cup with the same 7 cm rim that holds the same amount of water as the paper cone (3 cm), and explain why it is one third of the cone's height.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Four Solids and Their Volume Formulas
Drawn to one scale: a cylinder and a cone with r = 4 and h = 6, a square pyramid with base edge 8 and height 6, and a sphere with r = 4. The dashed segment in each pointed solid is the perpendicular height h, which is not the slant edge. Volumes: 96π, 32π, 128 and 256π/3 cubic units.
Diagram 2: Cone, Sphere and Cylinder in the Ratio 1 : 2 : 3
A cone, a sphere and a cylinder with radius 6 and height 12 (the sphere's diameter). Their volumes are 144π, 288π and 432π, and the bars are drawn to scale. The cone is one third of the cylinder and the sphere is two thirds of it.
04
Homework Assignment
~30 min
HSG.GMD.A.3 Homework: Solving Volume Problems
Directions: Sketch and label each solid. Write the formula before you substitute. Give the exact answer in terms of π where possible, then a decimal rounded as stated, with units.
Part 1: Cylinders and Cones (Problems 1-3)
A cylindrical rain barrel has an inside diameter of 60 cm and a height of 90 cm. (a) Find its volume in cubic centimeters, exactly and to the nearest whole number. (b) How many liters does it hold, to the nearest tenth? (1 L = 1000 cm³)
A cone-shaped paper cup has a rim diameter of 7 cm and holds 150 cm³ of water when full. Find the height of the cup to the nearest tenth of a centimeter. Show the step where you undo the ⅓.
A cone-shaped hopper with radius 1.5 m and height 2 m is full of grain. All of the grain is emptied into a cylindrical bin that also has radius 1.5 m. How deep is the grain in the bin? Explain why the answer is one third of the hopper's height.
Part 2: Pyramids, Spheres and Composite Solids (Problems 4-6)
A glass skylight is shaped like a square pyramid with base edge 3 m. The slant height of each triangular face is 2.5 m. Find the perpendicular height of the pyramid and the volume of air under the skylight.
A spherical water tank has an inside diameter of 14 m. (a) Find its volume, exactly and to the nearest tenth of a cubic meter. (b) A pump fills it at 50 m³ per hour. To the nearest tenth of an hour, how long does it take to fill the empty tank?
An ice cream cone has radius 2.5 cm and height 11 cm and is filled level with the rim. A hemisphere of ice cream with radius 2.5 cm sits on top. (a) Find the total volume of ice cream in terms of π and to the nearest tenth. (b) If only the hemisphere on top melted, would it fit inside the cone? Explain with volumes.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Formula Choice
Correct formula for every solid, written before substituting
One formula wrong or missing
Formulas missing or mostly wrong
Setup
Radius, perpendicular height and base area identified correctly
Minor setup error, such as a diameter used once
Setup does not match the solid
Solving Backward
Missing radius or height found with clear inverse steps
Correct idea with an algebra slip
No inverse steps shown
Accuracy and Units
Exact and rounded answers correct, with cubic units or liters
Small arithmetic or rounding errors
Many errors or no units
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Choose an answer for each multiple-choice question and work the short-answer questions on paper before opening the solution. Your score updates as you go, and Reset quiz starts over.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
What is the volume of a cylinder with radius 5 cm and height 8 cm?
Answer: C
V = πr²h = π(5²)(8) = 200π cm³. Choice A forgets to square the radius, choice B is the lateral surface area 2πrh, and choice D uses the diameter 10 as the radius.
Question 2 of 20 · Multiple Choice
What is the volume of a cone with radius 6 in and height 10 in?
Answer: A
V = ⅓πr²h = ⅓π(36)(10) = 120π in³. Choice B leaves out the ⅓ (that is the cylinder's volume), choice C does not square the radius, and choice D uses the diameter 12 as the radius.
Question 3 of 20 · Multiple Choice
What is the volume of a sphere with radius 3 in?
Answer: B
V = ⁴⁄₃π(3³) = ⁴⁄₃π(27) = 36π in³. Choice A squares the radius instead of cubing it, choice C leaves out the ⅓ in ⁴⁄₃, and choice D uses the diameter 6 as the radius.
Question 4 of 20 · Multiple Choice
A pyramid has a rectangular base 9 m by 4 m and a height of 7 m. What is its volume?
Answer: D
B = 9 · 4 = 36 m², so V = ⅓Bh = ⅓(36)(7) = 84 m³. Choice A leaves out the ⅓ (the prism's volume), and choice B uses ½ as if the solid were a triangle.
Question 5 of 20 · Multiple Choice
A cylindrical can holds 450π cm³ and has radius 5 cm. How tall is the can?
Answer: C
450π = π(5²)h, so h = 450 ÷ 25 = 18 cm. Choice A divides by 10², using the diameter, choice B uses the cone formula (3 · 450 ÷ 25), and choice D divides by 5 without squaring.
Question 6 of 20 · Multiple Choice
A cone has volume 64π cm³ and radius 4 cm. What is its height?
Answer: B
64π = ⅓π(4²)h gives 64 = 16h/3, so h = 192 ÷ 16 = 12 cm. Choice A forgets to undo the ⅓ (64 ÷ 16), and choice D multiplies by 3 but divides by 4 instead of 4².
Question 7 of 20 · Multiple Choice
A sphere has volume 972π cubic units. What is its radius?
Answer: A
⁴⁄₃πr³ = 972π gives r³ = 972 · ¾ = 729, so r = ∛729 = 9. Choice C stops at r³, choice B takes a square root instead of a cube root, and choice D is the diameter.
Question 8 of 20 · Multiple Choice
A cone and a cylinder have the same base and the same height. The cylinder holds 90 cm³. How much does the cone hold?
Answer: D
V(cone) = ⅓Bh and V(cylinder) = Bh with the same B and h, so the cone holds ⅓ · 90 = 30 cm³. Choice A multiplies by 3 instead of dividing, and choice B halves the volume.
Question 9 of 20 · Multiple Choice
A square pyramid has base edge 16 ft and slant height 17 ft. What is its volume?
Answer: C
The height, half the base edge and the slant height form a right triangle: h = √(17² - 8²) = √225 = 15 ft. Then V = ⅓(16²)(15) = 1280 ft³. Choice A uses the slant height 17 as the height, choice B leaves out the ⅓, and choice D does both.
Question 10 of 20 · Multiple Choice
A hemispherical bowl has inside radius 4.5 in. How much does it hold?
Answer: A
A hemisphere is half a sphere: V = ⅔πr³ = ⅔π(4.5)³ = ⅔π(91.125) = 243π/4 ≈ 190.9 in³. Choice B is the whole sphere, choice C uses the diameter 9 as the radius, and choice D squares the radius instead of cubing it.
Question 11 of 20 · Multiple Choice
The radius of a cylinder is doubled and its height stays the same. The new volume is how many times the old volume?
Answer: B
V = πr²h, and replacing r with 2r gives π(2r)²h = 4πr²h, so the volume is multiplied by 4. Choice A treats volume as if it grew like the radius, and choice D would be right only if the height also doubled.
Question 12 of 20 · Multiple Choice
The radius of a sphere is tripled. The new volume is how many times the old volume?
Answer: D
V = ⁴⁄₃πr³, and ⁴⁄₃π(3r)³ = 27 · ⁴⁄₃πr³, so the volume is multiplied by 3³ = 27. Choice B is the factor for surface area (3²), not volume.
Question 13 of 20 · Multiple Choice
A conical pile of gravel is 10 ft across at the bottom and 6 ft tall. To the nearest cubic foot, what is its volume?
Answer: A
r = 5 ft, so V = ⅓π(5²)(6) = 50π ≈ 157 ft³. Choice B leaves out the ⅓ (150π), choice C uses the diameter 10 as the radius (200π), and choice D drops π.
Question 14 of 20 · Multiple Choice
A cylindrical water heater tank has an inside diameter of 40 cm and an inside height of 120 cm. About how many liters does it hold? (1 L = 1000 cm³)
Answer: B
V = π(20²)(120) = 48000π ≈ 150,796 cm³, and 150,796 ÷ 1000 ≈ 150.8 L. Choice C uses the diameter 40 as the radius, and choices A and D move the decimal point in the conversion.
Question 15 of 20 · Short Answer
The attic of a building is shaped like a pyramid with a rectangular base 12 m by 10 m and a height of 4 m. Find the volume of the attic.
B = 12 · 10 = 120 m², so V = ⅓Bh = ⅓(120)(4) = 160 m³.
Question 16 of 20 · Short Answer
The glass globe of a gumball machine is a sphere with an inside diameter of 32 cm. Find its volume in cubic centimeters and in liters, to the nearest tenth.
r = 16 cm, so V = ⁴⁄₃π(16³) = 16384π/3 ≈ 17,157.3 cm³, which is about 17.2 L.
Question 17 of 20 · Short Answer
A propane tank is a cylinder 1.2 m long with a hemisphere on each end. The cylinder and the hemispheres all have radius 0.3 m. Find the volume of the tank in cubic meters (to the nearest thousandth) and in liters.
The two hemispheres make one sphere. Cylinder: π(0.3²)(1.2) = 0.108π. Sphere: ⁴⁄₃π(0.3³) = 0.036π. Total: 0.144π ≈ 0.452 m³, about 452 L.
Question 18 of 20 · Short Answer
A cone-shaped funnel has a rim diameter of 12 cm and must hold 300 cm³. How tall must it be, to the nearest tenth of a centimeter?
r = 6 cm. 300 = ⅓π(36)h = 12πh, so h = 300 ÷ 12π = 25/π ≈ 8.0 cm.
Question 19 of 20 · Short Answer
A square pyramid has volume 96 cm³ and height 8 cm. Find the length of a base edge.
96 = ⅓B(8), so B = 3 · 96 ÷ 8 = 36 cm². The base is a square, so the edge is √36 = 6 cm.
Question 20 of 20 · Short Answer
Three balls, each with radius 2 cm, are stacked in a cylindrical can. The balls just touch the side, the bottom and the lid. Find the volume of the empty space in the can, exactly and to the nearest tenth.
The can has radius 2 cm and height 3 · 4 = 12 cm, so V(can) = π(2²)(12) = 48π. The balls: 3 · ⁴⁄₃π(2³) = 32π. Empty space: 48π - 32π = 16π ≈ 50.3 cm³. The balls fill two thirds of the can.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.GMD.A.3 mean?
HSG.GMD.A.3 means students can use the volume formulas for cylinders, pyramids, cones and spheres to solve problems. That includes finding a volume, working backward to a missing radius or height, and finding the volume of objects built from several solids.
Is HSG.GMD.A.3 taught in Geometry or in middle school?
HSG.GMD.A.3 is usually taught in high school Geometry. Students first learn the cone, cylinder and sphere formulas in Grade 8 (8.G.C.9); the high school standard adds pyramids and asks for harder problem solving, such as composite solids, slant heights and solving for a missing dimension.
What are the volume formulas for a cylinder, pyramid, cone and sphere?
Cylinder V = πr²h, pyramid V = ⅓Bh, cone V = ⅓πr²h and sphere V = ⁴⁄₃πr³. In the pyramid formula, B is the area of the base and h is the perpendicular height. A hemisphere has half the volume of a sphere, ⅔πr³.
Why is a cone one third of a cylinder?
A cone holds exactly one third of the cylinder with the same base and height, and a pyramid holds one third of the matching prism. Pouring rice or water from a cone into a cylinder shows it; the informal arguments behind the formulas, including Cavalieri's principle, belong to HSG.GMD.A.1.
What is the difference between height and slant height?
The height is measured straight down from the apex to the base, at a right angle; the slant height runs along the side. Volume formulas always use the perpendicular height. When a problem gives the slant height, use the Pythagorean Theorem with half the base edge (for a square pyramid) or the radius (for a cone) to find the height.
How do you find the radius of a sphere from its volume?
Set ⁴⁄₃πr³ equal to the volume, multiply both sides by ¾, divide by π, and take the cube root. For example, a volume of 2304π gives r³ = 1728, so r = 12. Students often forget that the last step is a cube root, not a square root.
Should answers be left in terms of π?
Give the exact answer in terms of π when the problem asks for an exact value, and a rounded decimal when the context calls for a measurement, such as liters of water. Rounding π to 3.14 early can change the last digit of the answer, so keep π until the final step.
What mistakes do students make with volume formulas?
A frequent mistake is using the diameter in place of the radius. Others are leaving out the ⅓ for cones and pyramids, squaring instead of cubing the radius of a sphere, using the slant height as the height, and mixing units such as centimeters and meters in the same problem.
How do you find the volume of a composite solid?
Break the object into solids you know, find each volume, and add them, or subtract when a piece is removed. A silo is a cylinder plus a hemisphere; a capsule is a cylinder plus two hemispheres, which together make one sphere; a pipe is a large cylinder minus a smaller one.
Is volume of cones and spheres on the SAT?
Yes. Volume problems appear in the Geometry and Trigonometry domain of the digital SAT, and the reference sheet lists the formulas for these solids. Students still need to choose the right formula, use the radius and handle units, which is the work HSG.GMD.A.3 practices.
07
Related Standards
6 standards
These standards connect to HSG.GMD.A.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
8.G.C.9Prerequisite
Know the volume formulas for cones, cylinders and spheres and use them
