Find the volume of a cone with r = 3 cm and h = 8 cm.
- V = (1/3) × π × 3² × 8.
- The exact volume is 24π cm³.
- Using π ≈ 3.14 gives 75.36 cm³.
Grade 9 · Maths
A cone-shaped cup has a sloping side and a straight height down its middle. Choosing the sloping length for its volume makes the cup seem larger than it is.
The volume of a right circular cone is V = (1/3)πr²h, where h is the perpendicular height. The slant height l runs along the side; for a right cone, l² = r² + h².
Before you start: Circle area, Pythagoras and cubic units.
For a right circular cone, the tip is directly above the centre of its base. The perpendicular height runs from that tip to the centre. The slant height runs along the surface to the rim. Radius, perpendicular height and slant height form a right triangle in a central cross-section.
The volume formula uses perpendicular height because volume measures how the circular cross-sections stack above the base. Substituting the longer slant height overestimates the volume. Draw h inside the cone and l along its edge; the letters identify different measurements.
| Quantity | Formula for a right circular cone | Unit |
|---|---|---|
| Base area | πr² | square units |
| Volume | πr²h/3 | cubic units |
| Slant height | l = √(r²+h²) | length units |
| Curved surface area | πrl | square units |
A cone’s volume is one third of a cylinder with the same circular base and perpendicular height. For radius 3 cm and height 8 cm, the cylinder volume is 72π cm³ and the cone volume is 24π cm³. The comparison requires both dimensions to match; any randomly chosen cone and cylinder will not have this relationship.
For an ideal hollow container, volume gives capacity. Real wall thickness and how high the container is filled can change the usable capacity. State that you are modelling the inside shape when using a cone formula for a cup or funnel.
If radius is 5 cm and slant height is 13 cm, h²=13²−5²=144, so h=12 cm. Choose the positive root because a physical length is positive. Now V=π×25×12/3=100π cm³.
If volume is 48π cm³ and radius is 4 cm, rearrange 48π=π×16×h/3. Cancel π and multiply by 3: 144=16h, so h=9 cm. Substitution checks the result: 16×9/3=48.
Volume answers how much space is inside. Curved surface area answers how much material covers the sloping surface. For a closed cone, add the base area to the curved surface area. A funnel open at its base is a different surface question from a closed solid cone.
If all lengths scale by two, volume scales by 2³=8. If only the radius doubles and height stays fixed, volume scales by four. If only height doubles, volume doubles. State which dimensions change before applying a scale rule.
A radius of 20 mm and height of 6 cm cannot be inserted as 20 and 6 without conversion. Use radius 2 cm, giving V=π×4×6/3=8π cm³. One cubic centimetre equals 1,000 cubic millimetres; the factor is cubed because volume has three dimensions.
A diagram need not be to scale, so read labels rather than estimating lengths from its appearance. A claimed slant height shorter than the radius or perpendicular height cannot describe a right cone and should be checked before calculating.
Attempt a short group of questions before opening the solutions. Use a drawing or written method where helpful. Explain the answer, then try a similar question with a detail changed.
12π cm³
(1/3) × π × 4 × 9.
10 cm
√(36 + 64) = 10.
Cone is one third
Both use the same base area and height.
Volume doubles
The formula is directly proportional to h.
Volume becomes four times as large
The radius is squared.
48π cm³
π×16×9/3.
15 cm
√(289−64)=15.
5 cm
r²=3×75/9=25.
30π cm³
Use radius 3, not 6.
27
3³=27.
8π cm³
20 mm=2 cm.
No
The slant height is the hypotenuse and must exceed the radius.
Curved surface area
πrl uses l; volume uses perpendicular h.
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