Volume of a Cone

\( V=\tfrac{1}{3}\pi r^2 h \)
Geometry GCSE

\( Cone radius=3 cm, height=12 cm. Find volume. \)

Hint (H)
\( V=1/3 πr^2h. \)

Explanation

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Statement

The volume of a cone is given by:

\[ V = \tfrac{1}{3}\pi r^2 h \]

where \(r\) is the radius of the circular base and \(h\) is the perpendicular height of the cone.

Why it’s true

  • A cone is like a pyramid with a circular base.
  • For any pyramid, volume = \(\tfrac{1}{3}\)(base area)(height).
  • The base area of a cone is \(\pi r^2\), so the formula becomes \(\tfrac{1}{3}\pi r^2 h\).

Recipe (how to use it)

  1. Identify the radius \(r\) of the cone’s base.
  2. Find the perpendicular height \(h\) (not the slant height).
  3. Square the radius: \(r^2\).
  4. Multiply by \(\pi\), then by the height \(h\).
  5. Divide the result by 3.

Spotting it

Look for problems mentioning cones or pyramids with circular bases, often in 3D geometry questions.

Common pairings

  • Surface area of a cone (different formula).
  • Pythagoras used when only slant height is given, to find the perpendicular height.

Mini examples

  1. Given: \(r=3\), \(h=9\). Find: Volume. Answer: \(\tfrac{1}{3}\pi \cdot 9 \cdot 9 = 27\pi\).
  2. Given: \(r=5\), \(h=12\). Find: Volume. Answer: \(\tfrac{1}{3}\pi \cdot 25 \cdot 12 = 100\pi\).

Pitfalls

  • Using the slant height instead of the perpendicular height.
  • Forgetting to divide by 3 (cone is one third of a cylinder).
  • Not squaring the radius.

Exam strategy

  • Write down the formula before substituting values.
  • If height is missing but slant height is given, use Pythagoras with radius.
  • Leave answers in terms of \(\pi\) unless a decimal is requested.

Summary

The volume of a cone is one third of the volume of a cylinder with the same base and height. Always use the perpendicular height and take care with squaring the radius.

Worked examples

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  1. \( Find the volume of a cone with r=3 cm, h=9 cm. \)
    1. \( Base area=π×3^2=9π \)
    2. \( Multiply by height=9π×9=81π \)
    3. \( Divide by 3=27π \)
    Answer: 27π cm³
  2. \( Find the volume of a cone with r=5 cm, h=12 cm. \)
    1. \( Base area=π×25=25π \)
    2. \( Multiply by height=25π×12=300π \)
    3. \( Divide by 3=100π \)
    Answer: 100π cm³
  3. \( A cone has r=7 cm, h=10 cm. Find the volume. \)
    1. \( Base area=π×49=49π \)
    2. \( Multiply by height=49π×10=490π \)
    3. Divide by 3≈163.3π
    Answer: 490π/3 cm³
  4. \( Cone with r=4 cm, h=6 cm. Volume? \)
    1. \( Base area=π×16=16π \)
    2. \( Multiply by height=16π×6=96π \)
    3. \( Divide by 3=32π \)
    Answer: 32π cm³
  5. \( Cone with diameter=10 cm, h=15 cm. Find volume. \)
    1. \( Radius=5 cm \)
    2. \( Base area=π×25=25π \)
    3. \( Multiply by height=25π×15=375π \)
    4. \( Divide by 3=125π \)
    Answer: 125π cm³
  6. \( Find volume when r=2 cm, h=9 cm. \)
    1. \( Base area=π×4=4π \)
    2. \( Multiply by height=4π×9=36π \)
    3. \( Divide by 3=12π \)
    Answer: 12π cm³
  7. \( Cone radius=6 cm, slant height=10 cm. Find volume. \)
    1. \( Use Pythagoras: h=√(10^2-6^2)=√64=8 \)
    2. \( Base area=π×36=36π \)
    3. \( Multiply by height=36π×8=288π \)
    4. \( Divide by 3=96π \)
    Answer: 96π cm³
  8. \( Cone radius=9 cm, height=12 cm. Volume? \)
    1. \( Base area=π×81=81π \)
    2. \( Multiply by height=972π \)
    3. \( Divide by 3=324π \)
    Answer: 324π cm³
  9. \( Cone radius=1.5 cm, height=4 cm. Volume? \)
    1. \( Base area=π×(1.5^2)=2.25π \)
    2. \( Multiply by height=2.25π×4=9π \)
    3. \( Divide by 3=3π \)
    Answer: 3π cm³
  10. \( Find volume in terms of π, r=10 cm, h=24 cm. \)
    1. \( Base area=π×100=100π \)
    2. \( Multiply by height=2400π \)
    3. \( Divide by 3=800π \)
    Answer: 800π cm³