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📐 Mathematics  ·  Circles & Mensuration  ·  JEE

Volume of a frustum (truncated cone) with radii R, r and height h:

Answer: (1/3)πh(R² + Rr + r²).

  • A πh(R+r)/2, a simplified average radius form
  • B (1/3)πh(R² + Rr + r²)
  • C πh(R² + r²), omitting the cross term
  • D (2/3)πh(R+r)², an incorrectly squared form

Correct answer: B. (1/3)πh(R² + Rr + r²)

Explanation: Frustum volume = (1/3)πh(R² + Rr + r²). Derived by subtracting smaller cone from larger.

Cylinder, Cone, SphereV=πr²hCSA=2πrhV=⅓πr²hCSA=πrlV=4/3πr³SA=4πr²

Three standard 3D solids and their key formulas: a cylinder's volume scales with the full height, a cone's volume is exactly one-third of the cylinder with the same base and height, and a sphere's volume and surface area both follow distinct r-based formulas worth memorising separately.

Concept context

Perimeters, areas, and volumes of 2D and 3D shapes

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