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📐 Mathematics  ·  Circles & Mensuration  ·  JEE

If a sphere is cut by a plane at distance d from centre, the cross-section circle has radius:

Answer: √(r² - d²).

  • A r - d
  • B √(r² - d²)
  • C r + d
  • D √(r² + d²)

Correct answer: B. √(r² - d²)

Explanation: The centre O, the foot of perpendicular F (at distance d), and any point P on the cross-section circle form a right triangle. OP=r (radius), OF=d, FP=cross-section radius. Pythagoras: FP=√(r²−d²). Ans: √(r²−d²).

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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