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Class 10 Maths Notes — Chapter 12: Surface Areas and Volumes

Surface areas and volumes of solids and their combinations; conversion between shapes.

Detailed NCERT notes

  • Cylinder: CSA = 2πrh, TSA = 2πr(r + h), Volume = πr²h.
  • Cone: slant height l = √(r² + h²); CSA = πrl; TSA = πr(r + l); Volume = (1/3)πr²h.
  • Sphere: Surface area = 4πr²; Volume = (4/3)πr³.
  • Hemisphere: CSA = 2πr²; TSA = 3πr² (curved + flat circle); Volume = (2/3)πr³.
  • Frustum of a cone (with radii r₁, r₂ and height h): Volume = (1/3)πh(r₁² + r₂² + r₁r₂); slant height L = √(h² + (r₁ − r₂)²); CSA = π(r₁ + r₂)L; TSA = CSA + π(r₁² + r₂²).
  • Combinations: cone + hemisphere (ice-cream), cylinder + cone (tent), cylinder + hemispheres (capsule), sphere in cylinder.
  • For joined solids, TSA is NOT the sum of TSAs — subtract the overlapping (joined) face.
  • Conservation of volume: when a solid is melted/recast into another, volume is preserved.
  • Flow rate problems: volume flowing in time = cross-section area × speed × time.

Formulas & key results

  • Cylinder: CSA = 2πrh; TSA = 2πr(r + h); V = πr²h
  • Cone: CSA = πrl; TSA = πr(r + l); V = (1/3)πr²h; l = √(r² + h²)
  • Sphere: SA = 4πr²; V = (4/3)πr³
  • Hemisphere: CSA = 2πr²; TSA = 3πr²; V = (2/3)πr³
  • Frustum: V = (1/3)πh(r₁² + r₂² + r₁r₂); L = √(h² + (r₁ − r₂)²)

Mind map

  • Solids: cylinder, cone, sphere, hemisphere, frustum
  • Combinations of solids
  • Volume conservation (melting/recasting)
  • Flow-rate word problems

Tricks & shortcuts

  • For melted/recast problems, equate volumes.
  • For joined solid TSA: sum CSAs, add non-joined flat faces only.

Common mistakes to avoid

  • Adding TSAs of joined solids blindly.
  • Using h instead of l in cone CSA.

Competency-based questions & answers

  1. Q. A cone (r = 6, h = 8) is melted into a sphere. Find the sphere's radius.
    A. (1/3)π·36·8 = (4/3)πR³ ⇒ R³ = 72 ⇒ R = ∛72 ≈ 4.16 cm.
  2. Q. A capsule is 14 mm long with 5 mm diameter (cylinder + two hemispheres). Find its surface area.
    A. r = 2.5; cylindrical length = 14 − 5 = 9; SA = 2πrh + 4πr² = π(45 + 25) = 70π ≈ 220 mm².
  3. Q. Water flows from a pipe of radius 0.75 cm at 7 m/s into a tank. How much water in litres flows in 1 hour?
    A. Volume/s = π(0.75)²·700 cm³ = 1237.5 cm³/s; in 1 h = 4455000 cm³ = 4455 L.