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Surface Areas and Volumes — Mathematics Class 9 Notes (CBSE & HBSE)

Free NCERT Mathematics notes for Surface Areas and Volumes (Class 9) on Siksha Sarovar, aligned to CBSE and Haryana Board (HBSE). This chapter is broken into 3 topics with clear explanations, formulas, solved examples and board-pattern practice — free to read, no sign-up required.

Board exam focus — Surface Areas and Volumes (CBSE & HBSE)

High-mark formula-application chapter. CBSE and HBSE together pull 8–10 marks via word problems on cuboids, cones, spheres, hemispheres and conversions.

Cuboid & Cube

Cuboid (length l, breadth b, height h)

QuantityFormula
Lateral Surface Area2 (l + b) × h
Total Surface Area2 (lb + bh + hl)
Volumel × b × h
Diagonal√(l² + b² + h²)

Cube (side a)

QuantityFormula
LSA4 a²
TSA6 a²
Volume
Diagonala√3

Real-World Use

  • Boxes, bricks, rooms, tanks.
  • LSA is needed when painting the walls only (not the floor/ceiling).
  • TSA is needed when wrapping the whole shape.

Unit-Conversion Cheat Sheet

  • 1 m³ = 1000 L
  • 1 cm³ = 1 mL
  • 1 m = 100 cm; 1 m² = 10,000 cm²; 1 m³ = 1,000,000 cm³.
CBSE Tip: Always state units in every step. Mixing m and cm is the #1 source of lost marks in word problems.

Cylinder & Cone

Right Circular Cylinder (radius r, height h)

QuantityFormula
CSA (curved)2 π r h
TSA2 π r (r + h)
Volumeπ r² h
Hollow CSA2 π (R + r) h, where R = outer radius

Right Circular Cone (radius r, slant l, height h)

Relation: l² = r² + h².

QuantityFormula
CSAπ r l
TSAπ r (r + l)
Volume(1/3) π r² h

Common Applications

  • Open cylindrical tank ⇒ TSA = 2πrh + πr² (one circle missing).
  • Joker's cap = cone (no base) ⇒ area = πrl only.
  • Tent shaped as cone over cylinder ⇒ canvas required = CSA(cone) + CSA(cylinder).

π Value Choice

  • Use π = 22/7 if denominators contain 7.
  • Use π = 3.14 when answer must be in decimal.
HBSE Tip: Show l² = r² + h² substitution explicitly when slant length is not given directly.

Sphere & Hemisphere

Sphere (radius r)

QuantityFormula
Surface area4 π r²
Volume(4/3) π r³

Hemisphere (radius r)

QuantityFormula
CSA2 π r²
TSA3 π r² (includes flat circular base)
Volume(2/3) π r³

Hollow Hemispherical Bowl

Outer radius R, inner radius r:

  • Volume of material = (2/3) π (R³ − r³).
  • Inner CSA = 2 π r², outer CSA = 2 π R².

Standard Conversion Problems

Melting and recasting — volumes are conserved: Volume of source = Volume of new shape ⇒ solve for unknown.

Example: if a cone is melted to form a sphere of equal volume, set (1/3) π r_c² h = (4/3) π r_s³ and solve.

Quick Numerical Tip

Most problems work out cleanly with π = 22/7 when one of the linear dimensions is a multiple of 7. Always check.

CBSE Tip: In melting/recasting problems, state "Volume conserved" before setting up the equation. Method marks depend on this line.

Frequently asked questions

Are these Surface Areas and Volumes notes free?

Yes — the Surface Areas and Volumes notes for Mathematics (Class 9) on Siksha Sarovar are completely free to read, with no account required.

Do these notes follow CBSE and HBSE?

Yes. The Surface Areas and Volumes notes are NCERT-aligned and include guidance for both CBSE and Haryana Board (HBSE), with important questions and MCQs for revision.

What does the Surface Areas and Volumes chapter cover?

Concept explanations, key formulas and definitions, fully solved examples and board-pattern practice questions for Surface Areas and Volumes.