Chapter 13 — Surface Area & Volume | Class 9 CBSE Maths 2026-27

Section 13.1

Right Circular Cone

A cone has a circular base (radius r), perpendicular height h, and slant height l. Always find l before computing CSA.

🔺

Cone

Slant h: l = √(r²+h²)
CSA = πrl
TSA = πr(r+l)
Volume = (1/3)πr²h

🌐

Sphere

SA = 4πr²
Volume = (4/3)πr³

🔮

Hemisphere

CSA = 2πr²
TSA = 3πr²
Volume = (2/3)πr³

⚠️

Key: CSA of cone = πrl — uses SLANT height l, NOT perpendicular height h. Always compute l = √(r²+h²) first. TSA adds the circular base: πr² + πrl = πr(r+l).

Example 1 · 3M

Cone: r=6cm, h=8cm. Find l, CSA, TSA, Volume.

1

l = √(6²+8²) = √(36+64) = √100 = 10 cm

2

CSA = πrl = (22/7)×6×10 = 1320/7 ≈ 188.6 cm²

3

TSA = πr(r+l) = (22/7)×6×16 = 2112/7 ≈ 301.7 cm²

4

Volume = (1/3)πr²h = (1/3)×(22/7)×36×8 = 6336/21 = 301.7 cm³

l=10cm, CSA≈188.6cm², TSA≈301.7cm², Volume≈301.7cm³

Section 13.2

Sphere

Every point on a sphere is equidistant from the centre. Surface area = 4πr², Volume = (4/3)πr³.

Example 2 · 2M

Sphere: r=7cm. Find SA and Volume.

1

SA = 4πr² = 4×(22/7)×49 = 616 cm²

2

V = (4/3)×(22/7)×343 = 4312/3 ≈ 1437 cm³

SA=616cm², V≈1437cm³

Example 3 · 2M

SA of sphere = 154 cm². Find radius and volume.

1

4πr²=154 → r²=154×7/(4×22)=12.25 → r=3.5 cm

2

V=(4/3)×(22/7)×42.875=179.67 cm³

r=3.5cm, V≈179.67cm³

Section 13.3

Hemisphere

A hemisphere is half a sphere — curved surface (2πr²) plus a flat circular base (πr²), giving TSA = 3πr².

Example 4 · 3M

Hemisphere: r=10.5cm. Find CSA, TSA, Volume.

1

CSA = 2πr² = 2×(22/7)×110.25 = 693 cm²

2

TSA = 3πr² = 3×(22/7)×110.25 = 1039.5 cm²

3

V = (2/3)πr³ = (2/3)×(22/7)×1157.625 = 2425.5 cm³

CSA=693cm², TSA=1039.5cm², V=2425.5cm³

Section 13.4

Combined Shapes & Key Ratios

Example 5 · 5M

Cone of r=10cm, h=12cm surmounted on hemisphere of same radius. Find TSA and Volume.

1

l = √(100+144) = √244 ≈ 15.62 cm

2

TSA = CSA(cone) + CSA(hemisphere) = πrl + 2πr² = π×10×15.62 + 2π×100 = 156.2π + 200π = 356.2π ≈ 1119 cm²

3

V = V(cone)+V(hemisphere) = (1/3)πr²h + (2/3)πr³ = (πr²/3)(h+2r) = (π×100/3)(12+20) = 3200π/3 ≈ 3351 cm³

TSA ≈ 1119 cm², Volume ≈ 3351 cm³

⭐ Key Ratio — Same radius r, height r

Cone : Hemisphere : Cylinder
= (1/3)πr³ : (2/3)πr³ : πr³ = 1 : 2 : 3

This ratio is directly tested in CBSE.

Example 6 · 3M

A sphere of r=3cm is dropped into a cylinder of r=6cm containing water. By how much does water level rise?

1

V(sphere) = (4/3)π(27) = 36π cm³

2

πR²Δh = 36π → 36Δh = 36 → Δh = 1 cm

Water level rises by 1 cm.

NCERT Exercise · Solved

Exercise 13.1 & 13.2

Q1 · 3M

A metallic sphere of r=4.2cm is melted into small spheres of r=0.6cm. How many small spheres?

1

n = V(large)/V(small) = (4.2/0.6)³ = 7³ = 343

343 small spheres

Q2 · 2M

Volumes of two spheres are in ratio 64:27. Find ratio of surface areas.

1

(r₁/r₂)³=64/27 → r₁/r₂=4/3

2

SA ratio = (r₁/r₂)² = 16/9 → 16:9

SA ratio = 16:9

Smart Study

8 Essential Tips

1

l first, then CSA

Always calculate slant height l=√(r²+h²) before attempting CSA=πrl. Most errors occur by using h instead of l.

2

Hemisphere TSA = 3πr²

Curved part = 2πr², flat base = πr², total = 3πr². For an open bowl (no base): use CSA = 2πr² only.

3

Cone:Hemisphere:Cylinder = 1:2:3

Equal radii and heights. This ratio is directly asked in CBSE. Memorise it — saves calculation time.

4

Melting: volumes stay equal

When a shape is melted and recast, V₁ = V₂. Set the volume formulas equal and solve for the unknown.

5

Use r=7 with π=22/7

Choosing r=7 or multiples gives clean answers with π=22/7. Check if the question specifies which value of π to use.

6

Combined shapes: no shared face

For cone on hemisphere, TSA = CSA(cone) + CSA(hemi). The shared circular base is internal — don't include it.

7

Volume ratio ↔ radius ratio

If V₁:V₂ = a:b, then r₁:r₂ = a^(1/3):b^(1/3). Surface areas: SA₁:SA₂ = a^(2/3):b^(2/3).

8

Sphere in cylinder: water rise

Volume of sphere = rise in water volume. πR²Δh = (4/3)πr³ sphere. Solve for Δh.

Quick Reference

Chapter 13 — All Formulas

Shape	CSA	TSA	Volume
Cone	πrl	πr(r+l)	(1/3)πr²h
Sphere	4πr²		(4/3)πr³
Hemisphere	2πr²	3πr²	(2/3)πr³

l=slant height=√(r²+h²) | Hemisphere TSA includes flat base πr² | Sphere has no separate CSA/TSA

CBSE Practice

50 Practice Questions

MCQ · 1M · 2M · 3M · 5M · Case-Based — all with solutions.

Mensuration:Surface Area & Volume