What is the difference between CSA and TSA?

CSA (Curved Surface Area) is the area of only the curved part of a 3D shape — it excludes flat surfaces like the top, bottom or base. TSA (Total Surface Area) includes everything: the curved surface plus all flat surfaces. For a cylinder, CSA $= 2\pi rh$ while TSA $= 2\pi r(h+r)$.

How do you find the slant height of a cone?

Use the Pythagoras theorem. The slant height $l$, radius $r$ and height $h$ of a cone form a right triangle, so $l = \sqrt{r^2 + h^2}$. Always calculate the slant height first if it is not directly given.

Why is the volume of a cone one-third of a cylinder?

If a cone and a cylinder have the same base radius and height, exactly three cones of water can fill the cylinder. This can be demonstrated experimentally and is proved using calculus in higher classes. The result is: $V_{\text{cone}} = \frac{1}{3} \pi r^2 h$.

What formulas should I memorise for the Class 9 exam?

You must know CSA, TSA and Volume formulas for: cuboid, cube, cylinder, cone, sphere and hemisphere. Also remember the slant height formula $l = \sqrt{r^2 + h^2}$ for cones and the relationship between cone and cylinder volumes.

How do you solve problems where a shape is melted and recast?

The key principle is that volume is conserved — it does not change when a shape is melted and recast. Calculate the volume of the original shape, set it equal to the volume of the new shape, and solve for the unknown dimension.

Does a sphere have CSA and TSA?

A sphere has only one continuous surface with no flat face, so its CSA and TSA are the same: $4\pi r^2$. A hemisphere, however, has a curved part (CSA $= 2\pi r^2$) and a flat circular base, giving TSA $= 3\pi r^2$.

What is the difference between CSA and TSA?

CSA (Curved Surface Area) is the area of only the curved part of a 3D shape — it excludes flat surfaces like the top, bottom or base. TSA (Total Surface Area) includes everything: the curved surface plus all flat surfaces. For a cylinder, CSA $= 2\pi rh$ while TSA $= 2\pi r(h+r)$.

How do you find the slant height of a cone?

Use the Pythagoras theorem. The slant height $l$, radius $r$ and height $h$ of a cone form a right triangle, so $l = \sqrt{r^2 + h^2}$. Always calculate the slant height first if it is not directly given.

Why is the volume of a cone one-third of a cylinder?

If a cone and a cylinder have the same base radius and height, exactly three cones of water can fill the cylinder. This can be demonstrated experimentally and is proved using calculus in higher classes. The result is: $V_{\text{cone}} = \frac{1}{3} \pi r^2 h$.

What formulas should I memorise for the Class 9 exam?

You must know CSA, TSA and Volume formulas for: cuboid, cube, cylinder, cone, sphere and hemisphere. Also remember the slant height formula $l = \sqrt{r^2 + h^2}$ for cones and the relationship between cone and cylinder volumes.

How do you solve problems where a shape is melted and recast?

The key principle is that volume is conserved — it does not change when a shape is melted and recast. Calculate the volume of the original shape, set it equal to the volume of the new shape, and solve for the unknown dimension.

Does a sphere have CSA and TSA?

A sphere has only one continuous surface with no flat face, so its CSA and TSA are the same: $4\pi r^2$. A hemisphere, however, has a curved part (CSA $= 2\pi r^2$) and a flat circular base, giving TSA $= 3\pi r^2$.

Study Guide

Surface Areas & Volumes Class 9: All Formulas with Solved Examples

Your one-stop formula sheet and problem-solving guide for NCERT Chapter 13 — Surface Areas and Volumes.

CBSEClass 9

The SparkEd Authors (IITian & Googler)15 March 202614 min read

Why Surface Area & Volume Matters

$Surface Area and Volume is one of the most formula-heavy chapters in Class 9 CBSE Math. It covers 3D shapes that appear everywhere in real life — from water tanks (cylinders) to ice-cream cones to footballs (spheres).$

$The chapter comes from NCERT Chapter 13 and carries significant weight in exams. The good news? Once you memorise the formulas and practise applying them, the problems become very systematic.$

$Let's break down every formula, shape by shape, with solved examples.$

Cuboid: CSA, TSA & Volume

$l$

$Lateral / Curved Surface Area (CSA):$

\text{CSA} = 2h(l + b)

This is the area of the four walls (excludes top and bottom).

$Total Surface Area (TSA):$

\text{TSA} = 2(lb + bh + hl)

$Volume:$

V = l \times b \times h

$Diagonal of cuboid:$

d = \sqrt{l^2 + b^2 + h^2}

$8 \text{ m}$

$= \text{CSA} = 2h(l + b) = 2 \times 3 \times (8 + 6) = 6 \times 14 = 84 \text{ m}^2$

$= 84 \times 25 = \text{Rs } 2100$

Special Case: Cube

$l = b = h = a$

\text{CSA of cube} = 4a^2

\text{TSA of cube} = 6a^2

V = a^3

\text{Diagonal} = a\sqrt{3}

Right Circular Cylinder: CSA, TSA & Volume

$r$

$Curved Surface Area (CSA):$

\text{CSA} = 2\pi r h

Imagine unrolling the curved surface — you get a rectangle of width

2\pi r

and height

h

$Total Surface Area (TSA):$

\text{TSA} = 2\pi r h + 2\pi r^2 = 2\pi r(h + r)

(CSA + area of two circular bases)

$Volume:$

V = \pi r^2 h

$20 \text{ cm}$

$r = 20 \text{ cm} = 0.2 \text{ m}$

\text{CSA} = 2\pi r h = 2 \times \frac{22}{7} \times 0.2 \times 3.5 = 2 \times \frac{22}{7} \times 0.7 = 2 \times 2.2 = 4.4 \text{ m}^2

$= 4.4 \times 12.50 = \text{Rs } 55$

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Right Circular Cone: CSA, TSA & Volume

$r$

l = \sqrt{r^2 + h^2}

$Curved Surface Area (CSA):$

\text{CSA} = \pi r l

$Total Surface Area (TSA):$

\text{TSA} = \pi r l + \pi r^2 = \pi r(l + r)

$Volume:$

V = \frac{1}{3} \pi r^2 h

$\frac{1}{3}$

$7 \text{ cm}$

$Solution: First, find slant height:$

l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \text{ cm}

V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = \frac{1}{3} \times 22 \times 7 \times 24 = \frac{3696}{3} = 1232 \text{ cm}^3

\text{TSA} = \pi r(l + r) = \frac{22}{7} \times 7 \times (25 + 7) = 22 \times 32 = 704 \text{ cm}^2

Sphere: Surface Area & Volume

$r$

$Surface Area (only one surface — no separate CSA/TSA):$

\text{Surface Area} = 4\pi r^2

$Volume:$

V = \frac{4}{3} \pi r^3

$10.5 \text{ cm}$

$Solution:$

\text{Surface Area} = 4\pi r^2 = 4 \times \frac{22}{7} \times (10.5)^2 = 4 \times \frac{22}{7} \times 110.25 = 4 \times 346.5 = 1386 \text{ cm}^2

V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (10.5)^3 = \frac{4}{3} \times \frac{22}{7} \times 1157.625 = 4851 \text{ cm}^3

Hemisphere: CSA, TSA & Volume

$r$

$Curved Surface Area (CSA):$

\text{CSA} = 2\pi r^2

(Half the surface area of a full sphere.)

$Total Surface Area (TSA):$

\text{TSA} = 2\pi r^2 + \pi r^2 = 3\pi r^2

(CSA + flat circular base)

$Volume:$

V = \frac{2}{3} \pi r^3

$7 \text{ cm}$

$= \text{CSA} = 2\pi r^2 = 2 \times \frac{22}{7} \times 49 = 308 \text{ cm}^2$

$= 308 \times 2 = \text{Rs } 616$

Master Formula Table

$Keep this table for quick revision.$

$Shape$	$CSA$	$TSA$	$Volume$
$Cuboid$	$2h(l+b)$	$2(lb+bh+hl)$	$lbh$
$Cube$	$4a^2$	$6a^2$	$a^3$
$Cylinder$	$2\pi rh$	$2\pi r(h+r)$	$\pi r^2 h$
$Cone$	$\pi rl$	$\pi r(l+r)$	$\frac{1}{3}\pi r^2 h$
$Sphere$	$4\pi r^2$	$4\pi r^2$	$\frac{4}{3}\pi r^3$
$Hemisphere$	$2\pi r^2$	$3\pi r^2$	$\frac{2}{3}\pi r^3$

$= \frac{1}{3} \times$

More Solved Examples

$3 \text{ cm}$

$= \frac{4}{3}\pi (3)^3 = \frac{4}{3}\pi \times 27 = 36\pi \text{ cm}^3$

$= 0.2 \text{ cm}$

$Since volumes are equal:$

36\pi = 0.04\pi h

h = \frac{36}{0.04} = 900 \text{ cm} = 9 \text{ m}

Example 2: Combination of Shapes

$3.5 \text{ cm}$

$=$

$Slant height of cone:$

l = \sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + (12)^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 \text{ cm}

$=$

= \pi r l + 2\pi r^2

= \frac{22}{7} \times 3.5 \times 12.5 + 2 \times \frac{22}{7} \times (3.5)^2

= 11 \times 12.5 + 2 \times 11 \times 3.5

= 137.5 + 77 = 214.5 \text{ cm}^2

Example 3: Water in a Cylindrical Vessel

$6 \text{ cm}$

$= \frac{4}{3}\pi (3)^3 = 36\pi \text{ cm}^3$

$= h$

36\pi h = 36\pi

h = 1 \text{ cm}

$1 \text{ cm}$

Exam Strategy & Common Mistakes

$Mistake 1: Confusing CSA and TSA. Read the question carefully. If it says 'curved surface' or 'lateral surface', use CSA. If it says 'total surface area' or involves covering the entire shape, use TSA.$

$Mistake 2: Forgetting unit conversions. When radius is in cm and height in m (or vice versa), convert to the same unit before applying formulas.$

$Mistake 3: Using diameter instead of radius. Many problems give the diameter. Always halve it to get the radius before substituting.$

$l = \sqrt{r^2 + h^2}$

$Strategy: In 'melting and recasting' problems, the key principle is: volume remains constant. Equate the volume of the original shape to the volume of the new shape.$

$\pi = \frac{22}{7}$

Summary & Practice Resources

$Surface Areas and Volumes is a high-scoring chapter because the problems are formulaic — learn the formula, identify the shape, plug in the values, and solve. The challenge lies in: 1. Combination problems (cone on hemisphere, etc.) 2. Conversion problems (melting and recasting) 3. Unit conversions$

$Make sure you can write every formula from memory. Test yourself: cover the formula column in the table above and try to recall each one.$

$For interactive practice with instant feedback, head to the SparkEd Surface Areas and Volumes practice page . You can also use the SparkEd Math Solver to check your calculations step by step.$

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