MathsCBSE Class 10 · Surface Areas and Volumes

Surface Area and Volume Formulae Revision for Class 10

Updated 13 October 2026 · Fact-checked

This topic revises the curved surface area, total surface area and volume of cubes, cuboids, cylinders, cones, spheres and hemispheres. Every combination problem in the chapter is built from these formulas. The board tests them by hiding one measurement, such as the radius, height or slant height, so you must find it before you substitute.

Slant height l is the hypotenuse of the right triangle with r and h: curved surface area uses l, volume uses h, and a cone is one-third of its cylinder.

Illustration drawn for QuestionBuddy.

Understand Surface Area and Volume Formulae Revision

Every question in this chapter, however complicated it looks, is built from six basic solids: the cube, cuboid, cylinder, cone, sphere and hemisphere. If you know their formulas well and understand why each one works, combination problems become a matter of adding and subtracting pieces.

There are two kinds of measurement. Surface area is the amount of covering on the outside, in square units such as cm². Volume is the space inside, in cubic units such as cm³. A curved surface area (CSA) or lateral surface area leaves out the flat ends. A total surface area (TSA) includes every face. You need both, because in a combination some faces get hidden.

The formulas follow from simple ideas. A cuboid has six rectangles in three equal pairs, so its TSA is $2 (l b + bh + h l)$ . The side of a cylinder unrolls into a rectangle of length $2 π r$ and width $h$ , so its CSA is $2 π r h$ . Any prism-like solid has volume equal to base area times height, which gives $l bh$ and $π r^{2} h$ . A cone holds one-third of the matching cylinder, so its volume is $\frac{1}{3} π r^{2} h$ .

The cone needs extra care. Its height $h$ , radius $r$ and slant height $l$ form a right triangle, so $l = r^{2} + h^{2}$ . The CSA $π r l$ uses $l$ , while the volume uses $h$ . Mixing these up is one of the commonest ways to lose marks.

A sphere has surface area $4 π r^{2}$ and volume $\frac{4}{3} π r^{3}$ . A hemisphere is half a sphere. Its curved surface is $2 π r^{2}$ , and a solid hemisphere also has a flat circle, giving a TSA of $3 π r^{2}$ . Its volume is $\frac{2}{3} π r^{3}$ .

These formulas hold for right circular cylinders and cones, with all lengths in the same unit. Board questions rarely let you plug in directly. They give you a CSA or a volume and ask you to work backwards to $r$ or $h$ . They may also give a diameter instead of a radius, or ask for an answer in litres. Take $π = \frac{22}{7}$ unless told otherwise. Choose the form of $r$ , such as $\frac{21}{2}$ , that cancels neatly with 7.

Surface Area and Volume Formulae Revision: formulas to remember

Cuboid: total surface area: $2 (l b + bh + h l)$; All six faces of a closed box.
Cuboid: lateral surface area: $2 h (l + b)$; Four walls only, without the top and bottom.
Cuboid: volume: $V = l bh$; Divide cm³ by 1000 to get litres.
Cuboid: diagonal: $l^{2} + b^{2} + h^{2}$; The longest rod that fits inside a box.
Cube: TSA, LSA, volume: $6 a^{2}, 4 a^{2}, a^{3}$; $a$ is the edge.
Cylinder: curved surface area: $2 π r h$; The unrolled side wall.
Cylinder: total surface area: $2 π r (r + h)$; CSA plus two circular ends.
Cylinder: volume: $π r^{2} h$; Base area times height.
Cone: slant height: $l = r^{2} + h^{2}$; Pythagoras. Find this before any cone surface area.
Cone: curved surface area: $π r l$; Uses the slant height, never the vertical height.
Cone: total surface area: $π r (l + r)$; CSA plus the circular base.
Cone: volume: $\frac{1}{3} π r^{2} h$; Uses the vertical height $h$ .
Sphere: surface area: $4 π r^{2}$; No flat faces.
Sphere: volume: $\frac{4}{3} π r^{3}$; Doubling $r$ multiplies the volume by 8.
Hemisphere: curved surface area: $2 π r^{2}$; The inside of a bowl, or a dome joined to another solid.
Solid hemisphere: total surface area: $3 π r^{2}$; Curved $2 π r^{2}$ plus flat circle $π r^{2}$ .
Hemisphere: volume: $\frac{2}{3} π r^{3}$; Half the sphere.
Cone, hemisphere, cylinder (same r, height r): $V_{co n e} : V_{h e mi} : V_{cy l} = 1 : 2 : 3$; A quick MCQ result.
Units: $1 L = 1000 cm^{3}, 1 m^{3} = 1000 L$; Areas in square units, volumes in cubic units.

How to solve Surface Area and Volume Formulae Revision questions

Use this method on any question that revises a single solid, including questions that ask you to work backwards to a missing measurement.

1Name the solid and write down every measurement given. Convert a diameter to a radius straight away.
2Decide what is asked: CSA, TSA or volume. Check the words 'curved', 'total', 'open', 'capacity'.
3Write the formula in symbols before you put in any numbers.
4If a measurement is missing, set up an equation from the given CSA or volume and solve for $r$ or $h$ . For a cone, find $l = r^{2} + h^{2}$ separately.
5Substitute, cancel the 7 in $\frac{22}{7}$ early, and simplify.
6Write the final answer with the correct unit (cm², cm³ or litres), and check it with a quick reverse calculation.

Surface Area and Volume Formulae Revision in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Cylinder: curved surface area, total surface area and volume

Suppose your mother stores rice in a closed cylindrical steel dabba. Its capacity is 1.54 litres and its height is 10 cm. You want to stick a paper label all around its side, and you also want to know how much steel sheet a dabba like this needs.

Find (i) the radius of the dabba, (ii) the cost of a label that covers its curved side exactly, at Rs 0.50 per 100 cm², and (iii) the cost of the steel sheet needed for the closed dabba at Rs 5 per 100 cm². Take the inner and outer measurements as the same and ignore wastage.

Try it first, then show the solutionSolution

Capacity $= 1.54$ L $= 1.54 \times 1000 = 1540$ cm³
$π r^{2} h = 1540$ , so $\frac{22}{7} \times r^{2} \times 10 = 1540$
$r^{2} = \frac{1540 \times 7}{220} = 49$ , so $r = 7$ cm
Label area $= 2 π r h = 2 \times \frac{22}{7} \times 7 \times 10 = 440$ cm²
Label cost $= \frac{440}{100} \times 0.50 =$ Rs 2.20
Sheet area $= 2 π r (r + h) = 2 \times \frac{22}{7} \times 7 \times 17 = 44 \times 17 = 748$ cm²
Sheet cost $= \frac{748}{100} \times 5 =$ Rs 37.40

Answer: (i) 7 cm, (ii) Rs 2.20 for the label, (iii) Rs 37.40 for the steel sheet

Memory hook: Peel the label off the dabba and it is a rectangle: $2 π r$ long and $h$ wide.

Cone: slant height and curved surface area

Imagine you are making 30 conical paper caps for a birthday party. Each cap has base radius 7 cm and vertical height 24 cm. Coloured chart paper costs Rs 20 per m².

Find the paper needed for all 30 caps (caps are open at the base, ignore overlaps) in m², and the cost of the paper.

Try it first, then show the solutionSolution

Slant height $l = 7^{2} + 2 4^{2} = 625 = 25$ cm
CSA of one cap $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm²
For 30 caps: $30 \times 550 = 16500$ cm²
$16500 \div 10000 = 1.65$ m²
Cost $= 1.65 \times 20 =$ Rs 33

Answer: 1.65 m² of paper, costing Rs 33

Memory hook: A party cap has no base, so you only need the slant side: $π r l$ .

Tricks to remember Surface Area and Volume Formulae Revision

Short memory aids. Say each one aloud once, then use it on a question the same day.

Slant for skin, height for space
For a cone, the outside surface uses the slant height $l$ and the volume uses the vertical height $h$ .
Use it when: Before writing $π r l$ or $\frac{1}{3} π r^{2} h$ , check which height you need.
The 2-3-4 of round solids
Hemisphere curved surface $2 π r^{2}$ , solid hemisphere $3 π r^{2}$ , sphere $4 π r^{2}$ .
Use it when: Recall all three sphere-family areas in one breath.
One-third rule
A cone is one-third of a cylinder. A hemisphere with height $r$ is two-thirds of a cylinder. Cone : hemisphere : cylinder $= 1 : 2 : 3$ .
Use it when: Quick ratio MCQs and checking your volumes.
Volume over CSA
For a cylinder, $\frac{V}{CSA} = \frac{π r ^{2} h}{2 π r h} = \frac{r}{2}$ .
Use it when: When both the volume and the CSA are given, this finds $r$ in one line.

Surface Area and Volume Formulae Revision: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Cone CSA from its volume · 1 mark · MCQBoard level

A right circular cone has base radius 7 cm and volume 1232 cm³. Its curved surface area is
(a) 528 cm²
(b) 550 cm²
(c) 616 cm²
(d) 704 cm²

Try it first, then show the solutionSolution

$\frac{1}{3} \times \frac{22}{7} \times 49 \times h = 1232$ , so $\frac{154}{3} h = 1232$ and $h = 24$ cm
$l = 49 + 576 = 25$ cm
CSA $= \frac{22}{7} \times 7 \times 25 = 550$ cm²

Answer: (b) 550 cm²

Example 2: Assertion-reason: which height goes in the cone's CSA · 1 mark · Assertion-reasonBoard level

Assertion (A): The curved surface area of a cone of base radius 3 cm and vertical height 4 cm is $15 π$ cm².
Reason (R): The curved surface area of a cone of radius $r$ and vertical height $h$ is $π r h$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$l = 3^{2} + 4^{2} = 5$ cm, so CSA $= π \times 3 \times 5 = 15 π$ cm². A is true.
The CSA uses the slant height: $π r l$ , not $π r h$ . R is false (it would give $12 π$ ).

Answer: (c) A is true but R is false.

Example 3: Cube from its capacity in litres · 2 marks · Very short answerBoard level

A closed cubical container holds exactly 3.375 litres when full. Find the length of its edge and its total surface area.

Try it first, then show the solutionSolution

$3.375$ L $= 3.375 \times 1000 = 3375$ cm³ (½ mark)
$a^{3} = 3375 = 1 5^{3}$ , so $a = 15$ cm (½ mark)
TSA $= 6 a^{2} = 6 \times 225 = 1350$ cm² (1 mark)

Answer: The edge is 15 cm and the total surface area is 1350 cm².

Example 4: Sphere from its surface area · 3 marks · Short answerHarder board level

The surface area of a solid sphere is 1386 cm². Find its volume. Also find the total surface area of a solid hemisphere with the same radius. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$4 \times \frac{22}{7} \times r^{2} = 1386$ , so $r^{2} = \frac{1386 \times 7}{88} = 110.25$ (½ mark)
$r = 10.5 = \frac{21}{2}$ cm (½ mark)
$V = \frac{4}{3} \times \frac{22}{7} \times \frac{9261}{8} = 11 \times 441 = 4851$ cm³ (1 mark)
Hemisphere TSA $= 3 π r^{2} = \frac{3}{4} \times 1386 = 1039.5$ cm² (1 mark)

Answer: The volume is 4851 cm³ and the hemisphere's TSA is 1039.5 cm².

Example 5: Cylinder from its CSA and volume · 3 marks · Short answerHarder board level

A closed right circular cylinder has curved surface area 880 cm² and volume 6160 cm³. Find its radius, its height and its total surface area. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$\frac{V}{CSA} = \frac{π r ^{2} h}{2 π r h} = \frac{r}{2}$ (½ mark)
$\frac{r}{2} = \frac{6160}{880} = 7$ , so $r = 14$ cm (½ mark)
$2 \times \frac{22}{7} \times 14 \times h = 880$ , so $88 h = 880$ and $h = 10$ cm (1 mark)
TSA $= 2 π r (r + h) = 88 \times 24 = 2112$ cm² (1 mark)

Answer: r = 14 cm, h = 10 cm, TSA = 2112 cm²

Example 6: Cuboidal room: height, surface area and diagonal · 3 marks · Short answerHarder board level

A cuboidal room is 12 m long and 4 m wide. The cost of painting its four walls at Rs 20 per m² is Rs 1920. Find (i) the height of the room, (ii) the total inner surface area of the room (walls, floor and ceiling), and (iii) the length of the longest straight pole that can be placed inside the room.

Try it first, then show the solutionSolution

Area of four walls (LSA) $= \frac{1920}{20} = 96$ m² (½ mark)
$2 h (l + b) = 96$ , so $2 h (12 + 4) = 32 h = 96$ and $h = 3$ m (1 mark)
TSA $= 2 (l b + bh + h l) = 2 (48 + 12 + 36) = 192$ m² (½ mark)
Longest pole $=$ diagonal $= 1 2^{2} + 4^{2} + 3^{2} = 169 = 13$ m (1 mark)

Answer: (i) 3 m, (ii) 192 m², (iii) 13 m

Example 7: Comparing a cone, a hemisphere and a cylinder · 5 marks · Long answerBoard level

A solid cone and a solid cylinder each have base radius 21 cm and height 28 cm. A solid hemisphere has radius 21 cm. Find (i) the slant height of the cone, (ii) the curved surface area of each solid, (iii) the volume of each solid and the ratio of the volumes of the cone, hemisphere and cylinder. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

(i) $l = 2 1^{2} + 2 8^{2} = 1225 = 35$ cm (1 mark)
(ii) Cone CSA $= \frac{22}{7} \times 21 \times 35 = 2310$ cm² (½ mark)
Cylinder CSA $= 2 \times \frac{22}{7} \times 21 \times 28 = 3696$ cm² (½ mark)
Hemisphere CSA $= 2 \times \frac{22}{7} \times 441 = 2772$ cm² (½ mark)
(iii) Cylinder $V = \frac{22}{7} \times 441 \times 28 = 38808$ cm³ (½ mark)
Cone $V = \frac{1}{3} \times 38808 = 12936$ cm³ (½ mark)
Hemisphere $V = \frac{2}{3} \times \frac{22}{7} \times 9261 = 19404$ cm³ (½ mark)
Ratio $12936 : 19404 : 38808$ ; dividing by 6468 gives $2 : 3 : 6$ (1 mark)

Answer: l = 35 cm; the CSAs are 2310 cm², 2772 cm² and 3696 cm² (cone, hemisphere, cylinder); the volumes are 12936 cm³, 19404 cm³ and 38808 cm³, in the ratio 2 : 3 : 6.

Example 8: Cone and hemisphere with equal volumes · 5 marks · Long answerHarder board level

A solid cone and a solid hemisphere have the same base radius and the same volume. The slant height of the cone is $215$ cm. Find (i) the radius and the vertical height of the cone, (ii) the ratio of the curved surface area of the cone to that of the hemisphere, and (iii) the volume of each solid. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Equal volumes: $\frac{1}{3} π r^{2} h = \frac{2}{3} π r^{3}$ , so $h = 2 r$ (1 mark)
$l^{2} = r^{2} + h^{2} = r^{2} + 4 r^{2} = 5 r^{2}$ , so $l = r 5$ (½ mark)
$r 5 = 215$ , so $r = 21$ cm and $h = 42$ cm (½ mark)
Cone CSA $= π r l = \frac{22}{7} \times 21 \times 215 = 13865$ cm² (½ mark)
Hemisphere CSA $= 2 π r^{2} = 2 \times \frac{22}{7} \times 441 = 2772$ cm² (½ mark)
Ratio $= \frac{1386 5}{2772} = \frac{5}{2}$ , that is $5 : 2$ (1 mark)
Volume of each $= \frac{2}{3} \times \frac{22}{7} \times 9261 = 19404$ cm³; check: cone $= \frac{1}{3} \times \frac{22}{7} \times 441 \times 42 = 19404$ cm³ (1 mark)

Answer: (i) r = 21 cm, h = 42 cm; (ii) √5 : 2; (iii) each volume is 19404 cm³.

Example 9: Case study: packing a cylinder in a cube · 4 marks · Case-basedHOTS / competency

Suppose a craft shop carves the largest possible solid cylinder out of a wooden cube of edge 14 cm. The cylinder stands on one face of the cube. Take $π = \frac{22}{7}$ .
(i) Write the radius and height of the cylinder. (1 mark)
(ii) Find the volume of the cylinder. (1 mark)
(iii) Show that the ratio of the cylinder's TSA to the cube's TSA equals the ratio of their volumes, and find this ratio. (2 marks)

Try it first, then show the solutionSolution

(i) The diameter equals the edge, so $r = 7$ cm, and the height is $h = 14$ cm (1 mark)
(ii) $V = \frac{22}{7} \times 49 \times 14 = 2156$ cm³ (1 mark)
(iii) Cylinder TSA $= 2 π r (r + h) = 44 \times 21 = 924$ cm²; cube TSA $= 6 \times 196 = 1176$ cm² (½ mark)
TSA ratio $= \frac{924}{1176} = \frac{11}{14}$ (½ mark)
Volume ratio $= \frac{2156}{2744} = \frac{11}{14}$ (½ mark)
The two ratios are equal, both $11 : 14$ , which is $π : 4$ (½ mark)

Answer: (i) r = 7 cm, h = 14 cm; (ii) 2156 cm³; (iii) both ratios are 11 : 14.

Where students lose marks in Surface Area and Volume Formulae Revision

The slips examiners see most often, and what to write instead.

Mistake: Using the vertical height in the cone's CSA
$h$ and $l$ both look like 'height'.
Wrong working
$r = 7$ , $h = 24$ : CSA $= \frac{22}{7} \times 7 \times 24 = 528$ cm²
Right working
$l = 25$ , so CSA $= \frac{22}{7} \times 7 \times 25 = 550$ cm²
Do this: Always write $l = r^{2} + h^{2}$ on its own line first.
Usually costs: Usually all the calculation marks for that part.
Mistake: Confusing hemisphere CSA with TSA
Students forget the flat circle, or add it when it is not exposed.
Wrong working
Solid hemisphere, $r = 7$ : TSA $= 2 π r^{2} = 308$ cm²
Right working
TSA $= 3 π r^{2} = 462$ cm²
Do this: Ask whether the flat face is visible. If it is, use $3 π r^{2}$ . If not, use $2 π r^{2}$ .
Usually costs: 1 to 2 marks.
Mistake: Using the diameter as the radius
The question gives the diameter, and you copy it straight in.
Wrong working
Diameter 14 cm: $V = \frac{22}{7} \times 1 4^{2} \times 10 = 6160$ cm³
Right working
$r = 7$ : $V = \frac{22}{7} \times 49 \times 10 = 1540$ cm³
Do this: Underline the word 'diameter' and halve it before anything else.
Usually costs: Final answer marks, and errors carried into later parts.
Mistake: Wrong litre conversion
Dividing by 100, or converting m³ wrongly.
Wrong working
$4312$ cm³ $= 43.12$ litres
Right working
$4312 \div 1000 = 4.312$ litres
Do this: Remember 1 L $= 1000$ cm³ and 1 m³ $= 1000$ L.
Usually costs: ½ to 1 mark.

Exam tips

Write the formula in symbols before you substitute. The method mark is often given for this line.
Expect working backwards: the board often gives a CSA or a volume and asks for $r$ or $h$ first.
Keep radii as fractions, such as $\frac{21}{2}$ , so the 7 in $\frac{22}{7}$ cancels.
State units on every final line: cm² for area, cm³ or litres for volume.
In MCQs, check ratios such as $1 : 2 : 3$ and the scaling rules: doubling $r$ multiplies areas by 4 and volumes by 8.

Practice questions on Surface Area and Volume Formulae Revision

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
The volume of a solid hemisphere is 19404 cm³. Find its curved surface area. Take $π = \frac{22}{7}$ .
Hint
Solve $\frac{2}{3} π r^{3} = 19404$ for $r$ first.
Answer
$r^{3} = \frac{19404 \times 21}{44} = 9261$ , so $r = 21$ cm. CSA $= 2 π r^{2} = 2772$ cm².
Q2 · 3 marks · Short answerHarder board level
The radius and height of a solid cone are in the ratio $3 : 4$ , and its slant height is 35 cm. Find its volume and total surface area. Take $π = \frac{22}{7}$ .
Hint
Let $r = 3 k$ and $h = 4 k$ . Then $l = 5 k$ .
Answer
$5 k = 35$ , so $k = 7$ , $r = 21$ cm and $h = 28$ cm. $V = \frac{1}{3} \times \frac{22}{7} \times 441 \times 28 = 12936$ cm³. TSA $= π r (l + r) = \frac{22}{7} \times 21 \times 56 = 3696$ cm².
Q3 · 3 marks · Short answerHarder board level
The length, breadth and height of a closed cuboidal box are in the ratio $4 : 3 : 2$ . Its total surface area is 1300 cm². Find its dimensions and its capacity in litres.
Hint
Write the TSA in terms of $x$ : $2 (12 + 6 + 8) x^{2}$ .
Answer
$52 x^{2} = 1300$ , so $x = 5$ . The dimensions are 20 cm, 15 cm and 10 cm. $V = 3000$ cm³ $= 3$ litres.
Q4 · 5 marks · Long answerBoard level
A closed cylinder has total surface area 1848 cm², and the sum of its radius and height is 21 cm. Find (i) the radius, (ii) the height, (iii) the curved surface area, and (iv) the capacity in litres. Take $π = \frac{22}{7}$ .
Hint
TSA $= 2 π r (r + h)$ , and you already know $r + h$ .
Answer
$2 \times \frac{22}{7} \times r \times 21 = 132 r = 1848$ , so $r = 14$ cm and $h = 7$ cm. CSA $= 2 \times \frac{22}{7} \times 14 \times 7 = 616$ cm². $V = \frac{22}{7} \times 196 \times 7 = 4312$ cm³ $= 4.312$ litres.
Q5 · 3 marks · Short answerHOTS / competency
The radius of a right circular cylinder is doubled and its height is halved. Find the percentage change in (i) its volume and (ii) its curved surface area. Explain why the two answers are different.
Hint
Write the new volume as $π (2 r)^{2} (\frac{h}{2})$ and compare it with $π r^{2} h$ . Do the same for $2 π r h$ .
Answer
New volume $= π (2 r)^{2} \times \frac{h}{2} = 2 π r^{2} h$ , which is twice the old volume: a 100% increase. New CSA $= 2 π (2 r) \times \frac{h}{2} = 2 π r h$ : no change (0%). The volume depends on $r^{2}$ , so doubling $r$ multiplies it by 4 before halving $h$ brings it to 2. The CSA depends on $r$ only once, so doubling $r$ and halving $h$ cancel exactly.
Q6 · 4 marks · Case-basedHOTS / competency
Suppose a housing society installs a closed cylindrical water tank on a flat platform. The tank has radius 1.4 m and height 2 m. Its base rests on the platform, and the rest of the outer surface is to be painted. Take $π = \frac{22}{7}$ .
(i) Find the capacity of the tank in litres. (1 mark)
(ii) Find the curved surface area of the tank. (1 mark)
(iii) Find the area to be painted (curved surface and top only) and the cost of painting it at Rs 50 per m². (2 marks)
Hint
The base touches the platform, so it is not painted. Painted area $=$ CSA $+ π r^{2}$ .
Answer
(i) $V = \frac{22}{7} \times 1.96 \times 2 = 12.32$ m³ $= 12320$ litres. (ii) CSA $= 2 \times \frac{22}{7} \times 1.4 \times 2 = 17.6$ m². (iii) Top $= \frac{22}{7} \times 1.96 = 6.16$ m², so painted area $= 17.6 + 6.16 = 23.76$ m². Cost $= 23.76 \times 50 =$ Rs 1188.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Cone slant height?Show answer
$l = r^{2} + h^{2}$
2.Hemisphere CSA and TSA?Show answer
$2 π r^{2}$ and $3 π r^{2}$
3.Sphere volume?Show answer
$\frac{4}{3} π r^{3}$
4.Cylinder TSA?Show answer
$2 π r (r + h)$
5.Cuboid TSA?Show answer
$2 (l b + bh + h l)$
6.1 litre in cm³?Show answer
1000 cm³

Surface Area and Volume Formulae Revision: frequently asked questions

Do I need to memorise the Class 9 formulas for this chapter?

Yes. Every combination question uses the single-solid formulas. The board does not give them to you in the paper.

Should I take pi as 22/7 or 3.14?

Use 22/7 unless the question says otherwise. Questions usually choose radii like 7 or 10.5 so that 22/7 cancels neatly.

Is the frustum of a cone in the 2026-27 syllabus?

No. The frustum of a cone is not in the 2026-27 syllabus.

What is the difference between curved and total surface area?

Curved surface area covers only the curved or side surface. Total surface area adds every flat face as well, such as the circular ends of a cylinder.