Class 10 Maths Notes

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Surface Areas and Volumes Class 10 Maths Notes

Six solids, ten formulas, one distinction that decides every mark: curved surface area against total surface area. This page sets out CSA, LSA and volume for the cube, cuboid, sphere, hemisphere, cylinder and cone, explains where the board asks for each, and works a composite solid and a solid-metal problem with π = 22/7.

Class:10Subject:MathematicsUnit:VICovers:CBSE 2024-25

11 Key Formulas

DWritten byDeep Narayan

UpdatedOctober 2026

Key Concept Summary

What is the difference between CSA and TSA?

Curved surface area is the part of the surface you can hold in your palm, excluding the flat circular faces. Total surface area is the curved surface plus those flat circular faces. So for a cylinder, CSA = 2πrh and TSA = 2πrh + 2πr²; for a cone, CSA = πrl and TSA = πrl + πr². Decide which one the question wants before you substitute.

The 2024-25 Scope: Five Solids, Two at a Time

This chapter is Unit VI Mensuration, 10 of the 80 theory marks. The retained content is the surface areas and volumes of combinations of any two of cubes, cuboids, spheres, hemispheres and right circular cylinders and cones. Six solids appear, because the cylinder and the cone each count separately, and a question may combine at most two of them.Six solids means twelve surfaces to keep straight. Nothing here needs proof. The marks are lost on choosing CSA when TSA was asked, on forgetting one base circle, and on the combinations rule.

Cube, cuboid — 3D boxes with flat faces, no curved surface at all.
Sphere, hemisphere — no flat face on the curved part, so only surface area is asked.
Right circular cylinder — two flat circular bases plus one curved surface.
Right circular cone — one flat circular base plus one curved surface, and it needs a slant height.
Combinations are of any TWO of these solids. Nothing is ever combined with nothing, and never three at a time.

Out of scope for the 2024-25 board paper

Three things have been deleted and will not be set. The frustum of a cone is gone, so do not learn its formula. Problems that convert one type of metallic solid into another are gone, so there is no melted-solid question and no mass or density step. And any combination of more than two solids is gone, so take at most two. If a question seems to want three, you have misread it.

CSA and TSA: Which One Is Asked

Definition

Curved surface area (CSA)

The curved surface area of a solid is the area of the part of its surface that is curved, taken alone. For a cylinder it is the side of the tube, for a cone it is the sloping face, for a hemisphere it is the dome. The flat circular faces are excluded.

Definition

Lateral surface area (LSA)

The lateral surface area is the same idea for a solid made of flat faces. For a cuboid it is the four vertical faces only, with the top and bottom left out. For a cube it is the same thing, four equal squares.

Definition

Total surface area (TSA)

The total surface area of a solid is the curved or lateral surface together with every flat face that is exposed. So TSA = CSA + the area of the exposed circular bases, and TSA = LSA + 2 × (base area) for a cuboid.

Read the verb. Area of curved surface, curved surface area or area of the curved surface means CSA only.
Total surface area or TSA means add the flat faces. For a closed cylinder or cone that is one circle of area πr².
Lateral surface area of a cuboid means the four side faces only, so TSA = LSA + 2lb.
In a combined solid, a face that is glued to another solid is not exposed, so it is counted nowhere.
For a sphere and a hemisphere the board asks only the surface area, so never look for a volume question on them.

Cube and Cuboid

A cube has all six faces square and equal, so one measurement, the edge a, fixes everything. A cuboid has a length, a breadth and a height, and the three cross-products of those three numbers are exactly the three terms you need.

Cube with edge a

Cube: T S A = 6 a^{2}, L S A = 4 a^{2}, V = a^{3}

Cuboid with length l, breadth b, height h

Cuboid: T S A = 2 (l b + bh + h l), L S A = 2 h (l + b), V = l bh

a — the edge of the cube. All twelve edges are equal to a.
l, b, h — the length, breadth and height of the cuboid. lb is a face, bh is a face, hl is a face.
TSA of a cuboid counts every face, so each of the three face sizes appears twice — hence the factor 2.
LSA of a cuboid is the four faces that are not the top and the bottom, that is l and b on each of two opposite sides.
No π appears anywhere in this section. If a π is offered as an option, the solid is not a box.

The most common cuboid slip

Write 2(lb + bh + hl) in that order, not 2(lb + bh + hl) with bh and hl swapped by accident — the total is unchanged, so the error is invisible. More costly is using LSA when the question said total surface area, which loses the 2lb. Circle the word total before you start.

Sphere and Hemisphere

A sphere has no flat face and no edge, so its surface is entirely curved and its surface area is the whole story. A hemisphere is half a sphere with one flat circular face, so it has a curved part and a base, and that is why it has two surface areas instead of one.For both of these the 2024-25 board asks only the surface area. Learn the volume formulas as reading matter if you like, but do not expect to be tested on them.

Surface area of a sphere of radius r

Sphere: Surface area = 4 π r^{2}

Curved and total surface area of a hemisphere of radius r

Hemisphere: C S A = 2 π r^{2}, T S A = 3 π r^{2}

r — the radius of the sphere, and also the radius of the hemisphere's circular base.
Sphere surface area = 4πr². There is no CSA or TSA distinction because there is no flat face.
Hemisphere CSA = 2πr², which is exactly half of the sphere's surface area, as it must be.
Hemisphere TSA = 3πr², that is 2πr² of dome plus πr² of the flat circular base.
Ask yourself whether the hemisphere is being described as a dome or as a solid bowl. A closed solid bowl has TSA; an open one has CSA only.

A tank or a dome in the sky

A spherical gas tank is a sphere, so surface area = 4πr². A hemispherical bowl turned downwards with a lid is a closed solid hemisphere, so TSA = 3πr². Read the sentence, not the shape name, because the same dome can be either.

Cylinder and Cone

These two are the pair that generate most of the marks, because both are cylinders in disguise — a cone is a cylinder whose top has been squeezed to a point. Both have one or two flat circular bases, so the difference between CSA and TSA is always one circle of area πr².A right circular cone also needs a slant height l, which is the distance from the apex to the rim of the base measured along the face. It is a different measurement from the height h, and the board almost always gives it directly.

Right circular cylinder of radius r and height h

Cylinder: C S A = 2 π r h, T S A = 2 π r (r + h), V = π r^{2} h

Right circular cone of radius r, height h, slant height l

Cone: C S A = π r l, T S A = π r (l + r), V = \frac{1}{3} π r^{2} h

r — the radius of the circular base, and the same radius all the way up the cylinder.
h — the perpendicular height, base to top for a cylinder and base to apex for a cone.
l — the slant height of the cone, from apex to rim along the curved face. It is not on the diagram as a vertical.
Cylinder TSA = CSA + 2πr² because it has two bases; cone TSA = CSA + πr² because it has one.
The cone's volume is one third of the cylinder's on the same r and h, and that is the only difference between them.

If only the slant height is given

The syllabus does not give you a formula for l, but Pythagoras from Class 9 does. The height, the radius and the slant height form a right triangle, so h = √(l² − r²). Whenever a cone question gives l and r but not h, draw that right triangle and find h this way before you compute the volume.

Combinations of at Most Two Solids

When two solids are joined, the shared face disappears from the surface and from the volume counting, because it is measured twice and belongs to neither. The surviving faces are added. So a combination question is really two ordinary questions plus one subtraction of the hidden circle.

Total surface area of the combination = CSA of the lower part + CSA of the upper part + the exposed base circles only.
A circle that is glued to another solid contributes πr² to neither surface and neither volume.
Volume of the combination = volume of the lower solid + volume of the upper solid. Volumes simply add, so there is nothing to subtract.
If one solid is cut out of another, the surface area loses the area of the cut face and gains the area of the new inner face.
Take at most two solids. A three-solid combination will not be set for 2024-25.

Three deleted question types to refuse

If a question offers a frustum of a cone, reject it — that solid is out of scope. If it says a solid metal block is melted and recast as another solid, reject it — conversion problems are deleted, so there is no density or mass step anywhere in this chapter. And if you count three solids being joined, re-read, because a combination of more than two is not in the syllabus.

Worked Problem: A Cylinder Surmounted by a Cone

A right circular cylinder of radius 7 cm and height 12 cm stands on a circular base, and a right circular cone of radius 7 cm and slant height 25 cm is surmounted on it so that the two circular faces coincide. Find the total surface area and the volume of the solid so formed. Take π = 22/7.The joining circle is common to both solids, so it is hidden. The surface is therefore the cylinder's curved face, the cone's curved face, and the single circle at the bottom of the cylinder.

Height of the cone: the height h, radius r and slant height l form a right triangle, so h = √(25² − 7²) = √(625 − 49) = √576 = 24 cm.
CSA of cylinder = 2πrh = 2 × 22/7 × 7 × 12 = 2 × 22 × 12 = 528 cm².
CSA of cone = πrl = 22/7 × 7 × 25 = 22 × 25 = 550 cm².
Exposed bottom circle = πr² = 22/7 × 49 = 154 cm².
Total surface area = 528 + 550 + 154 = 1232 cm².
Volume of cylinder = πr²h = 22/7 × 49 × 12 = 154 × 12 = 1848 cm³.
Volume of cone = (1/3)πr²h = (1/3) × 22/7 × 49 × 24 = (1/3) × 3696 = 1232 cm³.
Total volume = 1848 + 1232 = 3080 cm³.

Why the answer has no second circle

There are two circular faces on the cylinder but only one is on the outside. The upper one is glued to the cone. Adding 2πr² by mistake gives 1386 cm² and loses the mark. Count the exposed faces on the finished solid, not the faces of the parts.

Worked Problems: Solid Metal and a Spherical Tank

Problem 2 — a solid metal billet is in the shape of a cuboid 120 cm long, 80 cm broad and 30 cm thick. Find its total surface area and its volume.Problem 3 — a spherical gas tank has radius 3.5 m. Take π = 22/7 and find the area to be painted on the outside. Then find the total area of the flat circular plates needed to close the tank if it is cut into two hemispheres.

Billet TSA = 2(lb + bh + hl) = 2(120 × 80 + 80 × 30 + 30 × 120) = 2(9600 + 2400 + 3600) = 2 × 15600 = 31200 cm².
Billet volume = lbh = 120 × 80 × 30 = 288000 cm³.
Spherical tank surface area = 4πr² = 4 × 22/7 × (3.5)² = 4 × 22/7 × 12.25 = 4 × 38.5 = 154 m².
Each new flat plate is a circle of radius 3.5 m, so its area = πr² = 22/7 × 12.25 = 38.5 m².
Two hemispheres need two plates, so total plate area = 2 × 38.5 = 77 m².

No π in a billet, no volume for a sphere

The billet problem uses no π at all, because every face of a cuboid is a rectangle. The tank problem needs π but no volume, because for a sphere the board asks only the surface area. Learning which formulas carry π and which do not saves a mark before you start.

How the Questions Are Asked

The 2024-25 paper is 80 marks in 3 hours with 38 questions. Section A has 18 MCQs and 2 assertion-reason questions of 1 mark each, Section B has 5 very short answer questions of 2 marks, Section C has 6 short answer questions of 3 marks, Section D has 4 long answer questions of 5 marks, and Section E has 3 case-study questions of 4 marks, with internal choice in two questions each of Sections B, C and D. Calculators are not allowed, so take π = 22/7 and pick dimensions that are multiples of 7.Internal assessment adds 20 marks, split 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical. There is no practical in this chapter, so the internal marks come from the chapters that do have one.

Find the curved surface area of a cylinder of radius 7 cm and height 10 cm. Answer 2πrh = 440 cm².
Find the total surface area of a cone of radius 7 cm, height 24 cm and slant height 25 cm. Answer πr(l + r) = 22 × 32 = 704 cm².
Find the volume of a solid metal sphere of radius 3.5 m — not asked this year, because only the surface area of a sphere is examined.
A cuboidal tank 8 m by 5 m by 4 m is full of water. Find the volume of water in litres. Volume = 160 m³ = 160,000 litres.
A hemisphere of radius 7 cm is surmounted on a cylinder of radius 7 cm and height 10 cm. Find the TSA. Answer 2πrh + 2πr² + πr² = 902 cm².
MCQ trap: a solid sphere of radius r and a solid hemisphere of radius r have surface areas in the ratio 4 : 3, not 2 : 1.
MCQ trap: the volume of a cone is one third of the cylinder on the same base and height, not half.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Cube

T S A = 6 a^{2}, L S A = 4 a^{2}, V = a^{3}

a is the edge. No π anywhere.

Cuboid

T S A = 2 (l b + bh + h l), L S A = 2 h (l + b), V = l bh

l length, b breadth, h height.

Sphere

Surface area = 4 π r^{2}

Only the surface area is asked for a sphere this year.

Hemisphere curved surface area

C S A = 2 π r^{2}

Exactly half of the sphere's surface area.

Hemisphere total surface area

T S A = 3 π r^{2}

2πr² of dome plus πr² of the flat circular base.

Cylinder curved surface area

C S A = 2 π r h

The side of the tube, with no base included.

Cylinder

T S A = 2 π r (r + h), V = π r^{2} h

Two circular bases, so add 2πr² to the curved surface.

Cone curved surface area

C S A = π r l

l is the slant height, which the board nearly always gives.

Cone

T S A = π r (l + r), V = \frac{1}{3} π r^{2} h

One circular base; the volume is one third of the cylinder on the same r and h.

Cone height from the slant height

h = l^{2} - r^{2}

Pythagoras from Class 9, used when a question gives l and r but not h.

Volume of a combination

V_{combination} = V_{1} + V_{2}

Volumes add; only surface areas need the hidden face subtracted.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

Decide CSA against TSA before substituting. Cylinder TSA = CSA + 2πr², cone TSA = CSA + πr², hemisphere TSA = CSA + πr².
In a combined solid the glued circle is exposed nowhere, so it appears in neither the surface area nor the volume working.
Take at most two solids; a combination of three is outside the 2024-25 syllabus.
The frustum of a cone and problems converting one metallic solid into another are both deleted — do not attempt them.
For a sphere and a hemisphere only the surface area is examined, so never hunt for a volume step.
A cone question that gives the slant height l and not the height h is testing h = √(l² − r²) from Pythagoras.
A cuboid question contains no π at all. If π is offered as an option, the solid is not a box.
Take π = 22/7 and choose radii that are multiples of 7, so (22/7) × 7² = 154 and every answer stays exact.
The volume of a cone is one third of the cylinder on the same base and height — the single most repeated fact comparison in the paper.
Unit VI Mensuration is 10 of the 80 theory marks, so a five-mark question here usually wants one solid's full set plus one combination.

FAQ

Frequently asked questions

What is the difference between curved surface area and total surface area?

Curved surface area is the curved part of the surface alone and leaves out every flat face, so a cylinder has CSA = 2πrh and a cone has CSA = πrl. Total surface area adds the exposed flat circular bases, so a cylinder has TSA = 2πr(r + h) and a cone has TSA = πr(l + r). A hemisphere has CSA = 2πr² and TSA = 3πr² for the same reason.

How do you find the total surface area when two solids are joined?

Add the curved surface areas of both solids, then add only the faces of the solid that are still exposed. For a cylinder of radius 7 cm and height 12 cm surmounted by a cone of radius 7 cm and slant height 25 cm, the answer is 2πrh + πrl + πr² = 528 + 550 + 154 = 1232 cm². The joining circle is hidden and is counted nowhere.

A cone has radius 7 cm and slant height 25 cm. Find its height.

The height, the radius and the slant height form a right triangle, because the radius is drawn from the centre of the base to the rim and is at right angles to the height. So by Pythagoras h = √(l² − r²) = √(625 − 49) = √576 = 24 cm. This is a Class 9 result being reused, not new syllabus.

Why is the volume of a cone one third of the cylinder on the same base and height?

The two solids have the same circular base of area πr² and the same height h, but a cone tapers to a point instead of staying full width. The volume formulas are V(cone) = (1/3)πr²h and V(cylinder) = πr²h, so the cone is exactly one third. This is a fact you use, not one you prove, in Class 10.

What has been removed from this chapter for 2024-25?

The retained content is the surface areas and volumes of combinations of any two of cubes, cuboids, spheres, hemispheres and right circular cylinders and cones. The frustum of a cone has been deleted, so has any problem that converts one type of metallic solid into another, and so has any combination of more than two solids. Solve questions that join at most two of the listed solids.

Next Chapters

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Class 10 Maths Notes Areas Related to Circles Class 10 Maths Notes Circles Class 10 Maths Notes Triangles Class 10 Maths Notes All Class 11 & 12 Notes

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