MathsCBSE Class 10 · Chapter 12

Surface Areas and Volumes Class 10: Combinations of Solids Explained

Q: Is the frustum of a cone in the Class 10 syllabus for 2026-27?

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

Q: Should I use 22/7 or 3.14 for pi?

Use 22/7 unless the question tells you to use 3.14. Board numbers are usually chosen so that 22/7 cancels neatly with radii such as 3.5, 7, 10.5 or 21 cm. If the question gives a value of pi, always use that one.

Q: How do I know which surfaces to include in the total surface area?

Imagine painting the object from outside. Any face you can paint counts, and any face where the two solids are joined does not. For hollowed solids, also count the inner wall of the cavity, because it is now exposed.

Q: Do I need to remember the Class 9 formulas for this chapter?

Yes. Every question in this chapter uses the basic formulas for cubes, cuboids, cylinders, cones, spheres and hemispheres. The new skill is combining them correctly, so learn the basic formulas thoroughly first.

Q: Why is my answer for a toy's surface area always too big?

You are probably using the total surface area of the cone and the hemisphere and adding them. That counts the joined circular bases, which are hidden. Use only the curved surface of each part: pi r l for the cone and 2 pi r squared for the hemisphere.

Q: Do I lose marks for leaving out units?

Often yes. The final answer mark usually expects the correct unit: square units for area, cubic units for volume, and litres or rupees when the question asks for them. Write the unit on the last line of every answer.

Updated 13 October 2026 · Fact-checked

Surface Areas and Volumes in Class 10 is about solids made by joining or cutting two basic shapes: cubes, cuboids, cylinders, cones, spheres and hemispheres. The board asks you to find the surface area of only the exposed faces, find volume by adding or subtracting parts, and solve real-life cost and capacity problems.

What Surface Areas and Volumes is about

This chapter takes the solids you already know (cube, cuboid, cylinder, cone, sphere and hemisphere) and puts two of them together. A toy might be a cone sitting on a hemisphere. A tent might be a cylinder with a cone on top. A block might have a hemisphere scooped out of it. Your job is to work out how much surface is on the outside and how much space is inside.

It builds directly on the surface area and volume formulas from Class 9. It uses Pythagoras' theorem to find the slant height of a cone. It also uses circle areas from Areas Related to Circles. If those formulas are solid, this chapter is mostly about careful thinking: which faces can you see, and which have disappeared where the two solids meet?

Board questions come in a few shapes. MCQs test a formula, a ratio or what happens when solids are joined. 2- and 3-mark questions ask for the surface area or volume of one combination. 5-mark questions often need an unknown height or radius first, then both surface area and volume. Case studies use real objects such as tents, containers, capsules and decorative blocks, often with a cost or capacity to work out.

Weightage: This chapter is part of Unit VI, Mensuration, which carries 10 of the 80 theory marks together with Areas Related to Circles.Source: CBSE Mathematics syllabus 2026-27, Unit VI Mensuration

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.2 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.6 marks per paper. See the questions

Key concepts

Revision: cuboid and cube

NCERT section 12.1 Introduction

A cuboid with length $l$ , breadth $b$ and height $h$ has six rectangular faces in three equal pairs. That is why its total surface area is $2 (l b + bh + h l)$ . The lateral (side) surface leaves out the top and bottom, giving $2 h (l + b)$ . Volume is base area times height, so $V = l bh$ .

A cube is a cuboid with all edges equal to $a$ . So the TSA is $6 a^{2}$ , the lateral surface is $4 a^{2}$ and the volume is $a^{3}$ .

Why does this matter here? When two equal cubes are joined face to face, the result is a cuboid. Two square faces press against each other and vanish from the outside. So the new surface area is less than the sum of the two cubes' areas, but the volume is exactly the sum. Keep this idea in mind: it is the heart of the whole chapter.

Worked example

Two identical cubes, each of volume 125 cm³, are joined end to end to form a cuboid. Find the surface area of the cuboid. By how much is it less than the total surface area of the two separate cubes?

Try it first, then show the solutionSolution

Edge of each cube: $a^{3} = 125 \Rightarrow a = 5$ cm
Joined end to end, the cuboid has $l = 10$ cm, $b = 5$ cm, $h = 5$ cm
TSA of cuboid $= 2 (l b + bh + h l) = 2 (50 + 25 + 50)$
$= 2 \times 125 = 250$ cm²
TSA of two separate cubes $= 2 \times 6 a^{2} = 2 \times 6 \times 25 = 300$ cm²
Difference $= 300 - 250 = 50$ cm², which equals the two hidden faces $2 \times 25$ cm²

Answer: Surface area of the cuboid = 250 cm². It is 50 cm² less than the two separate cubes.

Revision: cylinder, cone, sphere and hemisphere

NCERT section 12.1 Introduction

Unroll the curved surface of a cylinder and you get a rectangle of length $2 π r$ and width $h$ . So its curved surface area (CSA) is $2 π r h$ . Add two circular ends for the TSA: $2 π r (r + h)$ . Volume is $π r^{2} h$ .

In a cone, the height $h$ , radius $r$ and slant height $l$ form a right triangle. So $l = r^{2} + h^{2}$ . The CSA is $π r l$ . It uses the slant height, never the vertical height. A cone holds exactly one-third of a cylinder with the same base and height, so $V = \frac{1}{3} π r^{2} h$ .

A sphere has surface area $4 π r^{2}$ and volume $\frac{4}{3} π r^{3}$ . A hemisphere is half of it. Its curved surface is $2 π r^{2}$ . A solid hemisphere also has a flat circle, so its TSA is $3 π r^{2}$ . Its volume is $\frac{2}{3} π r^{3}$ . Many board questions hide one of $r$ , $h$ or $l$ , so you must find it first.

Worked example

The curved surface area of a right circular cone is 550 cm² and its slant height is 25 cm. Find the volume of the cone.

Try it first, then show the solutionSolution

CSA $= π r l$ , so $\frac{22}{7} \times r \times 25 = 550$
$r = \frac{550 \times 7}{22 \times 25} = \frac{3850}{550} = 7$ cm
Height: $h = l^{2} - r^{2} = 625 - 49 = 576 = 24$ cm
Volume $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24$
$= 22 \times 7 \times 8 = 1232$ cm³
Check: $π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm², which matches the given CSA

Answer: Volume of the cone = 1232 cm³

Surface area of a combination: count only exposed surfaces

NCERT section 12.2 Surface Area of a Combination of Solids

When two solids are joined, the face where they touch is sealed inside the new solid. Nobody can paint it or touch it, so it is not part of the surface area. The rule is simple: the surface area of a combination is the sum of the surfaces you can actually see.

Before substituting numbers, write the answer as a sum of named parts. For example: 'TSA = CSA of cylinder + CSA of hemisphere + area of base'. This line shows the examiner your method and can earn a mark even if your arithmetic slips.

Here is a useful check. When a hemisphere sits flat-side down on a cube face, the cube loses a circle of area $π r^{2}$ . The hemisphere adds its curved surface $2 π r^{2}$ . So the net change is $+ π r^{2}$ , and the TSA is $6 a^{2} + π r^{2}$ . Volume works differently: no space is lost when solids are joined, so volumes simply add.

Worked example

A decorative block is a cube of edge 14 cm. A solid hemisphere of radius 3.5 cm is fixed flat-side down at the centre of its top face. Find the total surface area of the block, and the cost of polishing it at ₹2 per 10 cm².

Try it first, then show the solutionSolution

Exposed surfaces: 6 faces of the cube, minus the circle covered by the hemisphere, plus the curved surface of the hemisphere
TSA $= 6 a^{2} - π r^{2} + 2 π r^{2} = 6 a^{2} + π r^{2}$
$6 a^{2} = 6 \times 196 = 1176$ cm²
$π r^{2} = \frac{22}{7} \times 3.5 \times 3.5 = 38.5$ cm²
TSA $= 1176 + 38.5 = 1214.5$ cm²
Cost $= \frac{1214.5}{10} \times 2 = 242.9$ , so the cost is ₹242.90

Answer: Total surface area = 1214.5 cm²; cost of polishing = ₹242.90

Cone mounted on a hemisphere (toys and similar shapes)

NCERT section 12.2 Surface Area of a Combination of Solids

This is the most common combination. A cone and a hemisphere have the same radius and are joined at their circular bases. Both flat circles are hidden, so only the two curved surfaces remain: TSA $= π r l + 2 π r^{2} = π r (l + 2 r)$ .

The trap is the height. The 'total height' of the toy is the cone's height plus the hemisphere's radius. So first find the cone height $h = H - r$ . Then find the slant height $l = r^{2} + h^{2}$ .

For volume, add the parts: $V = \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π r^{2} (h + 2 r)$ . Taking out common factors like this saves time and cuts down arithmetic errors.

The same thinking works for an ice-cream cone with a hemisphere of ice cream on top, a spinning top, or a cylinder with a hemispherical end.

Worked example

A toy is a right circular cone mounted on a hemisphere of the same radius, 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.

Try it first, then show the solutionSolution

Height of cone $h = 31 - 7 = 24$ cm (total height = cone height + radius of hemisphere)
Slant height $l = r^{2} + h^{2} = 49 + 576 = 625 = 25$ cm
CSA of cone $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm²
CSA of hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times 49 = 308$ cm²
TSA of toy $= 550 + 308 = 858$ cm² (the joined circular bases are hidden)
Check: $π r (l + 2 r) = 22 \times (25 + 14) = 22 \times 39 = 858$ cm², which matches

Answer: Total surface area of the toy = 858 cm²

Hollowing out: surface area when one solid is removed from another

NCERT section 12.2 Surface Area of a Combination of Solids

When a cavity is cut into a solid, two things happen to the surface. First, you lose the flat patch of outer surface where the hole opens. Second, you gain the inner wall of the cavity, which is now exposed.

So the rule is: remaining TSA = outer TSA − area of the opening(s) + inner curved surface of the cavity.

Common cases:

A conical cavity in a cylinder: lose one circle $π r^{2}$ , gain $π r l$ .
A hemisphere scooped from a cube face: lose $π r^{2}$ , gain $2 π r^{2}$ .
A cylindrical hole drilled right through a cuboid: lose two circles $2 π r^{2}$ , gain $2 π r h$ .

The remaining volume is always outer volume minus cavity volume. Many students only subtract the opening and forget the inner wall. That is why scooping out a hemisphere actually increases the surface area.

Worked example

A solid cylinder has radius 7 cm and height 30 cm. A conical cavity of the same radius and height 24 cm is hollowed out from its top face. Find the total surface area and the volume of the remaining solid.

Try it first, then show the solutionSolution

Surfaces left: curved surface of cylinder, bottom circle, inner curved surface of cone (the top circle is now the cone's opening)
Slant height of cone $l = 7^{2} + 2 4^{2} = 625 = 25$ cm
CSA of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 7 \times 30 = 1320$ cm²
Bottom base $= π r^{2} = \frac{22}{7} \times 49 = 154$ cm²
Inner CSA of cone $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm²
TSA $= 1320 + 154 + 550 = 2024$ cm²
Volume $= π r^{2} H - \frac{1}{3} π r^{2} h = 154 \times 30 - \frac{1}{3} \times 154 \times 24$
$= 4620 - 1232 = 3388$ cm³

Answer: TSA of remaining solid = 2024 cm²; volume = 3388 cm³

Capsules and depressions: a hemisphere at each end

NCERT section 12.2 Surface Area of a Combination of Solids

The syllabus example of a combination is a cylinder with a hemisphere at each end: the capsule shape. The cylinder and the two hemispheres have the same radius $r$ . Both flat ends of the cylinder are covered by the hemispheres, so only curved surfaces show: TSA $= 2 π r h + 2 π r^{2} + 2 π r^{2} = 2 π r h + 4 π r^{2}$ . The two ends together make one whole sphere, so $V = π r^{2} h + \frac{4}{3} π r^{3}$ .

The trap is the length. In these formulas $h$ is the cylindrical part only. The total length is $h + 2 r$ , because each end sticks out by one radius. So $h =$ total length $- 2 r$ .

The opposite case is a hemispherical depression scooped out of each end of a solid cylinder. Each flat end is lost, but a bowl of area $2 π r^{2}$ appears in its place. So the TSA is again $2 π r h + 4 π r^{2}$ , but now the volume is $π r^{2} h - \frac{4}{3} π r^{3}$ . This only works if $h \geq 2 r$ , so that the two bowls do not meet.

The same idea works for a hemispherical depression in a cube face: lose $π r^{2}$ of flat face, gain $2 π r^{2}$ of bowl, and subtract $\frac{2}{3} π r^{3}$ from the volume.

Worked example

A solid wooden cylinder has diameter 21 cm and height 30 cm. A hemispherical depression of the same diameter is scooped out of each end. Find the total surface area and the volume of the remaining solid. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$r = 10.5$ cm and $h = 30$ cm; since $2 r = 21 \leq 30$ , the two bowls do not meet
$π r = \frac{22}{7} \times 10.5 = 33$ cm and $π r^{2} = 33 \times 10.5 = 346.5$ cm²
Exposed surfaces: curved surface of the cylinder and the insides of the two hemispherical bowls (both flat ends are gone)
TSA $= 2 π r h + 4 π r^{2} = 2 \times 33 \times 30 + 4 \times 346.5 = 1980 + 1386 = 3366$ cm²
Volume $= π r^{2} h - \frac{4}{3} π r^{3} = π r^{2} (h - \frac{4 r}{3})$
$= 346.5 \times (30 - 14) = 346.5 \times 16 = 5544$ cm³
Check: cylinder $346.5 \times 30 = 10395$ cm³, two bowls $2 \times 2425.5 = 4851$ cm³, and $10395 - 4851 = 5544$ cm³

Answer: TSA of remaining solid = 3366 cm²; volume = 5544 cm³

Volume of a combination: add the parts

NCERT section 12.3 Volume of a Combination of Solids

Joining two solids does not create or destroy any space. So the volume of the combination is just the sum of the two volumes: $V = V_{1} + V_{2}$ . Unlike surface area, there is nothing hidden to subtract.

When both parts share the same radius, take $π r^{2}$ out as a common factor. For a cylinder of height $H$ with a cone of height $h$ on top, $V = π r^{2} (H + \frac{h}{3})$ . For a cylinder with a hemisphere, $V = π r^{2} (H + \frac{2 r}{3})$ .

For a hollow container, the volume of the space inside is its capacity. Board questions often ask for capacity in litres. Remember $1000$ cm³ = 1 litre, and $1$ m³ = 1000 litres.

A container question can also ask for the metal sheet used. That is a surface area, so the hidden-face rule applies again.

Worked example

A closed metal container is a cylinder of radius 7 cm and height 12 cm with a cone of the same radius on top. The total height of the container is 36 cm. Find its capacity in litres and the area of metal sheet used to make it (ignore thickness).

Try it first, then show the solutionSolution

Height of cone $h = 36 - 12 = 24$ cm
Volume $= π r^{2} H + \frac{1}{3} π r^{2} h = π r^{2} (H + \frac{h}{3})$
$= \frac{22}{7} \times 49 \times (12 + 8) = 154 \times 20 = 3080$ cm³
Capacity $= 3080 \div 1000 = 3.08$ litres
Slant height $l = 49 + 576 = 25$ cm
Sheet $=$ base + CSA of cylinder + CSA of cone (the circle where they join is hidden)
$= π r^{2} + 2 π r H + π r l = 154 + 528 + 550 = 1232$ cm²

Answer: Capacity = 3.08 litres; metal sheet used = 1232 cm²

Volume when a part is removed: subtract

NCERT section 12.3 Volume of a Combination of Solids

If a solid is scooped, drilled or carved out of another, the volume left is the outer volume minus the volume removed. The same idea answers 'how much water is left' questions. Suppose a solid is placed in a cylinder full of water. The water that stays is the cylinder's volume minus the solid's volume, because the solid pushes out exactly its own volume.

Watch for the word 'largest'. The largest hemisphere you can scoop from a cube face has diameter equal to the edge. The largest cylinder or cone that fits in a cube has diameter and height equal to the edge.

With $π = \frac{22}{7}$ , radii like 3.5, 7, 10.5 or 21 cm cancel neatly. Write the radius as a fraction such as $\frac{21}{2}$ and cancel before you multiply. This keeps the arithmetic short and accurate.

Worked example

From a solid cube of edge 21 cm, the largest possible hemisphere is scooped out of one face. Find the volume of the remaining solid and its total surface area.

Try it first, then show the solutionSolution

Largest hemisphere: diameter = edge, so $r = \frac{21}{2} = 10.5$ cm
Volume of cube $= 2 1^{3} = 9261$ cm³
Volume of hemisphere $= \frac{2}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2}$
$= \frac{2 \times 22 \times 441}{8} = \frac{19404}{8} = 2425.5$ cm³
Remaining volume $= 9261 - 2425.5 = 6835.5$ cm³
TSA $= 6 a^{2} - π r^{2} + 2 π r^{2} = 6 a^{2} + π r^{2}$
$= 6 \times 441 + \frac{22}{7} \times 110.25 = 2646 + 346.5 = 2992.5$ cm²

Answer: Remaining volume = 6835.5 cm³; total surface area = 2992.5 cm²

Real-life problems: canvas, paint, cost and capacity

NCERT section 12.2 Surface Area of a Combination of Solids

Case studies and long answers dress the same ideas in real objects. The skill is deciding which surfaces the real object actually uses.

A tent has no floor canvas, so count only curved surfaces.
An open vessel has no lid.
A painted toy is painted only on the outside.
The inside of a bowl is a curved surface, not a total surface.

Read every word: 'open', 'closed', 'inner', 'outer', 'floor not covered', 'extra for wastage'.

Then follow three steps. First find any missing length. Next write the area or volume as a sum of named parts. Finally convert to the unit asked: cost per m², litres, or rupees. Take a common factor (often $π r$ ) before multiplying. Give the final answer with units and, for money, in rupees.

Worked example

A circus tent is cylindrical up to a height of 4 m and conical above it. The radius of the base is 7 m and the slant height of the conical part is 10 m. The floor is not covered. Find the area of canvas needed if 10% extra is required for stitching and wastage. Also find the cost of the canvas at ₹50 per m².

Try it first, then show the solutionSolution

Canvas covers the CSA of the cylinder + CSA of the cone (no floor; the joining circle is open inside the tent)
Area $= 2 π r h + π r l = π r (2 h + l)$
$= \frac{22}{7} \times 7 \times (8 + 10) = 22 \times 18 = 396$ m²
Extra 10% $= 39.6$ m², so canvas needed $= 396 + 39.6 = 435.6$ m²
Cost $= 435.6 \times 50 = 21780$ , so the cost is ₹21,780

Answer: Canvas needed = 435.6 m²; cost = ₹21,780

Maths in real life: Surface Areas and Volumes

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.

Revision: cuboid and cube

Suppose your family plans to repaint your bedroom before Diwali. The room is a cuboid 5 m long, 4 m wide and 3 m high. You will paint the four walls and the ceiling but not the floor. The room has one door 2 m high and 1 m wide, which will not be painted.

If the painter charges ₹25 per m², find the total cost of painting the room.

Try it first, then show the solutionSolution

Area of four walls = lateral surface of the cuboid $= 2 h (l + b)$
$= 2 \times 3 \times (5 + 4) = 54$ m²
Area of ceiling $= l \times b = 5 \times 4 = 20$ m²
Walls + ceiling $= 54 + 20 = 74$ m²
Area of door $= 2 \times 1 = 2$ m², which is not painted
Area to be painted $= 74 - 2 = 72$ m²
Cost $= 72 \times 25 =$ ₹1800

Answer: ₹1800 (area painted is 72 m²)

Memory hook: Walls are the lateral surface, the ceiling is one lid, and the floor stays bare.

Surface area of a combination: count only exposed surfaces

Imagine your art class makes a decorative Diwali stand: a solid wooden cube of edge 14 cm with a solid hemispherical dome of diameter 7 cm glued flat-side down at the centre of its top face. The whole outer surface, including the bottom, is to be coated with golden paint.

Find the total surface area to be painted. If the paint costs ₹10 per 100 cm², find the cost of painting one stand.

Try it first, then show the solutionSolution

TSA = TSA of cube − base circle of hemisphere + CSA of hemisphere
Net change from the hemisphere $= - π r^{2} + 2 π r^{2} = + π r^{2}$ , so TSA $= 6 a^{2} + π r^{2}$
$r = \frac{7}{2} = 3.5$ cm and $a = 14$ cm
$6 a^{2} = 6 \times 196 = 1176$ cm²
$π r^{2} = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 38.5$ cm²
TSA $= 1176 + 38.5 = 1214.5$ cm²
Cost $= \frac{1214.5}{100} \times 10 =$ ₹121.45

Answer: Area painted = 1214.5 cm²; cost = ₹121.45

Memory hook: The dome hides one circle but shows two circles' worth of curve: net gain of one $π r^{2}$ .

Cone mounted on a hemisphere (toys and similar shapes)

Suppose a sweet shop sells ice cream in a cone of radius 3.5 cm. The cone is filled completely, and a hemisphere of ice cream of the same radius sits on top. The total height from the tip of the cone to the top of the ice cream is 17.5 cm.

Find the volume of ice cream in one cone. How many such cones can be served from a tub holding 10.78 litres of ice cream?

Try it first, then show the solutionSolution

Height of cone $h = 17.5 - 3.5 = 14$ cm (total height minus hemisphere radius)
Volume $= \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π r^{2} (h + 2 r)$
$π r^{2} = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 38.5$ cm²
$h + 2 r = 14 + 7 = 21$ cm
Volume $= \frac{1}{3} \times 38.5 \times 21 = 38.5 \times 7 = 269.5$ cm³
Tub $= 10.78$ L $= 10780$ cm³ (since 1 L = 1000 cm³)
Number of cones $= \frac{10780}{269.5} = 40$

Answer: 269.5 cm³ per cone; 40 cones

Memory hook: Ice-cream cone: subtract the scoop's radius from the total height before you find the cone.

Hollowing out: surface area when one solid is removed from another

Imagine your woodwork class makes a candle holder from a cuboidal block 10 cm long, 10 cm wide and 7 cm high. A cylindrical hole of diameter 7 cm is drilled right through the block from the top face to the bottom face. Every surface, including the inside of the hole, is to be varnished.

Find the total surface area of the holder after drilling. If varnish costs ₹2 per 10 cm², find the cost of varnishing it.

Try it first, then show the solutionSolution

Remaining TSA = TSA of cuboid − 2 circles + CSA of cylindrical hole
TSA of cuboid $= 2 (10 \times 10 + 10 \times 7 + 7 \times 10) = 2 (240) = 480$ cm²
$r = 3.5$ cm, hole length $h = 7$ cm
Two circles $= 2 π r^{2} = 2 \times 38.5 = 77$ cm²
CSA of hole $= 2 π r h = 2 \times \frac{22}{7} \times \frac{7}{2} \times 7 = 154$ cm²
Remaining TSA $= 480 - 77 + 154 = 557$ cm²
Cost $= \frac{557}{10} \times 2 =$ ₹111.40

Answer: 557 cm²; cost of varnish = ₹111.40

Memory hook: A drilled hole steals two coins of surface but gives back a whole tube inside.

Volume of a combination: add the parts

Suppose your school canteen uses a steel tea boiler shaped like a cylinder with a hemispherical bottom. Its inner radius is 21 cm and the total inner height, from the lowest point of the bottom to the top rim, is 57 cm. Tea is served in cups of 150 mL.

Find the capacity of the boiler in litres and the number of cups of tea it can fill when full.

Try it first, then show the solutionSolution

Height of cylindrical part $H = 57 - 21 = 36$ cm
Volume $= π r^{2} H + \frac{2}{3} π r^{3} = π r^{2} (H + \frac{2 r}{3})$
$π r^{2} = \frac{22}{7} \times 21 \times 21 = 1386$ cm²
$H + \frac{2 r}{3} = 36 + 14 = 50$ cm
Volume $= 1386 \times 50 = 69300$ cm³
Capacity $= \frac{69300}{1000} = 69.3$ litres
Cups $= \frac{69300}{150} = 462$ (1 mL = 1 cm³)

Answer: 69.3 litres; 462 cups

Memory hook: Tea boiler: take out $π r^{2}$ once, then add the heights $H$ and $\frac{2 r}{3}$ .

Real-life problems: canvas, paint, cost and capacity

Imagine your school trip to a hill camp uses a tent shaped like a cylinder with a conical top. The base diameter is 5.6 m, the cylindrical wall is 2.5 m high, and the total height of the tent is 4.6 m. The floor is not covered with canvas.

Find the area of canvas needed for the tent. If canvas costs ₹500 per m², find the total cost.

Try it first, then show the solutionSolution

$r = 2.8$ m; cone height $h = 4.6 - 2.5 = 2.1$ m
Slant height $l = 2. 8^{2} + 2. 1^{2} = 7.84 + 4.41 = 12.25 = 3.5$ m
Canvas = CSA of cylinder + CSA of cone (no floor, joined circle hidden)
CSA of cylinder $= 2 π r H = 2 \times \frac{22}{7} \times 2.8 \times 2.5 = 44$ m²
CSA of cone $= π r l = \frac{22}{7} \times 2.8 \times 3.5 = 30.8$ m²
Canvas $= 44 + 30.8 = 74.8$ m²
Cost $= 74.8 \times 500 =$ ₹37400

Answer: 74.8 m² of canvas; cost ₹37400

Memory hook: A tent has no floor and no hidden circle: only the two curved walls need canvas.

Topics in this chapter

Surface Areas and Volumes formula sheet

Cuboid: total surface area: $2 (l b + bh + h l)$; All six faces. For a closed box or a block.
Cuboid: lateral surface area: $2 h (l + b)$; Four walls only, as in a room or an open-top tank without the base.
Cuboid: volume: $V = l bh$; Capacity of tanks and boxes; convert cm³ to litres by dividing by 1000.
Cube: TSA, LSA, volume: $6 a^{2}, 4 a^{2}, a^{3}$; Two equal cubes joined face to face give a cuboid of TSA $10 a^{2}$ .
Cylinder: curved surface area: $2 π r h$; Side wall of pipes, tents, containers and drilled holes.
Cylinder: total surface area: $2 π r (r + h)$; Closed solid cylinder only. In combinations, add the bases one by one instead.
Cylinder: volume: $V = π r^{2} h$; Base area times height.
Cone: slant height: $l = r^{2} + h^{2}$; Pythagoras in the right triangle of height, radius and slant height. Needed before any cone CSA.
Cone: curved surface area: $π r l$; Always uses the slant height $l$ , never the vertical height $h$ .
Cone: total surface area: $π r (l + r)$; Solid cone on its own. In a toy the base is hidden, so use only $π r l$ .
Cone: volume: $V = \frac{1}{3} π r^{2} h$; One-third of the cylinder with the same base and height.
Sphere: surface area and volume: $4 π r^{2}, \frac{4}{3} π r^{3}$; Balls, spherical shells; a sphere has no flat face to hide.
Hemisphere: curved surface area: $2 π r^{2}$; Use this when the flat face is joined to another solid or for the inside of a bowl.
Solid hemisphere: total surface area: $3 π r^{2}$; Curved $2 π r^{2}$ plus flat circle $π r^{2}$ ; only when the flat face is exposed.
Hemisphere: volume: $V = \frac{2}{3} π r^{3}$; Bowls and hemispherical ends of toys and vessels.
Combination: surface area rule: $TSA = sum of exposed surfaces only$; Leave out every face where the two solids touch.
Combination: volume rule: $V = V_{1} + V_{2} (joined), V = V_{1} - V_{2} (removed)$; Volume never loses a hidden face; it only adds or subtracts.
Cone on hemisphere (same r): $TSA = π r (l + 2 r), V = \frac{1}{3} π r^{2} (h + 2 r)$; Toys and ice-cream cones. Cone height $h =$ total height $- r$ .
Cylinder with cone on top (same r): $TSA = π r^{2} + 2 π r H + π r l, V = π r^{2} (H + \frac{h}{3})$; Closed containers and rockets. For a tent with no floor, drop the $π r^{2}$ .
Cylinder with hemisphere on one end (same r): $TSA = π r^{2} + 2 π r H + 2 π r^{2}, V = π r^{2} (H + \frac{2 r}{3})$; Vessels and bottles; for an open vessel, drop the $π r^{2}$ lid.
Cylinder with a hemisphere at each end (capsule, same r): $TSA = 2 π r h + 4 π r^{2}, V = π r^{2} h + \frac{4}{3} π r^{3}$; $h$ is the length of the cylindrical part only; the total length is $h + 2 r$ . The two ends together make one full sphere.
Cylinder with a hemisphere scooped from each end (same r): $TSA = 2 π r h + 4 π r^{2}, V = π r^{2} h - \frac{4}{3} π r^{3}$; Both flat ends are lost and two bowls of $2 π r^{2}$ are gained, so the TSA matches the capsule, but the volume is subtracted. Needs $h \geq 2 r$ .
Hemisphere on, or scooped from, a cube face: $TSA = 6 a^{2} + π r^{2}$; The same result either way: lose $π r^{2}$ of flat face, gain $2 π r^{2}$ of curved surface. Needs $2 r \leq a$ .
Cylindrical hole drilled through a cuboid: $TSA = 2 (l b + bh + h l) - 2 π r^{2} + 2 π r h$; $h$ is the length of the hole; both circular openings are lost, the inner wall is gained.
Units and constants: $1 L = 1000 cm^{3}, 1 m^{3} = 1000 L, π = \frac{22}{7}$; Use $π = \frac{22}{7}$ unless told otherwise; area in square units, volume in cubic units.

How the board asks Surface Areas and Volumes

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Total surface area of two joined solids	A toy, block or vessel made of a cone on a hemisphere, a hemisphere on a cube or a cone on a cylinder; find the TSA or the cost of painting it.	List the exposed surfaces in words first, then write the formula as a sum of named parts. Leave out every face where the solids meet, and add a flat base only if it is really outside.
Volume or capacity of a combination	Find the volume of a toy, or the capacity in litres of a container shaped like a cylinder with a cone or hemisphere.	Add the volumes of the parts. Take $π r^{2}$ out as a common factor if the radii are equal. Convert cm³ to litres by dividing by 1000, and state the unit.
Find a missing dimension first	The total height, total volume or a curved surface area is given, and you must find the cone height, radius or slant height before anything else.	Subtract the hemisphere radius from the total height, or set up an equation from the given volume or CSA and solve for r or h. Then get l from Pythagoras. Write each found value on its own line with its unit.
Hollowing out, scooping or drilling	A cone cut from a cylinder, a hemisphere scooped from a cube or a cylindrical hole drilled through a block; find the remaining TSA and/or volume.	For area: outer TSA minus opening(s) plus inner curved surface. For volume: outer minus removed. Check the cavity fits (diameter not more than the face, height not more than the solid).
Cost and material problems	Canvas for a tent, paint for a toy, metal sheet for a container, often with a rate per m² and sometimes extra for wastage.	Decide which surfaces the real object uses (no tent floor, no lid on an open vessel). Find the area, add any percentage extra, then multiply by the rate and give the answer in rupees.
Water left or space left	A solid toy is placed in a cylinder full of water, or a solid sits inside a box; find the water that remains or the empty space.	Water left = volume of container − volume of solid. Work out the solid's volume as a combination first, then subtract.
MCQ and assertion-reason on joining solids	What is the surface area of two cubes joined end to end? Is the TSA of a combination equal to the sum of the TSAs? Which ratio holds for a cone, hemisphere and cylinder of equal radius and height?	Use the hidden-face idea: surface area drops by twice the area of the joined face, while volume stays the sum. Use the 1 : 2 : 3 volume ratio for a cone, hemisphere and cylinder with height equal to radius.
Case study (Section E)	A real object such as a tent, ice-cream cone, capsule-shaped container or decorative block, with sub-parts asking for a length, a surface area, a volume and a cost or capacity.	Read the whole description, mark r, h and l on a rough sketch, and use answers from earlier parts in later ones. Each sub-part needs its own clear final line with units.

Tricks to remember Surface Areas and Volumes

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

Glue hides, volume adds
Wherever two solids are glued, that face is hidden, so subtract it from the surface area. Volume never hides anything, so it just adds.
Use it when: Say this before every combination question to decide which surfaces to count.
1 : 2 : 3
A cone, a hemisphere and a cylinder with the same radius $r$ and height $r$ have volumes in the ratio $1 : 2 : 3$ , that is $\frac{1}{3} π r^{3} : \frac{2}{3} π r^{3} : π r^{3}$ .
Use it when: Quick MCQ answers and a check that your cone and hemisphere volumes look sensible.
Slant-height triples
Learn the right-triangle triples 3-4-5, 5-12-13, 8-15-17, 7-24-25 and 20-21-29 (and their multiples, like 6-8-10).
Use it when: Board numbers are usually chosen so l comes out whole. If r = 7 and h = 24, you know at once that l = 25.
Factor before you multiply
Take $π r$ or $π r^{2}$ out as a common factor first, then cancel 7 with $\frac{22}{7}$ . For example, $π r (l + 2 r)$ with $r = 7$ becomes $22 (l + 14)$ .
Use it when: Cuts long multiplications and reduces careless slips in 3- and 5-mark answers.
Sense check on the answer
Joined solids: TSA must be less than the sum of the separate TSAs. Scooping a hemisphere from a cube: TSA goes up by $π r^{2}$ . Area answers are in square units, volumes in cubic units.
Use it when: A 10-second check before moving on catches the most common method errors.

Surface Areas and Volumes: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: TSA of a cylinder topped by a hemisphere · 1 mark · MCQBoard level

A solid is made by placing a solid hemisphere of radius $r$ on top of a solid cylinder of radius $r$ and height $r$ . What is the total surface area of the solid?
(a) $4 π r^{2}$
(b) $5 π r^{2}$
(c) $6 π r^{2}$
(d) $3 π r^{2}$

Try it first, then show the solutionSolution

Hidden faces: the top of the cylinder and the flat face of the hemisphere.
Exposed: bottom base $π r^{2}$ , curved surface of cylinder $2 π r \times r = 2 π r^{2}$ , curved surface of hemisphere $2 π r^{2}$ .
Total $= π r^{2} + 2 π r^{2} + 2 π r^{2} = 5 π r^{2}$ .

Answer: (b) $5 π r^{2}$

Example 2: Area lost when two solids are joined · 1 mark · Assertion-reasonBoard level

Assertion (A): A solid hemisphere of radius 3.5 cm is joined by its flat face to one end of a solid cylinder of the same radius. The total surface area of the combined solid is 77 cm² less than the sum of the total surface areas of the two separate solids.
Reason (R): When the two solids are joined, two circular faces, each of area $π r^{2}$ , are hidden.
(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

When joined, the flat face of the hemisphere and one end of the cylinder touch, so two circles of area $π r^{2}$ each are hidden. R is true.
Area lost $= 2 π r^{2} = 2 \times \frac{22}{7} \times 3.5 \times 3.5 = 77$ cm². A is true.
The 77 cm² is exactly the two hidden circles, so R explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 3: Find the cylinder height from the volume, then TSA · 3 marks · Short answerHarder board level

A solid model of a rocket is a right circular cylinder with a right circular cone of the same radius on top. The common radius is 7 cm, the height of the cone is 24 cm and the volume of the model is 2772 cm³. Find the height of the cylinder and the total surface area of the model. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = 1232$ cm³. [½]
Volume of cylinder $= 2772 - 1232 = 1540$ cm³, so $154 H = 1540$ , giving $H = 10$ cm. [1]
Slant height $l = 7^{2} + 2 4^{2} = 625 = 25$ cm. Exposed surfaces: bottom base, curved surface of cylinder, curved surface of cone. [½]
TSA $= π r^{2} + 2 π r H + π r l = 154 + 440 + 550 = 1144$ cm². [1]
Check: $154 \times 10 + 1232 = 2772$ cm³, which matches the given volume.

Answer: Height of the cylinder = 10 cm; total surface area = 1144 cm².

Example 4: Capsule-shaped tank: a hemisphere at each end · 3 marks · Short answerHarder board level

Suppose the steel tank on a water tanker is a right circular cylinder with a hemisphere of the same radius at each end. The diameter of the tank is 2.1 m and its total length, end to end, is 6.1 m. Find (i) the outer surface area of the tank and (ii) its capacity in litres. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$r = 1.05$ m. Length of the cylindrical part $h = 6.1 - 2 \times 1.05 = 4$ m, since each hemisphere adds one radius to the length. [½]
$π r = \frac{22}{7} \times 1.05 = 3.3$ m. (i) TSA $= 2 π r h + 4 π r^{2} = 2 π r (h + 2 r) = 6.6 \times 6.1 = 40.26$ m² (both flat ends of the cylinder are hidden). [1]
$π r^{2} = 3.3 \times 1.05 = 3.465$ m². Volume $= π r^{2} h + \frac{4}{3} π r^{3} = π r^{2} (h + \frac{4 r}{3}) = 3.465 \times (4 + 1.4) = 3.465 \times 5.4 = 18.711$ m³. [1]
(ii) Capacity $= 18.711 \times 1000 = 18711$ litres, since $1$ m³ $= 1000$ L. [½]
Check: cylinder $3.465 \times 4 = 13.86$ m³ and sphere $\frac{4}{3} \times 3.465 \times 1.05 = 4.851$ m³; $13.86 + 4.851 = 18.711$ m³.

Answer: (i) 40.26 m² (ii) 18711 litres

Example 5: Water left after a cone is lowered into a full vessel · 3 marks · Short answerHarder board level

A cylindrical vessel of radius 7 cm and height 20 cm is full of water. A solid cone of base radius 3.5 cm and slant height 12.5 cm is lowered gently into the vessel until it rests on the bottom, fully under water. Find the volume of water left in the vessel and the percentage of the water that overflows. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Height of cone $h = 12. 5^{2} - 3. 5^{2} = 156.25 - 12.25 = 144 = 12$ cm, which is less than 20 cm, so the cone is fully under water. [½]
Volume of water at the start $= π r^{2} H = \frac{22}{7} \times 49 \times 20 = 3080$ cm³. [½]
Volume of cone $= \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 12 = \frac{1}{3} \times 38.5 \times 12 = 154$ cm³. This much water overflows. [1]
Water left $= 3080 - 154 = 2926$ cm³. [½]
Percentage overflow $= \frac{154}{3080} \times 100 = 5%$ . [½]

Answer: Water left = 2926 cm³ (2.926 L); 5% of the water overflows.

Example 6: Hemisphere scooped from a cylinder: volume and change in area · 5 marks · Long answerHarder board level

From one end of a solid wooden cylinder of diameter 21 cm and height 25 cm, a hemispherical depression of the same diameter is scooped out. Find (i) the volume of wood left, (ii) the total surface area of the remaining solid, and (iii) by how much the total surface area has increased compared with the original cylinder. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$r = 10.5$ cm, $h = 25$ cm, $π r^{2} = \frac{22}{7} \times 10.5 \times 10.5 = 346.5$ cm², $π r = 33$ cm. Volume of cylinder $= 346.5 \times 25 = 8662.5$ cm³. [1]
Volume of hemisphere $= \frac{2}{3} π r^{3} = \frac{2}{3} \times 346.5 \times 10.5 = 2425.5$ cm³. [½]
(i) Volume left $= 8662.5 - 2425.5 = 6237$ cm³. [½]
(ii) Exposed surfaces: the flat bottom $π r^{2}$ , the curved surface of the cylinder $2 π r h$ , and the inside of the depression $2 π r^{2}$ . The top circle is gone. [½]
TSA $= 346.5 + 2 \times 33 \times 25 + 2 \times 346.5 = 346.5 + 1650 + 693 = 2689.5$ cm². [1½]
(iii) Original TSA $= 2 π r^{2} + 2 π r h = 693 + 1650 = 2343$ cm². [½]
Increase $= 2689.5 - 2343 = 346.5$ cm². This equals $π r^{2}$ , because a flat circle $π r^{2}$ was removed and a curved bowl $2 π r^{2}$ was added. [½]

Answer: (i) 6237 cm³ (ii) 2689.5 cm² (iii) The area increases by 346.5 cm².

Example 7: Village water tank with a conical roof · 5 marks · Long answerHarder board level

A village water tank is a closed right circular cylinder of diameter 4.2 m and height 5 m, with a conical roof of the same diameter. The vertical height of the roof is 2.8 m. Find (i) the slant height of the roof, (ii) the total capacity of the tank including the roof space, in litres, and (iii) the cost of painting the outer curved surface of the cylinder and the roof at ₹40 per m². Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$r = 2.1$ m. (i) $l = 2. 1^{2} + 2. 8^{2} = 4.41 + 7.84 = 12.25 = 3.5$ m. [1]
$π r^{2} = \frac{22}{7} \times 4.41 = 13.86$ m². Volume of cylinder $= 13.86 \times 5 = 69.3$ m³. [1]
Volume of cone $= \frac{1}{3} \times 13.86 \times 2.8 = 12.936$ m³. [½]
(ii) Total volume $= 69.3 + 12.936 = 82.236$ m³ $= 82236$ L, since $1$ m³ $= 1000$ L. [½]
Area to paint $= 2 π r h + π r l$ , with $π r = 6.6$ m: $2 \times 6.6 \times 5 + 6.6 \times 3.5 = 66 + 23.1 = 89.1$ m². [1]
(iii) Cost $= 89.1 \times 40 =$ ₹3564. [1]

Answer: (i) 3.5 m (ii) 82236 litres (iii) ₹3564

Example 8: Find the error: a hole drilled partway into a cube · 3 marks · Short answerHOTS / competency

A pen holder is a solid cube of edge 10 cm. A cylindrical hole of radius 3.5 cm and depth 7 cm is drilled into it from the centre of the top face. The hole does not go right through. A student finds the total surface area like this:

TSA $= 6 a^{2} - π r^{2} + 2 π r h = 600 - 38.5 + 154 = 715.5$ cm².

Find the error, give the correct total surface area, and find the volume of wood left. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Error: the hole has a flat circular bottom of area $π r^{2}$ inside the cube, because it stops 3 cm above the base. The student removed the top circle but forgot to add this bottom circle. [1]
Correct TSA $= 6 a^{2} - π r^{2} + 2 π r h + π r^{2} = 6 a^{2} + 2 π r h = 600 + 2 \times \frac{22}{7} \times 3.5 \times 7 = 600 + 154 = 754$ cm². [1]
Volume left $= a^{3} - π r^{2} h = 1000 - 38.5 \times 7 = 1000 - 269.5 = 730.5$ cm³. [1]

Answer: The bottom of the hole was missed. Correct TSA = 754 cm²; volume of wood left = 730.5 cm³.

Example 9: Packing a globe lamp: cube box or cylindrical tin? · 4 marks · Case-basedHOTS / competency

A shop sells a spherical globe lamp of diameter 21 cm. It can be packed in a cubical box of edge 21 cm, or in a cylindrical tin of radius 10.5 cm and height 21 cm. In each case the lamp just fits. Take $π = \frac{22}{7}$ .
(i) Find the volume of the lamp. (1 mark)
(ii) Find the empty space left in the cubical box. (1 mark)
(iii) Find, as a fraction of the container's volume, the empty space in each container, and state which container wastes less space. (2 marks)

Try it first, then show the solutionSolution

(i) $r = 10.5$ cm. Volume of lamp $= \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times 1157.625 = 4851$ cm³. [1]
(ii) Volume of cube $= 2 1^{3} = 9261$ cm³. Empty space $= 9261 - 4851 = 4410$ cm³. [1]
(iii) Fraction empty in cube $= \frac{4410}{9261} = \frac{10}{21}$ , dividing top and bottom by 441. [½]
Volume of tin $= π r^{2} h = 346.5 \times 21 = 7276.5$ cm³. Empty space $= 7276.5 - 4851 = 2425.5$ cm³. [½]
Fraction empty in tin $= \frac{2425.5}{7276.5} = \frac{1}{3}$ . [½]
Since $\frac{1}{3} = \frac{7}{21} < \frac{10}{21}$ , the cylindrical tin wastes less space. [½]

Answer: (i) 4851 cm³ (ii) 4410 cm³ (iii) Cube: 10/21 empty; tin: 1/3 empty. The cylindrical tin wastes less space.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Channapatna lacquered toy rockets

Channapatna in Karnataka is known for its lacquered wooden toys. A craftsman makes a toy rocket from a single piece of wood. It is a solid cylinder of base diameter 7 cm and height 10 cm, with a solid cone of the same base fixed on top. The vertical height of the cone is 12 cm. The whole outer surface of the rocket, including the flat bottom, is coated with coloured lacquer. The workshop receives an order for 50 such rockets. The lacquer work costs ₹5 for every 100 cm² of surface. Take $π = \frac{22}{7}$ .

(i) · 1 mark
Find the slant height of the conical part.
Answer
Answer: 12.5 cm
$r = 3.5$ cm, $h = 12$ cm. $l = 3. 5^{2} + 1 2^{2} = 12.25 + 144 = 156.25 = 12.5$ cm.
(ii) · 1 mark
Find the volume of wood in one rocket.
Answer
Answer: 539 cm³
$π r^{2} = \frac{22}{7} \times 12.25 = 38.5$ cm². Volume $= π r^{2} H + \frac{1}{3} π r^{2} h = 38.5 \times 10 + \frac{1}{3} \times 38.5 \times 12 = 385 + 154 = 539$ cm³.
(iii) · 2 marks
Find the area lacquered on one rocket, and the total cost of lacquering all 50 rockets.
Answer
Answer: 396 cm² per rocket; total cost ₹990
Exposed surfaces: bottom base, curved surface of cylinder and curved surface of cone. Area $= π r^{2} + 2 π r H + π r l = 38.5 + 220 + 137.5 = 396$ cm². For 50 rockets: $396 \times 50 = 19800$ cm². Cost $= \frac{19800}{100} \times 5 =$ ₹990.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

A toy is in the form of a cone of radius 7 cm and height 10 cm mounted on a hemisphere of the same radius. Find the volume of the toy. (2 marks)

Step	Marks
Writes volume = volume of cone + volume of hemisphere $= \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π r^{2} (h + 2 r)$	0.5
Substitutes correctly: $\frac{1}{3} \times \frac{22}{7} \times 49 \times (10 + 14)$	0.5
Simplifies: $\frac{1}{3} \times 154 \times 24 = 154 \times 8$	0.5
Final answer with units: volume $= 1232$ cm³	0.5

A cuboidal wooden block measures 14 cm by 10 cm by 7 cm. A cylindrical hole of radius 3.5 cm is drilled right through the block, from the centre of one 14 cm by 10 cm face to the centre of the opposite face, so the hole is 7 cm long. Find the total surface area of the remaining block. (3 marks)

Step	Marks
TSA of cuboid $= 2 (14 \times 10 + 10 \times 7 + 7 \times 14) = 2 (140 + 70 + 98) = 616$ cm²	1
Area of the two circular openings removed $= 2 π r^{2} = 2 \times \frac{22}{7} \times 3.5 \times 3.5 = 77$ cm²	0.5
Inner curved surface of the hole added $= 2 π r h = 2 \times \frac{22}{7} \times 3.5 \times 7 = 154$ cm²	1
TSA of remaining block $= 616 - 77 + 154 = 693$ cm²	0.5

A vessel is in the form of a hollow hemisphere surmounted by a hollow cylinder of the same radius. The diameter of the hemisphere is 21 cm and the total height of the vessel is 25.5 cm. The vessel is open at the top. Find (i) the capacity of the vessel in litres and (ii) its inner surface area. (5 marks)

Step	Marks
Radius $r = \frac{21}{2} = 10.5$ cm	0.5
Height of cylinder $H = 25.5 - 10.5 = 15$ cm	0.5
Capacity $= π r^{2} H + \frac{2}{3} π r^{3} = π r^{2} (H + \frac{2 r}{3})$	0.5
$= \frac{22}{7} \times 110.25 \times (15 + 7) = 346.5 \times 22 = 7623$ cm³	1
Capacity $= 7623 \div 1000 = 7.623$ litres	0.5
Inner surface area $= 2 π r H + 2 π r^{2} = 2 π r (H + r)$ (open top, joined circle hidden)	0.5
$= 2 \times \frac{22}{7} \times 10.5 \times (15 + 10.5) = 66 \times 25.5$	1
Inner surface area $= 1683$ cm²	0.5

Where students lose marks in Surface Areas and Volumes

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

Mistake: Adding the full TSA of both solids in a combination
Students recall the TSA formulas from Class 9 and add them without asking which faces are still visible after joining.
Wrong working
TSA of toy $= π r (l + r) + 3 π r^{2}$
Right working
TSA of toy $= π r l + 2 π r^{2}$ (both joined circular bases are hidden)
Do this: Before writing any formula, list the exposed surfaces in words. Use CSA, not TSA, for each part that is joined.
Usually costs: Usually 1 to 2 marks, because the method is wrong from the first line.
Mistake: Using vertical height instead of slant height for the cone's CSA
Both are called 'height' and the formula pi r l looks like pi r h when written in a hurry.
Wrong working
CSA of cone $= π r h = \frac{22}{7} \times 7 \times 24 = 528$ cm²
Right working
$l = 7^{2} + 2 4^{2} = 25$ cm, so CSA $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm²
Do this: Whenever a cone's curved surface is needed, write the line for l first, even if you think you know it.
Usually costs: ½ to 1 mark in a short answer; it also spoils every later step that uses the area.
Mistake: Treating the diameter as the radius
Questions often give the diameter, and r is cubed in volume formulas, so the error is huge but easy to miss.
Wrong working
Diameter = 14 cm, so volume of hemisphere $= \frac{2}{3} π (14)^{3}$
Right working
$r = \frac{14}{2} = 7$ cm, so volume $= \frac{2}{3} π (7)^{3}$
Do this: Write 'r = ...' as a separate first line every time, halving the diameter where needed.
Usually costs: Usually ½ to 1 mark, plus the final answer mark.
Mistake: Forgetting the inner surface when a solid is scooped or drilled
Students think 'something was removed, so subtract', and forget that a new inner wall is now exposed.
Wrong working
TSA after scooping a hemisphere from a cube $= 6 a^{2} - π r^{2}$
Right working
TSA $= 6 a^{2} - π r^{2} + 2 π r^{2} = 6 a^{2} + π r^{2}$
Do this: For any cavity use the three-part rule: outer TSA, minus opening, plus inner curved surface.
Usually costs: 1 mark or more in a 3-mark question.
Mistake: Using the total height of the toy as the cone's height
The question gives 'total height', and students plug it straight into Pythagoras.
Wrong working
Total height 31 cm, r = 7 cm, so $l = 3 1^{2} + 7^{2}$
Right working
$h = 31 - 7 = 24$ cm (subtract the hemisphere's radius), so $l = 2 4^{2} + 7^{2} = 25$ cm
Do this: Picture the toy: the total height is the cone's height plus the hemisphere's radius. Find h on its own line.
Usually costs: 1 to 2 marks, since l and every area or volume after it come out wrong.
Mistake: Wrong unit conversion or no units in the final answer
Capacity is asked in litres but worked in cm³, and students divide by 100 instead of 1000 or leave out units.
Wrong working
Capacity $= 7623$ cm³ $= 76.23$ litres
Right working
Capacity $= 7623 \div 1000 = 7.623$ litres
Do this: Remember 1 litre = 1000 cm³ and 1 m³ = 1000 litres. Write the unit on every final line.
Usually costs: ½ mark for the final answer.

NCERT exercise guide: Surface Areas and Volumes

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 12.1

What it tests: Surface area of solids made by joining two of the basic solids: a cone on a hemisphere, a cylinder with a hemisphere or cone at one end, a hemisphere on a cube, and two cones or a cylinder and cone joined together. Some questions ask for the cost of painting or polishing, or ask you to compare the area of the combination with the areas of the separate pieces.

How to approach it: Sketch the solid and mark every face that is hidden where the two parts touch. List only the exposed faces, such as curved surfaces and any flat faces still outside, and add their areas. If a dimension like cylinder height or slant height is not given, find it first from the total height or with $l = r^{2} + h^{2}$ .

Like the questions in Exercise 12.1 on a cylinder joined to a hemisphere

A wooden paperweight is made of a solid hemisphere of radius 7 cm. A solid cylinder of the same radius and height 8 cm stands on the flat face of the hemisphere, so that the flat face is completely covered. Find the total surface area of the paperweight. Take $π = \frac{22}{7}$ .

Show the solutionSolution

The flat face of the hemisphere and the bottom face of the cylinder touch, so both are hidden.
Exposed surfaces: the curved surface of the hemisphere, the curved surface of the cylinder and the top circular face of the cylinder.
TSA $= 2 π r^{2} + 2 π r h + π r^{2} = 3 π r^{2} + 2 π r h = π r (3 r + 2 h)$ .
$π r = \frac{22}{7} \times 7 = 22$ and $3 r + 2 h = 21 + 16 = 37$ .
TSA $= 22 \times 37 = 814$ cm².
Check: $3 π r^{2} = 462$ and $2 π r h = 352$ , and $462 + 352 = 814$ .

Answer: The total surface area of the paperweight is 814 cm².

Exercise 12.2

What it tests: Volume or capacity of solids made by joining two basic solids, such as a cylinder with a hemispherical or conical end, a cone on a hemisphere, or a cube or cuboid with a cylinder. Some questions need the answer in litres, and some need a missing height found first.

How to approach it: Volume does not depend on which faces are hidden. Add the volumes of the two parts that are joined, and subtract the volume of any part that is removed. Write the expression with $π r^{2}$ taken out as a common factor where you can, because this keeps the arithmetic short. Convert units at the end: $1000$ cm³ $= 1$ L and $1$ m³ $= 1000$ L.

Like the capacity questions in Exercise 12.2 on a cylinder with a hemispherical end

A water storage tank is a right circular cylinder of radius 1.05 m and height 2 m, closed at the bottom by a hemisphere of the same radius. Find the capacity of the tank in litres. Take $π = \frac{22}{7}$ .

Show the solutionSolution

Capacity $=$ volume of cylinder $+$ volume of hemisphere $= π r^{2} h + \frac{2}{3} π r^{3}$ .
$π r^{2} = \frac{22}{7} \times 1.05 \times 1.05 = 3.465$ m².
Volume of cylinder $= 3.465 \times 2 = 6.93$ m³.
Volume of hemisphere $= \frac{2}{3} \times 3.465 \times 1.05 = \frac{2}{3} \times 3.63825 = 2.4255$ m³.
Total volume $= 6.93 + 2.4255 = 9.3555$ m³.
Since $1$ m³ $= 1000$ L, capacity $= 9.3555 \times 1000 = 9355.5$ L.

Answer: The tank holds 9355.5 litres.

Surface Areas and Volumes: previous year questions

Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.2 marks per paper; 47 questions in all, counting OR alternatives. In 4 of 4 sample papers.
Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.6 marks per paper; 54 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Surface Areas and Volumes

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026 Basic, set 430/1/1, Q15 · 1 mark · MCQ

For which of the following solids is the lateral / curved surface area and total surface area the same ?

(A)Cube
(B)Cuboid
(C)Hemisphere
(D)Sphere

Try it first, then show our solutionSolution

A sphere has no flat faces, so its whole surface is curved: CSA = TSA $= 4 π r^{2}$ .
Cube, cuboid and hemisphere all have extra faces (top/bottom or flat base), so their TSA exceeds their LSA/CSA. Hence option (D).
1 mark

Answer: Sphere, option (D)

Where the marks are

Identifies sphere as CSA = TSA = $4 π r^{2}$ (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q34(a) · 5 marks · Long answer

A toy is in the form of a cone surmounted on a hemisphere. The cone and hemisphere have the same radii. The height of the conical part of the toy is equal to the diameter of its base. If the radius of the conical part is 5 cm, find the volume of the toy.

Try it first, then show our solutionSolution

Radius $r = 5$ cm. Height of cone $h$ = diameter $= 10$ cm. Volume of toy = volume of hemisphere + volume of cone $= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h$ .
Volume of hemisphere $= \frac{2}{3} \times \frac{22}{7} \times 5 \times 5 \times 5 = \frac{5500}{21}$ cm³
2 marks
Volume of cone $= \frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 10 = \frac{5500}{21}$ cm³
2 marks
Volume of toy $= \frac{5500}{21} + \frac{5500}{21} = \frac{11000}{21}$ cm³ $\approx 523.81$ cm³
1 mark

Answer: 11000/21 cm³ ≈ 523.81 cm³

Where the marks are

Correct volume of the hemisphere, $\frac{5500}{21}$ cm³ (2 marks)
Correct volume of the cone with $h = 10$ cm, $\frac{5500}{21}$ cm³ (2 marks)
Total volume $\frac{11000}{21} \approx 523.81$ cm³ (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/4/1, Q38 · 4 marks · Case-basedCompetency-based

'Gilli Danda' is a very popular traditional game of India which is played with two wooden sticks – the larger one is called 'Danda' and smaller one 'Gilli'.

'Danda' – It is cylindrical in shape with diameter $4$ cm and length $42$ cm.

Gilli – It is cylindrical in middle with identical conical ends of same radius $1.5$ cm and length $2.8$ cm. The length of cylindrical part is $7$ cm.

Based on the above, answer the following questions :

(i)
Find the volume of wood used in making both the conical parts of Gilli.
1 mark
(ii)
Find the volume of wood used in making cylindrical part of Gilli.
1 mark
(iii)
(a) A cylindrical log of wood of radius $1.5$ cm and length $14$ cm is used to make Gilli. Find the volume of the wood scrapped.
2 marks
OR
(iii)
(b) Find the total surface area of 'Danda'.
2 marks

Figure in the paper: Left: a long cylindrical stick (Danda) and a short stick (Gilli). Right: Danda is a long cylinder with diameter 4 cm and length 42 cm. Gilli is a cylinder of length 7 cm with a cone attached at each end; the cones have radius 1.5 cm and slant/height dimension 2.8 cm marked. See it in the official paper.

Try it first, then show our solutionSolution

(i) Each cone has $r = 1.5$ cm and $h = 2.8$ cm.
Volume of both cones $= 2 \times \frac{1}{3} π r^{2} h = 2 \times \frac{1}{3} \times \frac{22}{7} \times 1.5 \times 1.5 \times 2.8$
½ mark
$= \frac{2}{3} \times 19.8 = 13.2$ cm³
½ mark
(ii) The cylindrical part has $r = 1.5$ cm and $h = 7$ cm.
Volume $= π r^{2} h = \frac{22}{7} \times 1.5 \times 1.5 \times 7$
½ mark
$= 49.5$ cm³
½ mark
(iii)(a) Volume of the log $= π r^{2} h = \frac{22}{7} \times 1.5 \times 1.5 \times 14 = 99$ cm³
1 mark
Volume of the Gilli $= 13.2 + 49.5 = 62.7$ cm³
½ mark
Volume of wood scrapped $= 99 - 62.7 = 36.3$ cm³
½ mark
(iii)(b) The Danda has $r = \frac{4}{2} = 2$ cm and $h = 42$ cm.
TSA of a cylinder $= 2 π r^{2} + 2 π r h = 2 \times \frac{22}{7} \times 2 \times 2 + 2 \times \frac{22}{7} \times 2 \times 42$
1 mark
$= \frac{88}{7} \times (2 + 42) = \frac{3872}{7}$ cm² $\approx 553.14$ cm²
1 mark

Answer: (i) 13.2 cm³; (ii) 49.5 cm³; (iii)(a) 36.3 cm³; (iii)(b) $\frac{3872}{7}$ cm² ≈ 553.14 cm²

Where the marks are

Applies $2 \times \frac{1}{3} π r^{2} h$ with $r = 1.5$ , $h = 2.8$ (½ mark) and gets 13.2 cm³ (½ mark)
Applies $π r^{2} h$ with $h = 7$ (½ mark) and gets 49.5 cm³ (½ mark)
Finds the volume of the log as 99 cm³ (1 mark)
Finds the volume of the Gilli as 62.7 cm³ (½ mark)
Finds the wood scrapped as 36.3 cm³ (½ mark)
Uses TSA $= 2 π r^{2} + 2 π r h$ with $r = 2$ , $h = 42$ (1 mark)
Gets $\frac{3872}{7}$ cm² ≈ 553.14 cm² (1 mark)

See it in the full 2026 board paper, set 430/4/1

Every Surface Areas and Volumes question, by year

Standard (59 questions, sample papers included)

For 2027

Sample paper 2026-27: Q11 (1 mark, solved), Q34(A) (5 marks, solved), Q34(B) (5 marks, solved)

2026

Set 30/1/1: Q16 (1 mark, solved), Q30 (3 marks, solved)

Set 30/2/1: Q16 (1 mark, solved), Q19 (1 mark, solved), Q38 (4 marks, solved)

Set 30/3/1: Q7 (1 mark, solved), Q25 (2 marks, solved), Q38 (4 marks, solved)

Set 30/4/1: Q13 (1 mark, solved), Q16 (1 mark, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Set 30/5/1: Q7 (1 mark, solved), Q12 (1 mark, solved), Q37 (4 marks, solved)

Sample paper 2025-26: Q11 (1 mark, solved), Q34(A) (5 marks, solved), Q34(B) (5 marks, solved)

2025

Set 30/1/1: Q20 (1 mark, solved), Q30 (3 marks, solved)

Set 30/2/1: Q38 (4 marks, solved)

Set 30/3/1: Q12 (1 mark, solved), Q35 (5 marks, solved)

Set 30/4/1: Q11 (1 mark, solved), Q34 (5 marks, solved)

Set 30/5/1: Q15 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 30/6/1: Q14 (1 mark, solved), Q34 (5 marks, solved)

Sample paper 2024-25: Q5 (1 mark, solved), Q13 (1 mark, solved), Q38 (4 marks, solved)

2024

Set 30/1/1: Q13 (1 mark, solved), Q15 (1 mark, solved), Q31 (3 marks, solved)

Set 30/2/1: Q20 (1 mark, solved), Q37 (4 marks, solved)

Set 30/3/1: Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 30/4/1: Q9 (1 mark, solved), Q38 (4 marks, solved)

Set 30/5/1: Q35 (5 marks, solved)

Sample paper 2023-24: Q19 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

2023

Set 30/1/1: Q10 (1 mark, solved), Q12 (1 mark, solved), Q31(a) (3 marks, solved), Q31(b) (3 marks, solved)

Set 30/2/1: Q10 (1 mark, solved), Q37 (4 marks, solved)

Set 30/4/1: Q34 (5 marks, solved)

Set 30/5/1: Q36 (4 marks, solved)

Set 30/6/1: Q12 (1 mark, solved), Q33 (5 marks, solved)

Basic (62 questions, sample papers included)

For 2027

Sample paper 2026-27: Q18 (1 mark, solved), Q34 (5 marks, solved)

2026

Set 430/1/1: Q15 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 430/2/1: Q15 (1 mark, solved), Q34 (5 marks, solved)

Set 430/3/1: Q15 (1 mark, solved), Q34 (5 marks, solved)

Set 430/4/1: Q3 (1 mark, solved), Q11 (1 mark, solved), Q38 (4 marks, solved)

Set 430/5/1: Q4 (1 mark, solved), Q18 (1 mark, solved), Q37 (4 marks, solved)

Sample paper 2025-26: Q35 (5 marks, solved)

2025

Set 430/1/1: Q15 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 430/2/1: Q15 (1 mark, solved), Q34 (5 marks, solved)

Set 430/3/1: Q15 (1 mark, solved), Q34 (5 marks, solved)

Set 430/4/1: Q6 (1 mark, solved), Q15 (1 mark, solved), Q37 (4 marks, solved)

Set 430/5/1: Q10 (1 mark, solved), Q19 (1 mark, solved), Q30(a) (3 marks, solved), Q30(b) (3 marks, solved)

Set 430/6/1: Q4 (1 mark, solved), Q18 (1 mark, solved), Q37 (4 marks, solved)

Sample paper 2024-25: Q14 (1 mark, solved), Q38 (4 marks, solved)

2024

Set 430/1/1: Q12 (1 mark, solved), Q14 (1 mark, solved), Q30(A) (3 marks, solved), Q30(B) (3 marks, solved)

Set 430/2/1: Q13 (1 mark, solved), Q35(a) (5 marks, solved), Q35(b) (5 marks, solved)

Set 430/3/1: Q10 (1 mark, solved), Q16 (1 mark, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 430/4/1: Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Set 430/5/1: Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Sample paper 2023-24: Q17 (1 mark, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

2023

Set 430/1/1: Q13 (1 mark, solved), Q33 (5 marks, solved)

Set 430/2/1: Q8 (1 mark, solved), Q11 (1 mark, solved), Q38 (4 marks, solved)

Set 430/4/1: Q35 (5 marks, solved)

Set 430/5/1: Q20 (1 mark, solved), Q37 (4 marks, solved)

Set 430/6/1: Q34 (5 marks, solved)

Practice questions on Surface Areas and Volumes

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
Two identical solid cones, each of base radius 3.5 cm and height 12 cm, are joined base to base. Find the total surface area of the solid formed. Take $π = \frac{22}{7}$ .
Hint
The two bases are hidden, so only the two curved surfaces remain. Find the slant height first.
Answer
275 cm². Slant height $l = 3. 5^{2} + 1 2^{2} = 12.5$ cm. TSA $= 2 π r l = 2 \times \frac{22}{7} \times 3.5 \times 12.5 = 275$ cm².
Q2 · 1 mark · MCQBoard level
A cone of base radius $r$ and height $h$ is placed on top of a cylinder of base radius $r$ and height $h$ . The volume of the combined solid is:
(a) $\frac{4}{3} π r^{2} h$
(b) $\frac{2}{3} π r^{2} h$
(c) $2 π r^{2} h$
(d) $\frac{5}{3} π r^{2} h$
Hint
Add the two volumes and take $π r^{2} h$ common.
Answer
(a) $\frac{4}{3} π r^{2} h$ . Volume $= π r^{2} h + \frac{1}{3} π r^{2} h = \frac{4}{3} π r^{2} h$ .
Q3 · 1 mark · Assertion-reasonBoard level
Assertion (A): The largest sphere that can be carved out of a solid cube of edge 14 cm has volume $1437 \frac{1}{3}$ cm³.
Reason (R): The diameter of the largest sphere that fits inside a cube is equal to the edge of the cube.
(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.
Hint
Use R to get the radius, then calculate the volume of the sphere.
Answer
(a). By R, $r = 7$ cm. Volume $= \frac{4}{3} \times \frac{22}{7} \times 343 = \frac{4312}{3} = 1437 \frac{1}{3}$ cm³, so A is true and R explains it.
Q4 · 3 marks · Short answerHarder board level
A cuboidal tank has a base 33 cm long and 21 cm wide, and is 30 cm high. It contains water to a depth of 20 cm. A solid sphere of diameter 21 cm is lowered into the tank and is fully under water. Find the rise in the water level and the new depth of water. Take $π = \frac{22}{7}$ .
Hint
The water displaced equals the volume of the sphere. Spread this volume over the base of the tank.
Answer
Rise = 7 cm; new depth = 27 cm. Volume of sphere $= \frac{4}{3} \times \frac{22}{7} \times 10. 5^{3} = 4851$ cm³. Rise $= \frac{4851}{33 \times 21} = \frac{4851}{693} = 7$ cm. New depth $= 27$ cm, which is less than 30 cm, so no water spills.
Q5 · 3 marks · Short answerHarder board level
A toy is a solid cylinder of radius 3.5 cm with a solid hemisphere of the same radius fixed on top. The total height of the toy is 15.5 cm. Find the total surface area of the toy. Take $π = \frac{22}{7}$ .
Hint
Height of the cylinder = total height minus the radius of the hemisphere.
Answer
379.5 cm². Cylinder height $= 15.5 - 3.5 = 12$ cm. TSA $= π r^{2} + 2 π r h + 2 π r^{2} = π r (3 r + 2 h) = 11 \times (10.5 + 24) = 11 \times 34.5 = 379.5$ cm².
Q6 · 5 marks · Long answerHarder board level
A school trophy has a solid cuboidal base 20 cm long, 16 cm wide and 5 cm high. A solid cylinder of radius 7 cm and height 10 cm stands on the centre of the top face of the base. Find (i) the total surface area of the trophy, (ii) its volume, and (iii) the cost of painting the whole trophy at ₹5 per 100 cm². Take $π = \frac{22}{7}$ .
Hint
The cylinder covers a circle on the cuboid, but its own top face is a circle of the same size, so these two cancel out.
Answer
(i) 1440 cm² (ii) 3140 cm³ (iii) ₹72. TSA of cuboid $= 2 (320 + 80 + 100) = 1000$ cm². TSA $= 1000 - π r^{2} + π r^{2} + 2 π r h = 1000 + 440 = 1440$ cm². Volume $= 1600 + 154 \times 10 = 3140$ cm³. Cost $= \frac{1440}{100} \times 5 =$ ₹72.
Q7 · 3 marks · Short answerHarder board level
A solid cube has volume 2744 cm³. A conical depression of base radius 3.5 cm and depth 12 cm is cut out from the centre of its top face. Find the total surface area and the volume of the remaining solid. Take $π = \frac{22}{7}$ .
Hint
Find the edge of the cube first. The cut removes a flat circle and adds the curved surface of the cone.
Answer
TSA = 1275 cm²; volume = 2590 cm³. Edge $= 32744 = 14$ cm, $l = 12.5$ cm. TSA $= 6 \times 196 - 38.5 + \frac{22}{7} \times 3.5 \times 12.5 = 1176 - 38.5 + 137.5 = 1275$ cm². Volume $= 2744 - \frac{1}{3} \times 38.5 \times 12 = 2744 - 154 = 2590$ cm³.
Q8 · 3 marks · Short answerHarder board level
A solid wooden cylinder of diameter 21 cm and height 35 cm has a hemispherical depression of the same diameter scooped out of each end. Find (i) the total surface area of the remaining solid, (ii) the volume of wood left, and (iii) the cost of polishing all its surfaces at ₹5 per 100 cm². Take $π = \frac{22}{7}$ .
Hint
Both flat ends are lost and two bowls of $2 π r^{2}$ each are gained. Use $π r = 33$ cm and $π r^{2} = 346.5$ cm².
Answer
(i) 3696 cm² (ii) 7276.5 cm³ (iii) ₹184.80. $r = 10.5$ cm and $2 r = 21 \leq 35$ , so the bowls do not meet. TSA $= 2 π r h + 4 π r^{2} = 2 \times 33 \times 35 + 4 \times 346.5 = 2310 + 1386 = 3696$ cm². Volume $= π r^{2} (h - \frac{4 r}{3}) = 346.5 \times (35 - 14) = 7276.5$ cm³. Cost $= \frac{3696}{100} \times 5 =$ ₹184.80.
Q9 · 5 marks · Long answerHarder board level
A planetarium building is a cylinder of diameter 21 m and height 8 m with a hemispherical dome of the same diameter on top. Find (i) the volume of air inside the building, and (ii) the cost of painting the outer curved wall and the dome at ₹20 per m². Take $π = \frac{22}{7}$ .
Hint
Use $π r^{2} = 346.5$ m² and $π r = 33$ m with $r = 10.5$ m. The floor and the hidden circle are not painted.
Answer
(i) 5197.5 m³ (ii) ₹24420. Volume $= 346.5 \times 8 + \frac{2}{3} \times 346.5 \times 10.5 = 2772 + 2425.5 = 5197.5$ m³. Area $= 2 π r h + 2 π r^{2} = 528 + 693 = 1221$ m². Cost $= 1221 \times 20 =$ ₹24420.
Q10 · 3 marks · Short answerHOTS / competency
A solid sphere of radius $r$ fits exactly inside a closed cylinder, touching its base, its top and its curved surface. Show that the curved surface area of the cylinder equals the surface area of the sphere. Then find the ratio of the volume of the cylinder to the volume of the sphere, and the fraction of the cylinder left empty.
Hint
If the sphere touches the top and the base, what is the height of the cylinder in terms of $r$ ?
Answer
Height $= 2 r$ . CSA of cylinder $= 2 π r \times 2 r = 4 π r^{2}$ , which is the surface area of the sphere. Volumes: $2 π r^{3} : \frac{4}{3} π r^{3} = 3 : 2$ . Empty fraction $= 1 - \frac{2}{3} = \frac{1}{3}$ .
Q11 · 3 marks · Short answerHOTS / competency
A kulfi seller offers two servings. Serving A is a cone of radius 3.5 cm and height 14 cm filled with kulfi, with a hemisphere of kulfi of the same radius on top. Serving B is a cylindrical cup of radius 3.5 cm and height 7 cm filled level to the brim. The seller says Serving A gives more kulfi. Is he right? Justify with calculation. Take $π = \frac{22}{7}$ .
Hint
Find both volumes using $π r^{2} = 38.5$ cm², then compare.
Answer
He is not right: both hold 269.5 cm³. A: $\frac{1}{3} \times 38.5 \times 14 + \frac{2}{3} \times 38.5 \times 3.5 = \frac{1}{3} \times 38.5 \times (14 + 7) = 269.5$ cm³. B: $38.5 \times 7 = 269.5$ cm³.

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.When two solids are joined, which surfaces are left out of the total surface area?Show answer
The faces where the two solids touch. They are sealed inside, so they cannot be seen or painted.
2.Formula for the slant height of a cone?Show answer
$l = r^{2} + h^{2}$
3.How does the volume of a cone compare with a cylinder of the same base and height?Show answer
It is one-third: $\frac{1}{3} π r^{2} h$ compared with $π r^{2} h$ .
4.Curved surface area and total surface area of a solid hemisphere?Show answer
CSA $= 2 π r^{2}$ ; TSA $= 3 π r^{2}$ .
5.Two cubes of edge $a$ are joined end to end. What is the surface area of the cuboid formed?Show answer
$10 a^{2}$ (12 faces minus the 2 hidden ones).
6.How many cm³ make 1 litre, and how many litres make 1 m³?Show answer
1 litre = 1000 cm³; 1 m³ = 1000 litres.
7.Volume ratio of a cone, a hemisphere and a cylinder, each of radius $r$ and height $r$ ?Show answer
1 : 2 : 3
8.A hemisphere of radius $r$ is scooped out of a face of a cube of edge $a$ . What is the new TSA?Show answer
$6 a^{2} + π r^{2}$ : lose $π r^{2}$ of flat face, gain $2 π r^{2}$ of curved inner surface.

Best order to study it

Revise all Class 9 formulas for cube, cuboid, cylinder, cone, sphere and hemisphere until you can write them from memory, including l = √(r² + h²).
Learn the hidden-face idea with simple cases: two cubes joined, then a hemisphere on a cube.
Practise surface area of common combinations: cone on hemisphere, cone on cylinder, hemisphere on cylinder, always listing exposed surfaces first.
Move to volume of combinations, both adding (joined solids) and subtracting (scooped, drilled or water-left questions), including litre conversions.
Practise multi-step problems where a height, radius or slant height must be found first, then both surface area and volume.
Finish with real-life cost, canvas and capacity problems and case-study sets, timing yourself on 5-mark answers.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write the CSA, TSA and volume of a cylinder, cone and hemisphere without looking?
Can you find the slant height of a cone with r = 7 cm and h = 24 cm in your head?
Can you write the TSA of a cone mounted on a hemisphere of the same radius as a single factorised expression?
Given the total height of a toy, can you find the cone's height correctly?
Can you find the surface area of the cuboid formed by joining two equal cubes of edge a?
Can you write the TSA left after scooping a hemisphere from a cube, and explain why it increases?
Can you find the TSA of a cuboid with a cylindrical hole drilled right through it?
Can you write the volume of a cylinder with a cone on top using a common factor?
Can you convert 3080 cm³ into litres, and 2.5 m³ into litres?
Can you decide which surfaces count for a tent with no floor and for an open vessel?

Surface Areas and Volumes: frequently asked questions

Is the frustum of a cone in the Class 10 syllabus for 2026-27?

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

Should I use 22/7 or 3.14 for pi?

Use 22/7 unless the question tells you to use 3.14. Board numbers are usually chosen so that 22/7 cancels neatly with radii such as 3.5, 7, 10.5 or 21 cm. If the question gives a value of pi, always use that one.

How do I know which surfaces to include in the total surface area?

Imagine painting the object from outside. Any face you can paint counts, and any face where the two solids are joined does not. For hollowed solids, also count the inner wall of the cavity, because it is now exposed.

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

Yes. Every question in this chapter uses the basic formulas for cubes, cuboids, cylinders, cones, spheres and hemispheres. The new skill is combining them correctly, so learn the basic formulas thoroughly first.

Why is my answer for a toy's surface area always too big?

You are probably using the total surface area of the cone and the hemisphere and adding them. That counts the joined circular bases, which are hidden. Use only the curved surface of each part: pi r l for the cone and 2 pi r squared for the hemisphere.

Do I lose marks for leaving out units?

Often yes. The final answer mark usually expects the correct unit: square units for area, cubic units for volume, and litres or rupees when the question asks for them. Write the unit on the last line of every answer.

Based on: CBSE Mathematics syllabus 2026-27; NCERT Mathematics, Textbook for Class X (Reprint 2026-27)