MathsCBSE Class 9 · Chapter 14

Math of Space: Surface Area and Volume Class 9 Formulas and Solved Questions

Updated 15 October 2026 · Fact-checked

Math of Space: Surface Area and Volume in Class 9 gives the surface areas and volumes of cuboids, cubes, cylinders, right circular cones, pyramids, spheres and hemispheres. You see why a cone holds one-third of a cylinder, why a sphere's surface is 4πr², and use the formulas in packing, painting, melting and estimation problems.

What Math of Space: Surface Area and Volume is about

This last chapter of *Ganita Manjari* Part II is CBSE's Surface Area and Volume (Unit V, Mensuration). The CBSE content names spheres (including hemispheres) and right circular cones, with competencies covering cuboids, cubes, cylinders, cones, pyramids and spheres.

The chapter builds each formula from an idea you can check:

Cuboids and cubes: the same volume can have different surface areas (a cube needs the least wrapping), which matters in packaging and heat loss.
Cylinders: the curved surface unrolls into a rectangle of sides $2 π r$ and h.
Cones: the curved surface unrolls into a sector, giving $π r l$ ; filling a cone with salt three times fills a cylinder of the same base and height, so the volume is $\frac{1}{3} π r^{2} h$ .
Pyramids: volume $\frac{1}{3} \times$ base area $\times$ height; a cone is a pyramid with a circular base.
Spheres and hemispheres: Archimedes showed the surface area is $4 π r^{2}$ (string wound on a ball fills four circles) and the volume $\frac{4}{3} π r^{3}$ .
Guesstimates: sensible estimates (golgappas from a vessel of pani, tennis balls in a classroom) using assumptions and approximations.

Most questions are numerical. Write the formula, substitute with units, use the value of π given, and give area in square units and volume in cubic units (1000 cm³ = 1 litre).

Weightage: This chapter is CBSE's Surface Area and Volume, part of Unit V, Mensuration, which carries 14 of the 80 marks in the school annual exam, shared with Area and Perimeter.Source: CBSE Mathematics syllabus, Class IX 2026-27, course structure, Unit V Mensuration

Key concepts

Cuboids and cubes

NCERT section 14.1 Cuboids and Cubes

A cuboid with length l, width w and height h has six rectangular faces:

total surface area $= 2 (l w + w h + h l)$
volume $= l w h$

A cube of side a is a cuboid with $l = w = h = a$ : surface area $6 a^{2}$ , volume $a^{3}$ .

Same volume, different surface area: a cube of side 6 cm and a 9 × 6 × 4 cuboid both hold 216 cm³, but the cuboid has more surface. For a fixed volume, a cube needs the least covering material, while a flatter shape exposes more area (useful for cooling, wasteful for packaging).

Worked example

Compare a cube of side 8 cm with a cuboid 16 cm × 8 cm × 4 cm: their volumes and their surface areas.

Try it first, then show the solutionSolution

Cube: $V = 8^{3} = 512$ cm³; $S A = 6 \times 64 = 384$ cm².
Cuboid: $V = 16 \times 8 \times 4 = 512$ cm³.
$S A = 2 (128 + 32 + 64) = 448$ cm².
Same volume; the cuboid needs 64 cm² more material to cover.

Answer: Both 512 cm³; surface areas 384 cm² (cube) and 448 cm² (cuboid).

Right circular cylinder

NCERT section 14.2 Right Circular Cylinder

Cut a cylinder (radius r, height h) along a straight line and unroll it: the curved surface becomes a rectangle with sides $2 π r$ (the circumference) and h.

curved surface area $= 2 π r h$
total surface area $= 2 π r h + 2 π r^{2} = 2 π r (r + h)$
volume $=$ base area $\times$ height $= π r^{2} h$

An open cylinder (a tank without a lid) has only one circular face.

Worked example

A closed cylinder has radius 7 cm and height 10 cm. Find its curved surface area, total surface area and volume (use $π \approx \frac{22}{7}$ ).

Try it first, then show the solutionSolution

CSA $= 2 \times \frac{22}{7} \times 7 \times 10 = 440$ cm².
TSA $= 2 \times \frac{22}{7} \times 7 \times (7 + 10) = 748$ cm².
$V = \frac{22}{7} \times 49 \times 10 = 1540$ cm³.

Answer: CSA 440 cm², TSA 748 cm², volume 1540 cm³.

Right circular cone: slant height and surface area

NCERT section 14.3 Cones

Rotating a right triangle about one of its perpendicular sides makes a right circular cone: radius r, height h, and slant height l with
$l = r^{2} + h^{2}$

Cut along a slant edge and the curved surface opens into a sector of radius l whose arc is the base circumference $2 π r$ . Its area is

curved surface area $= π r l$
total surface area $= π r l + π r^{2} = π r (l + r)$

Use l (not h) for surface area.

Worked example

A cone has radius 7 cm and height 24 cm. Find its slant height, curved surface area and total surface area.

Try it first, then show the solutionSolution

$l = 49 + 576 = 625 = 25$ cm.
CSA $= \frac{22}{7} \times 7 \times 25 = 550$ cm².
TSA $= 550 + \frac{22}{7} \times 49 = 550 + 154 = 704$ cm².

Answer: $l = 25$ cm, CSA 550 cm², TSA 704 cm².

Volume of a cone: one-third of a cylinder

NCERT section 14.3 Cones

Fill a cone with fine salt and pour it into a cylinder of the same base radius and height: it takes exactly three cones to fill the cylinder. So
$V_{cone} = \frac{1}{3} π r^{2} h$
You can also see it with clay: a clay cylinder can be remoulded into exactly three such cones. Archimedes (c. 225 BCE) first found this formula.

Worked example

A cone and a cylinder have the same radius 7 cm and the same height 24 cm. Find both volumes and check the 1 : 3 relation.

Try it first, then show the solutionSolution

Cylinder: $\frac{22}{7} \times 49 \times 24 = 3696$ cm³.
Cone: $\frac{1}{3} \times 3696 = 1232$ cm³.
$1232 \times 3 = 3696$ .

Answer: Cone 1232 cm³, cylinder 3696 cm³: ratio 1 : 3.

Pyramids

NCERT section 14.4 Pyramidal Shapes

A pyramid has a flat base (triangle, square or any shape) with every point of the base joined to one apex outside the base's plane.

volume $= \frac{1}{3} \times$ base area $\times$ height (perpendicular height of the apex above the base)
surface area $=$ the sum of the areas of all the faces (no special formula needed)

A cone is a pyramid with a circular base, which is why both have the factor $\frac{1}{3}$ . For a square pyramid, find the slant height of a triangular face with the Baudhāyana–Pythagoras theorem.

Worked example

A pyramid has a square base of side 10 m and height 12 m. Find its volume and its total surface area.

Try it first, then show the solutionSolution

$V = \frac{1}{3} \times 100 \times 12 = 400$ m³.
Slant height of each triangular face: from the centre of the base to the middle of a side is 5 m, so $1 2^{2} + 5^{2} = 13$ m.
Each face $= \frac{1}{2} \times 10 \times 13 = 65$ m²; four faces $= 260$ m².
Total $= 260 + 100 = 360$ m².

Answer: Volume 400 m³; total surface area 360 m².

Sphere

NCERT section 14.5 Spheres and Hemispheres

A sphere is the set of all points in space at distance r (the radius) from a centre.

surface area $= 4 π r^{2}$
volume $= \frac{4}{3} π r^{3}$

Why (Archimedes): the sphere's surface equals the curved surface of the cylinder that just fits round it ( $r$ and $h = 2 r$ ): $2 π r \times 2 r = 4 π r^{2}$ . Hands-on check: string wound over a ball exactly fills four circles of the ball's radius.

Volume from area: cut the surface into tiny pieces and join each to the centre; you get thin cones of height r, so $V = \frac{1}{3} \times 4 π r^{2} \times r = \frac{4}{3} π r^{3}$ .

Worked example

Find the surface area and volume of a sphere of radius 21 cm, and check that volume $= \frac{1}{3} \times$ surface area $\times$ radius.

Try it first, then show the solutionSolution

$S A = 4 \times \frac{22}{7} \times 441 = 5544$ cm².
$V = \frac{4}{3} \times \frac{22}{7} \times 9261 = 38808$ cm³.
Check: $\frac{1}{3} \times 5544 \times 21 = 38808$ .

Answer: 5544 cm² and 38808 cm³.

Hemisphere

NCERT section 14.5 Spheres and Hemispheres

A hemisphere is half a sphere, cut by a plane through the centre. It has one curved surface and one flat circular base.

curved surface area $= 2 π r^{2}$
total surface area (solid) $= 2 π r^{2} + π r^{2} = 3 π r^{2}$
volume $= \frac{2}{3} π r^{3}$

A bowl or a dome has only the curved surface (no base), so use $2 π r^{2}$ for painting a dome.

Worked example

A hemispherical bowl has inner radius 10.5 cm. How much water can it hold, in litres?

Try it first, then show the solutionSolution

$V = \frac{2}{3} \times \frac{22}{7} \times 10. 5^{3} = \frac{2}{3} \times \frac{22}{7} \times 1157.625$ .
$= 2425.5$ cm³.
1000 cm³ = 1 L, so about 2.43 L.

Answer: 2425.5 cm³, about 2.4 litres.

Scaling: what happens when lengths grow

If every length of a solid is multiplied by k:

every area is multiplied by $k^{2}$
every volume is multiplied by $k^{3}$

So a 10% larger radius ( $k = 1.1$ ) gives $1.21$ times the surface area (21% more) and $1.331$ times the volume (33.1% more). This is why cutting a cube into small cubes keeps the volume but multiplies the surface area, which matters for how quickly things dissolve, cook or cool.

Worked example

The radius of a sphere increases by 10%. By what percentage do its surface area and volume increase?

Try it first, then show the solutionSolution

$k = 1.1$ .
Surface area factor $= 1. 1^{2} = 1.21$ : an increase of 21%.
Volume factor $= 1. 1^{3} = 1.331$ : an increase of 33.1%.

Answer: Surface area up 21%, volume up 33.1%.

Guesstimates

NCERT section 14.6 Areas and Volumes Around Us

A guesstimate is a reasonable estimate when exact data are not available. The method:
1. Make a quick intuitive guess.
2. Model the situation with shapes (a vessel as a cylinder, a ball as a sphere).
3. State your assumptions (size, how full, how tightly packed).
4. Calculate with sensible approximations, and compare with your first guess.

Dividing a big volume by a small one gives only an upper bound for packing spheres, because spheres leave gaps. Different assumptions give different but equally sensible answers; the reasoning is what matters.

Worked example

Estimate how many tennis balls of radius 3 cm fit in a cupboard 1.8 m × 0.9 m × 0.6 m if they are stacked in straight rows and columns.

Try it first, then show the solutionSolution

Each ball needs a 6 cm cube of space in this packing.
Along the length: $180 \div 6 = 30$ ; width: $90 \div 6 = 15$ ; height: $60 \div 6 = 10$ .
$30 \times 15 \times 10 = 4500$ balls.
Staggered packing would fit a few more; volume division would wrongly suggest more than 8000.

Answer: About 4500 tennis balls.

Maths in real life: Math of Space: Surface Area and Volume

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.

Same volume, less surface

Suppose a sweet shop wants a box holding 512 cm³ of sweets and can choose a cube of side 8 cm or a 16 × 8 × 4 cm box.

Which uses less cardboard?

Try it first, then show the solutionSolution

Cube: $6 \times 64 = 384$ cm².
Cuboid: $2 (128 + 32 + 64) = 448$ cm².

Answer: The cube, by 64 cm².

Memory hook: For the same volume, the cube wraps tightest.

Cone volume

Suppose a street vendor's paper cone for roasted peanuts has radius 3.5 cm and height 12 cm.

How many cubic centimetres of peanuts does a full cone hold (level with the top)?

Try it first, then show the solutionSolution

$V = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 12$ .
$= 154$ cm³.

Answer: 154 cm³

Memory hook: A cone is a third of its cylinder.

Sphere surface

Suppose a football has radius 11 cm and its outer leather costs 2 paise per cm².

Estimate the cost of the leather (use $π \approx 3.14$ ).

Try it first, then show the solutionSolution

$S A = 4 \times 3.14 \times 121 \approx 1519.8$ cm².
Cost $\approx 1519.8 \times 0.02 \approx ₹30.40$ .

Answer: About ₹30

Memory hook: Four circles' worth of leather covers a ball.

Cylinder capacity

Suppose your building's cylindrical overhead tank has radius 0.7 m and height 1.5 m.

How many litres can it hold?

Try it first, then show the solutionSolution

$V = \frac{22}{7} \times 0.49 \times 1.5 = 2.31$ m³.
$= 2310$ L.

Answer: 2310 litres

Memory hook: 1 cubic metre = 1000 litres.

Guesstimate

Suppose a golgappa seller has a cylindrical vessel of pani of radius 10 cm, filled to 50 cm, and each golgappa holds about 10 cm³ of pani.

Roughly how many golgappas can he serve?

Try it first, then show the solutionSolution

Pani $= π \times 100 \times 50 \approx 15700$ cm³.
$15700 \div 10 \approx 1570$ .

Answer: About 1500 to 1600, depending on how full each golgappa is.

Memory hook: Guess, model, assume, calculate, compare.

Topics in this chapter

Math of Space: Surface Area and Volume formula sheet

Cuboid: $T S A = 2 (l w + w h + h l), V = l w h$
Cube: $T S A = 6 a^{2}, V = a^{3}$
Cylinder: $C S A = 2 π r h, T S A = 2 π r (r + h), V = π r^{2} h$
Slant height of a cone: $l = r^{2} + h^{2}$
Cone surface: $C S A = π r l, T S A = π r (l + r)$
Cone volume: $V = \frac{1}{3} π r^{2} h$; One-third of the cylinder with the same base and height.
Pyramid: $V = \frac{1}{3} \times base area \times height$; Surface area = sum of the face areas.
Sphere: $S A = 4 π r^{2}, V = \frac{4}{3} π r^{3}$
Hemisphere: $C S A = 2 π r^{2}, T S A = 3 π r^{2}, V = \frac{2}{3} π r^{3}$
Sphere volume from area: $V = \frac{1}{3} \times S A \times r$
Scaling: $lengths \times k \Rightarrow areas \times k^{2}, volumes \times k^{3}$
Units: $1000 cm^{3} = 1 L, 1 m^{3} = 1000 L$
Melting and recasting: $number of new solids = \frac{volume of original}{volume of one new solid}$; Volume is unchanged when a solid is reshaped.

How the school exam asks Math of Space: Surface Area and Volume

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

Question type	How it is asked	Method
Direct formula questions	Find CSA, TSA or volume of a cylinder, cone, sphere or hemisphere from given dimensions.	Write the formula, substitute with units, simplify, and give the right unit (cm² or cm³).
Find a missing dimension	Volume or surface area is given; find the radius, height or slant height.	Substitute the known values into the formula and solve; for a cone use $l^{2} = r^{2} + h^{2}$ to move between l and h.
Cost and capacity	Cost of painting a dome or tin for a cone; litres in a tank; time to fill at a given rate.	Use the right surface (curved only for domes and caps) or volume; convert units (cm³ to litres, cm² to m²) before multiplying by the rate.
Combinations and packing	Spheres packed in a box with liquid filling the rest; two cubes joined; an ice-cream cone with a hemispherical top.	Add or subtract volumes; for joined solids, remove the faces that are hidden where they touch.
Melting and recasting, scaling	How many small solids from a large one; percentage change in area or volume when lengths change.	Equate volumes; for scaling use $k^{2}$ for areas and $k^{3}$ for volumes.
Guesstimates	Estimate scoops in an ice-cream box, balls in a room, golgappas from a vessel.	Guess first, model with simple shapes, state assumptions, calculate roughly, and comment on how assumptions change the answer.

Tricks to remember Math of Space: Surface Area and Volume

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

Three cones fill one cylinder
Same base, same height: pour three cones of salt to fill the cylinder.
Use it when: Never forgetting the $\frac{1}{3}$ in the cone and pyramid volume.
Four circles cover a ball
The string that covers a ball fills four circles of the same radius: $4 π r^{2}$ .
Use it when: Sphere surface area.
Area squares, volume cubes
Double the lengths: area ×4, volume ×8.
Use it when: Scaling questions.
Slant for skin, height for volume
Use the slant height l for a cone's surface and the vertical height h for its volume.
Use it when: Avoiding the most common cone slip.
Hemisphere: two for the dome, three for the solid
Curved $2 π r^{2}$ ; add the lid $π r^{2}$ to get $3 π r^{2}$ .
Use it when: Bowls, domes and solid hemispheres.

Math of Space: Surface Area and Volume: exam-level solved questions

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

Example 1: Cube from its surface area · 1 mark · MCQExam level

A cube has total surface area 150 cm². Its volume is
(a) 25 cm³
(b) 125 cm³
(c) 150 cm³
(d) 216 cm³

Try it first, then show the solutionSolution

$6 a^{2} = 150$ gives $a = 5$ .
$V = 125$ cm³.

Answer: (b)

Example 2: Cone and cylinder · 1 mark · MCQExam level

A cone and a cylinder have equal base radii and equal heights. The ratio of their volumes is
(a) 1 : 1
(b) 1 : 2
(c) 1 : 3
(d) 3 : 1

Try it first, then show the solutionSolution

$\frac{1}{3} π r^{2} h : π r^{2} h = 1 : 3$ .

Answer: (c)

Example 3: Solid hemisphere (assertion-reason) · 1 mark · Assertion-reasonExam level

Assertion (A): The total surface area of a solid hemisphere of radius r is $3 π r^{2}$ .
Reason (R): A solid hemisphere has a curved surface of area $2 π r^{2}$ and a flat circular base of area $π r^{2}$ .
(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

$2 π r^{2} + π r^{2} = 3 π r^{2}$ , exactly as R explains.

Answer: (a)

Example 4: Curved surface of a cone · 2 marks · Very short answerExam level

Find the curved surface area of a cone of radius 7 cm and slant height 10 cm.

Try it first, then show the solutionSolution

CSA $= π r l = \frac{22}{7} \times 7 \times 10$ .
$= 220$ cm².

Answer: 220 cm²

Example 5: Two cubes joined · 3 marks · Short answerHarder exam level

Two cubes, each of volume 125 cm³, are joined end to end. Find the surface area of the new solid, and explain why it is not $2 \times 150$ cm².

Try it first, then show the solutionSolution

Each cube has side 5 cm; together they make a 10 × 5 × 5 cuboid.
$S A = 2 (50 + 25 + 50) = 250$ cm².
Two faces of 25 cm² are pressed together and hidden: $300 - 50 = 250$ .

Answer: 250 cm²

Example 6: Radius from volume · 3 marks · Short answerHarder exam level

A cone of height 24 cm has volume 1232 cm³. Find its radius, slant height and curved surface area.

Try it first, then show the solutionSolution

$\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 24 = 1232$ , so $r^{2} = 49$ and $r = 7$ cm.
$l = 49 + 576 = 25$ cm.
CSA $= \frac{22}{7} \times 7 \times 25 = 550$ cm².

Answer: $r = 7$ cm, $l = 25$ cm, CSA 550 cm²

Example 7: Painting a dome · 3 marks · Short answerHarder exam level

A hemispherical dome has a base circumference of 44 m. Find the area to be painted and the cost at ₹12 per m².

Try it first, then show the solutionSolution

$2 π r = 44$ , so $r = 7$ m.
Curved surface $= 2 π r^{2} = 308$ m² (no base on a dome).
Cost $= 308 \times 12 = ₹3, 696$ .

Answer: 308 m²; ₹3,696

Example 8: Spheres in a box · 3 marks · Short answerHarder exam level

Eight metal spheres of radius 3 cm are packed in a cubical box of internal side 12 cm, and the rest of the box is filled with oil. Find the volume of oil (use $π \approx \frac{22}{7}$ ), to the nearest cm³.

Try it first, then show the solutionSolution

Box $= 1 2^{3} = 1728$ cm³.
Spheres $= 8 \times \frac{4}{3} \times \frac{22}{7} \times 27 \approx 905.14$ cm³.
Oil $\approx 1728 - 905.14 = 822.86$ , about 823 cm³.

Answer: About 823 cm³

Example 9: Canvas for a tent · 5 marks · Long answerHarder exam level

A tent is a cylinder of radius 6 m and height 3 m topped by a cone of height 8 m on the same base. Find the canvas needed (no floor) and its cost at ₹50 per m² (use $π \approx 3.14$ ).

Try it first, then show the solutionSolution

Cone slant height $= 36 + 64 = 10$ m.
Cylinder curved surface $= 2 π \times 6 \times 3 = 36 π$ .
Cone curved surface $= π \times 6 \times 10 = 60 π$ .
Total $= 96 π \approx 301.44$ m².
Cost $= 301.44 \times 50 = ₹15, 072$ .

Answer: 301.44 m²; ₹15,072

Example 10: Growing a sphere · 3 marks · Short answerHOTS / competency

The radius of a balloon increases by 10% as it is blown up. By what percentage do its surface area and volume increase? Explain why the volume grows faster.

Try it first, then show the solutionSolution

Area factor $1. 1^{2} = 1.21$ : up 21%.
Volume factor $1. 1^{3} = 1.331$ : up 33.1%.
Area depends on $r^{2}$ and volume on $r^{3}$ , so a change in r is multiplied once more for volume.

Answer: Area up 21%, volume up 33.1%.

Example 11: Melting a sphere into cones · 4 marks · Case-basedHOTS / competency

A metal sphere of radius 6 cm is melted and recast into cones of radius 3 cm and height 4 cm. How many cones are made? Is the total surface area of the cones more or less than that of the sphere?

Try it first, then show the solutionSolution

Sphere volume $= \frac{4}{3} π \times 216 = 288 π$ ; cone volume $= \frac{1}{3} π \times 9 \times 4 = 12 π$ .
Number of cones $= \frac{288 π}{12 π} = 24$ .
Each cone: $l = 5$ , TSA $= π \times 3 \times (5 + 3) = 24 π$ ; all 24: $576 π$ .
Sphere: $4 π \times 36 = 144 π$ . The cones have 4 times the surface, though the volume is unchanged.

Answer: 24 cones; their total surface ( $576 π$ cm²) is more than the sphere's ( $144 π$ cm²).

Case-study practice (Section E style)

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

The school's water storage

A school's water tank is a cylinder of radius 1.4 m and height 2 m, closed by a hemispherical dome of the same radius on top. The dome is to be painted to reflect sunlight. Use $π \approx \frac{22}{7}$ .

(i) · 1 mark
Find the volume of the cylindrical part.
Answer
Answer: 12.32 m³
$\frac{22}{7} \times 1.96 \times 2 = 12.32$ m³.
(ii) · 1 mark
Find the volume of the hemispherical dome.
Answer
Answer: About 5.75 m³
$\frac{2}{3} \times \frac{22}{7} \times 2.744 \approx 5.75$ m³.
(iii) · 2 marks
Find the total capacity in litres and the area of the dome to be painted.
Answer
Answer: About 18,070 L; 12.32 m²
Total $\approx 12.32 + 5.75 = 18.07$ m³ $\approx 18, 070$ L (1 mark). Dome curved surface $= 2 π r^{2} = 2 \times \frac{22}{7} \times 1.96 = 12.32$ m² (1 mark).

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.

Find the surface area of a sphere of radius 7 cm. (2 marks)

Step	Marks
$S A = 4 π r^{2} = 4 \times \frac{22}{7} \times 7 \times 7$	1
$= 616$ cm²	1

A cone has radius 7 cm and height 24 cm. Find its slant height, curved surface area and volume. (3 marks)

Step	Marks
$l = 7^{2} + 2 4^{2} = 25$ cm	1
CSA $= \frac{22}{7} \times 7 \times 25 = 550$ cm²	1
$V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = 1232$ cm³	1

A tent is a cylinder of radius 6 m and height 3 m with a conical top of height 8 m on the same base. Find the area of canvas needed (no floor) and its cost at ₹50 per m² (use $π \approx 3.14$ ). (5 marks)

Step	Marks
Slant height of the cone: $l = 6^{2} + 8^{2} = 10$ m	1
Cylinder CSA $= 2 π \times 6 \times 3 = 36 π$ m²	1
Cone CSA $= π \times 6 \times 10 = 60 π$ m²	1
Canvas $= 96 π \approx 96 \times 3.14 = 301.44$ m²	1
Cost $= 301.44 \times 50 = ₹15, 072$	1

Where students lose marks in Math of Space: Surface Area and Volume

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

Mistake: Using height instead of slant height in the cone's surface
h and l are mixed up.
Wrong working
Cone $r = 7$ , $h = 24$ : CSA $= π \times 7 \times 24$
Right working
$l = 25$ , so CSA $= π \times 7 \times 25 = 550$ cm²
Do this: Find l first whenever a cone's surface is asked.
Usually costs: Most of the marks.
Mistake: Counting the base of a dome or bowl
The solid hemisphere formula is used for an open shape.
Wrong working
Paint for a dome $= 3 π r^{2}$
Right working
A dome has no base to paint: $2 π r^{2}$
Do this: Ask which surfaces are actually there.
Usually costs: 1 to 2 marks.
Mistake: Mixing units
Radius in cm and height in m, or volume in cm³ converted wrongly to litres.
Wrong working
$V = π \times 14 0^{2} \times 2$ with r in cm and h in m
Right working
Use r = 1.4 m and h = 2 m: $V = 12.32$ m³ $= 12320$ L
Do this: Convert everything to one unit before substituting.
Usually costs: The final answer.
Mistake: Adding all faces when solids are joined
Hidden faces are counted.
Wrong working
Two 5 cm cubes joined: SA $= 2 \times 150 = 300$ cm²
Right working
The new cuboid is 10 × 5 × 5: SA $= 2 (50 + 25 + 50) = 250$ cm² (two faces of 25 cm² are hidden)
Do this: Find the new shape's dimensions and use its formula.
Usually costs: 1 to 2 marks.
Mistake: Forgetting to cube when scaling volume
The percentage is applied directly.
Wrong working
Radius up 10%, so volume up 10%.
Right working
Volume factor $1. 1^{3} = 1.331$ : up 33.1%.
Do this: Areas use $k^{2}$ , volumes $k^{3}$ .
Usually costs: The whole answer.

NCERT exercise guide: Math of Space: Surface Area and Volume

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 Set 14.1

What it tests: Cube surface area from volume, packing small cubes in a big box, boxes in a godown, two cubes joined, cutting a cube into small cubes, a cuboid from its face areas, and (starred) painted cubes and cuboids with a given surface area.

How to approach it: Use $V = l w h$ and $S A = 2 (l w + w h + h l)$ ; for packing, divide each dimension by the small size and multiply.

Like the face-areas question in Exercise Set 14.1

The three faces of a cuboid that meet at a corner have areas 12 cm², 20 cm² and 15 cm². Find its volume.

Show the solutionSolution

$l w = 12$ , $w h = 20$ , $h l = 15$ .
Multiply: $(l w h)^{2} = 12 \times 20 \times 15 = 3600$ .
$V = l w h = 60$ cm³.

Answer: 60 cm³

Exercise Set 14.2

What it tests: Curved and total surface areas and volumes of cylinders, including models and containers.

How to approach it: Unroll the curved surface into a rectangle $2 π r \times h$ ; add circles as needed; $V = π r^{2} h$ .

Like the cylinder questions in Exercise Set 14.2

A cylindrical water tank has radius 1.4 m and height 2 m. How many litres does it hold, and how long does a pump delivering 40 L per minute take to fill it?

Show the solutionSolution

$V = \frac{22}{7} \times 1.96 \times 2 = 12.32$ m³.
$12.32 \times 1000 = 12320$ L.
Time $= 12320 \div 40 = 308$ minutes (5 h 8 min).

Answer: 12,320 L; 308 minutes

Exercise Set 14.3

What it tests: Cone surface areas from slant height or height, radius from volume or curved surface, a joker's cap, and a cone half-filled with water.

How to approach it: Find l or h from $l^{2} = r^{2} + h^{2}$ ; CSA $= π r l$ ; $V = \frac{1}{3} π r^{2} h$ .

Like the radius-from-volume question in Exercise Set 14.3

A cone of height 24 cm has volume 1232 cm³. Find its radius and curved surface area.

Show the solutionSolution

$\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 24 = 1232$ gives $r^{2} = 49$ , $r = 7$ cm.
$l = 49 + 576 = 25$ cm.
CSA $= \frac{22}{7} \times 7 \times 25 = 550$ cm².

Answer: $r = 7$ cm; CSA 550 cm²

Exercise Set 14.4

What it tests: Surface areas of spheres and hemispheres, ratios of surface areas, spheres packed in a box with liquid, percentage change in volume, and painting a dome.

How to approach it: Use $4 π r^{2}$ , $\frac{4}{3} π r^{3}$ and the hemisphere formulas; for ratios use $k^{2}$ and $k^{3}$ .

Like the dome-painting question in Exercise Set 14.4

The base of a hemispherical dome has circumference 44 m. Find the cost of painting the dome at ₹12 per m².

Show the solutionSolution

$2 π r = 44$ gives $r = 7$ m.
Curved surface $= 2 π r^{2} = 2 \times \frac{22}{7} \times 49 = 308$ m².
Cost $= 308 \times 12 = ₹3, 696$ .

Answer: ₹3,696

End-of-chapter Exercises

What it tests: Guesstimates (ice-cream scoops, objects in a room), unit cubes needed to grow a cube, and mixed surface area and volume problems.

How to approach it: Guess first, model with simple shapes, state assumptions, then calculate; check how the answer depends on the assumptions.

Like the cube-growing question at the end of the chapter

A cube of side a is built from unit cubes. How many unit cubes must be added to make a cube of side $a + 1$ ? Check for $a = 3$ .

Show the solutionSolution

$(a + 1)^{3} - a^{3} = 3 a^{2} + 3 a + 1$ .
For $a = 3$ : $27 + 9 + 1 = 37$ , and $64 - 27 = 37$ .

Answer: $3 a^{2} + 3 a + 1$ unit cubes (37 when $a = 3$ ).

Practice questions on Math of Space: Surface Area and Volume

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

Q1 · 1 mark · MCQExam level
The total surface area of a cube of side 4 cm is
(a) 64 cm²
(b) 96 cm²
(c) 16 cm²
(d) 48 cm²
Hint
$6 a^{2}$
Answer
(b) 96 cm²
Q2 · 1 mark · MCQExam level
The slant height of a cone of radius 6 cm and height 8 cm is
(a) 14 cm
(b) 10 cm
(c) 7 cm
(d) 100 cm
Hint
$r^{2} + h^{2}$
Answer
(b) 10 cm
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): A cube and a cuboid with the same volume can have different surface areas.
Reason (R): A cube of side 6 cm and a 9 cm × 6 cm × 4 cm cuboid both have volume 216 cm³ but surface areas 216 cm² and 228 cm².
(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
Check R's numbers.
Answer
(a). R gives a valid example that shows A.
Q4 · 2 marks · Very short answerExam level
Find the volume of a hemispherical bowl of radius 2.1 cm.
Hint
$\frac{2}{3} π r^{3}$
Answer
$\frac{2}{3} \times \frac{22}{7} \times 9.261 = 19.404$ cm³
Q5 · 2 marks · Very short answerHarder exam level
The faces of a cuboid meeting at a corner have areas 12 cm², 20 cm² and 15 cm². Find its volume.
Hint
Multiply the three areas.
Answer
$12 \times 20 \times 15 = 3600 = 60$ cm³
Q6 · 2 marks · Very short answerHarder exam level
How many cubes of side 25 cm can be packed in a cubical box of side 2 m?
Hint
Divide each edge first.
Answer
$(200 \div 25)^{3} = 8^{3} = 512$
Q7 · 3 marks · Short answerHarder exam level
The curved surface area of a cone is 550 cm² and its slant height is 25 cm. Find its radius, height and volume.
Hint
$π r l = 550$ .
Answer
$r = 7$ cm, $h = 625 - 49 = 24$ cm, $V = 1232$ cm³.
Q8 · 2 marks · Very short answerHarder exam level
The radii of two spheres are in the ratio 2 : 3. Find the ratio of their surface areas and of their volumes.
Hint
Square and cube the ratio.
Answer
Surface areas 4 : 9; volumes 8 : 27.
Q9 · 3 marks · Short answerHarder exam level
An ice-cream cone has radius 3.5 cm and height 12 cm and is filled with ice cream, with a hemisphere of ice cream on top. Find the total volume of ice cream.
Hint
Cone + hemisphere.
Answer
Cone $= \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 12 = 154$ cm³; hemisphere $= \frac{2}{3} \times \frac{22}{7} \times 42.875 \approx 89.83$ cm³; total $\approx 243.83$ cm³.
Q10 · 5 marks · Long answerHarder exam level
A cylindrical tank of radius 1.4 m and height 2 m is to be filled by a pump delivering 40 litres a minute. Find its capacity in litres and the time to fill it. If the radius were doubled, how would the time change?
Hint
1 m³ = 1000 L; volume scales with $r^{2}$ here.
Answer
$V = 12.32$ m³ $= 12320$ L; time $= 308$ min. Doubling r multiplies the volume by 4, so the time becomes 1232 min.
Q11 · 3 marks · Short answerHOTS / competency
A 4 cm cube is painted on all faces and then cut into 1 cm cubes. How many small cubes have exactly 3, 2, 1 and 0 painted faces? Check that the counts add up.
Hint
Corners, edges, faces, inside.
Answer
3 faces: 8 corners; 2 faces: $12 \times 2 = 24$ ; 1 face: $6 \times 4 = 24$ ; 0 faces: $2^{3} = 8$ . Total $8 + 24 + 24 + 8 = 64 = 4^{3}$ .
Q12 · 3 marks · Short answerHOTS / competency
Guesstimate: about how many tennis balls of radius 3 cm would fit in a cupboard 1.8 m × 0.9 m × 0.6 m? State your assumptions, and explain why dividing the volumes gives too many.
Hint
Each ball needs a 6 cm cube in straight stacking.
Answer
Straight rows: $30 \times 15 \times 10 = 4500$ balls. Dividing volumes ( $\approx 972000 \div 113 \approx 8600$ ) ignores the gaps between spheres, so it is only an upper limit.

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.Write the total surface area of a cuboid.Show answer
$2 (l w + w h + h l)$
2.Write the curved surface area of a cylinder.Show answer
$2 π r h$
3.How is the slant height of a cone found?Show answer
$l = r^{2} + h^{2}$
4.What is the curved surface area of a cone?Show answer
$π r l$
5.How does a cone's volume compare with a cylinder of the same base and height?Show answer
It is one-third.
6.What is the volume of a pyramid?Show answer
$\frac{1}{3} \times$ base area $\times$ height
7.Who found the surface area and volume of a sphere?Show answer
Archimedes.
8.Write the surface area and volume of a sphere.Show answer
$4 π r^{2}$ and $\frac{4}{3} π r^{3}$
9.What is the total surface area of a solid hemisphere?Show answer
$3 π r^{2}$
10.If all lengths are doubled, what happens to the volume?Show answer
It becomes 8 times as large.

Best order to study it

Revise cuboid and cube formulas and work the same-volume, different-area example (Exercise Set 14.1).
Learn the cylinder formulas by unrolling a paper cylinder (Exercise Set 14.2).
Learn slant height, the cone's surface (unrolled sector) and its volume (three cones fill a cylinder) (Exercise Set 14.3).
Add pyramids: one-third base times height, faces added for surface area.
Learn sphere and hemisphere formulas with Archimedes' reasoning (Exercise Set 14.4).
Practise scaling, melting and packing problems, then guesstimates and the end-of-chapter exercises.

Quick revision checklist

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

Can you find the surface area of a cube whose volume is 216 cm³?
Can you find the volume of a cuboid whose three face areas are given?
Can you find the TSA of a closed cylinder?
Can you find a cone's slant height and use it for the curved surface?
Can you explain why a cone holds one-third of a cylinder?
Can you find the volume of a square pyramid?
Can you find the surface area and volume of a sphere of radius 3.5 cm?
Can you find the paint needed for a hemispherical dome?
Can you say how volume changes when the radius increases by 20%?
Can you make a sensible guesstimate and state your assumptions?

Math of Space: Surface Area and Volume: frequently asked questions

Which solids are in the Class 9 Surface Area and Volume syllabus for 2026-27?

CBSE's content names spheres (including hemispheres) and right circular cones, and its competencies include cuboids, cubes, cylinders, cones, pyramids and spheres. The NCERT chapter covers all of these.

Do I need to prove the sphere formulas?

No formal proof is expected. Know the reasoning: Archimedes' cylinder for 4\pi r^2, the string-and-four-circles activity, and the thin-cones idea giving V = \frac{1}{3} \times SA \times r.

Which value of π should I use?

The one the question gives (\frac{22}{7} or 3.14). Choose \frac{22}{7} when the radius is a multiple of 7, as it cancels neatly.

Are frustums or combinations of solids in Class 9?

Frustums are not in the chapter. Simple combinations (two cubes joined, a cone on a cylinder, spheres in a box, an ice-cream cone with a hemispherical top) are good practice and use only the formulas on this page.

What is a guesstimate question and how is it marked?

It asks for a sensible estimate using assumptions. Marks go to clear reasoning: the model, the assumptions, the calculation and a sensible final number, not to one exact answer.

Based on: CBSE Mathematics syllabus, Class IX 2026-27; NCERT Ganita Manjari, Part II, Mathematics Textbook for Grade 9 (First Edition 2026)