Class 10 Mathematics Chapter 12 Notes: Surface Areas and Volumes

Recognise a combination of solids: an object made by joining two basic solids (cuboid, cube, cylinder, cone, sphere, hemisphere, which is half a sphere)
Find the total surface area (TSA) of such an object by adding only the surfaces you can actually see
Find the volume of such an object by adding (or subtracting) the volumes of its parts
Spot the ring that appears when two joined solids have different radii (radii is the plural of radius), and include its area

combination of solidscurved surface area (CSA)total surface area (TSA)
volumeslant heighthemispherecapacity

1 Introduction

In Class IX you learnt the surface areas and volumes of the cuboid, cone, cylinder and sphere.
Many everyday objects are made by joining two or more of these basic solids:

a truck container carrying oil or water: you may guess that it is made of a cylinder with two hemispheres as its ends
a test tube in the science lab: a cylinder and a hemisphere
big buildings and monuments: made of combinations of such solids

No single formula fits such objects, so we break them into basic solids we already know.

Figure 1: The basic solids from Class IX that every combined solid is built from — **Figure 1**: The basic solids from Class IX that every combined solid is built from

Learn by heartDefinition 1

CSA (Curved Surface Area) is the area of only the curved part of a solid’s surface, leaving out its flat top and bottom. A cuboid or cube has no curved part, so for them we use the lateral surface area: the area of the 4 side walls only.

Learn by heartDefinition 2

TSA (Total Surface Area) is the area of the whole outer surface of a solid: the curved part and the flat faces together.

Class IX formulas this chapter builds on

Solid	CSA (lateral area for cuboid, cube)	Total surface area (TSA)	Volume
Cuboid (length $l$ , breadth $b$ , height $h$ )	$2h(l+b)$ (4 side walls)	$2(lb+bh+hl)$	$lbh$
Cube (edge $a$ )	$4a^2$ (4 side walls)	$6a^2$	$a^3$
Cylinder (radius $r$ , height $h$ )	$2\pi rh$	$2\pi r(r+h)$	$\pi r^2 h$
Cone (radius $r$ , height $h$ , slant height $l$ )	$\pi rl$	$\pi r(l+r)$	$\tfrac13 \pi r^2 h$
Sphere (radius $r$ )	$4\pi r^2$ (the whole surface is curved)		$\tfrac43 \pi r^3$
Hemisphere (radius $r$ )	$2\pi r^2$	$3\pi r^2$	$\tfrac23 \pi r^3$

The slant height

l

of a cone is the length of its sloping side, from the tip to the edge of the base. Its height

h

, radius

r

and

l

form a right triangle, so by Pythagoras theorem

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

Which value of π?

Both NCERT exercises of this chapter (12.1 and 12.2) begin with “Unless stated otherwise, take $\pi = \frac{22}{7}$ “. Use $3.14$ only
when the question says so. When a radius is a multiple of 0.7 (like 0.7, 1.4, 3.5, 7), the 7 in $\frac{22}{7}$ divides into it and the working stays simple.

2 Surface Area of a Combination of Solids

Take the truck container, seen as a cylinder with a hemisphere stuck at each end. When the pieces
are joined, the flat circular faces where they touch go inside the object. From outside you can
see only three surfaces: the curved surface of each hemisphere and the curved surface of the cylinder.

The container

TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere

Learn by heartRule 1: surface area of a combination

The total surface area of a combination of solids is the sum of the areas of only those surfaces
that remain on the outside. It is not the sum of the TSAs of the parts, because some surface
disappears inside when the parts are joined.

The same idea works for a toy made from a cone and a hemisphere. We keep the base radius of the cone
equal to the radius of the hemisphere so that the toy has a smooth surface. Their flat faces are pressed
together and vanish, so:

The toy

Total surface area of the toy = CSA of hemisphere + CSA of cone

Common Mistake

Adding the TSAs of the parts counts the hidden joining faces. Always ask: which faces went inside?

2.1 Example 1: cone and hemisphere (Rasheed’s playing top)

Rasheed’s playing top (lattu) is a cone surmounted by (topped with) a hemisphere. The whole top is 5 cm high and
its diameter is 3.5 cm. Find the area he has to colour ( $\pi=\frac{22}{7}$ ).

Figure 3: The top: TSA = CSA of hemisphere + CSA of cone (same radius r) — **Figure 3**: The top: TSA = CSA of hemisphere + CSA of cone (same radius r)

Solved Example 1

$r=\frac{3.5}{2}=1.75$ cm. Height of cone $h=$ height of the whole top $–$ radius of hemisphere (a hemisphere is as tall as its radius) $=5-1.75=3.25$ cm
Slant height $l=\sqrt{r^2+h^2}=\sqrt{(1.75)^2+(3.25)^2}\approx 3.7$ cm

CSA of hemisphere $=\tfrac12(4\pi r^2)=2\pi r^2$ and CSA of cone $=\pi rl$
TSA $=2\pi r^2+\pi rl=\pi r(2r+l)=\dfrac{22}{7}\times1.75\times(3.5+3.7)=\dfrac{22}{7}\times1.75\times7.2\approx$
39.6 cm²

2.2 Example 2: cube and hemisphere (the decorative block)

A decorative block is a cube of edge 5 cm with a hemisphere of diameter 4.2 cm fixed on top. Find
its total surface area ( $\pi=\frac{22}{7}$ ).

Figure 4: The block: the circle under the hemisphere is hidden, and the hemisphere's curved surface is added — **Figure 4**: The block: the circle under the hemisphere is hidden, and the hemisphere’s curved surface is added

Solved Example 2

TSA of cube $=6\times(\text{edge})^2=6\times5\times5=150$ cm². The part of the cube’s top covered by the
hemisphere (a circle of area $\pi r^2$ ) is no longer on the outside, but the hemisphere adds its curved surface $2\pi r^2$ . Here $r=2.1$ cm.

Surface area of block $=$ TSA of cube $–$ base area of hemisphere $+$ CSA of hemisphere
$=150-\pi r^2+2\pi r^2=150+\pi r^2=150+\dfrac{22}{7}\times2.1\times2.1=150+13.86=$ 163.86 cm²

2.3 Example 3: wide cone on a narrower cylinder (the toy rocket and the ring)

A wooden toy rocket is a cone mounted on a cylinder: total height 26 cm, cone height 6 cm, cone base
diameter 5 cm, cylinder base diameter 3 cm. The cone is painted orange and the cylinder yellow. Find
each painted area ( $\pi=3.14$ ).

Figure 5: The cone's base is wider than the cylinder, so a ring of the cone's base stays visible and is painted — **Figure 5**: The cone’s base is wider than the cylinder, so a ring of the cone’s base stays visible and is painted

Solved Example 3

Cone: $r=2.5$ cm, $h=6$ cm, $l=\sqrt{r^2+h^2}=\sqrt{2.5^2+6^2}=\sqrt{42.25}=6.5$ cm.
Cylinder: $r'=1.5$ cm, $h'=26-6=20$ cm (the dash in $r'$ , $h'$ just marks the cylinder’s values).

Orange $=$ CSA of cone $+$ base area of cone $–$ base area of cylinder $=\pi rl+\pi r^2-\pi r'^2$
$=3.14\times[(2.5\times6.5)+(2.5)^2-(1.5)^2]=3.14\times20.25=$ 63.585 cm²

Yellow $=$ CSA of cylinder $+$ area of one base of cylinder $=2\pi r'h'+\pi r'^2=\pi r'(2h'+r')$
$=(3.14\times1.5)(2\times20+1.5)=4.71\times41.5=$ 195.465 cm²

Common Mistake

When the two radii are different, part of the wider base stays outside as a flat ring. Its area is
$\pi r^2-\pi r'^2=\pi(r^2-r'^2)$ , and it must be added. Students who assume equal radii leave this area out and get the wrong total.

2.4 Example 4: cylinder with a hemispherical depression (Mayank’s bird-bath)

Mayank’s bird-bath is a cylinder, 1.45 m high with radius 30 cm, with a hemispherical depression
(a bowl-shaped hollow) at one end. Find its total surface area ( $\pi=\frac{22}{7}$ ).

Solved Example 4

The hollow has the same radius as the cylinder, so the cylinder’s top face is gone and the curved inside of
the hemisphere is seen in its place. The book’s solution counts only the cylinder’s curved surface and the hemisphere’s curved surface, so the flat base of the cylinder is not included.
Take $h=1.45$ m $=145$ cm (same unit as $r$ ).

TSA $=$ CSA of cylinder $+$ CSA of hemisphere $=2\pi rh+2\pi r^2=2\pi r(h+r)$
$=2\times\dfrac{22}{7}\times30\times(145+30)=33000$ cm² $=$ 3.3 m² (as $1\text{ m}^2=10000\text{ cm}^2$ )

2.5 What stays on the outside: common shapes at a glance

Object	Surfaces that count	TSA
Two cubes (edge $a$ ) joined end to end	TSA of both cubes minus the two hidden faces	$12a^2-2a^2=10a^2$
Hollow hemisphere with hollow cylinder on it (inside of a container)	inner CSA of cylinder + inner CSA of hemisphere	$2\pi rh+2\pi r^2$
Cone on hemisphere (toy, top)	CSA of cone + CSA of hemisphere	$\pi rl+2\pi r^2$
Cylinder with two hemispherical ends (capsule, container)	CSA of cylinder + 2 hemisphere CSAs	$2\pi rh+4\pi r^2$
Cylinder with a conical (cone-shaped) top, open base (tent)	CSA of cylinder + CSA of cone	$2\pi rh+\pi rl$ (canvas area; the open base is not counted, Ex 12.1 Q7)
Hemisphere of radius $r$ on a cube of edge $a$ , or a hemispherical depression cut into it (Ex 12.1 Q5)	cube TSA $–$ circle $+$ hemisphere CSA	$6a^2+\pi r^2$
Conical cavity (a cone-shaped hollow) cut from a cylinder of the same $r$ , $h$	CSA of cylinder + bottom base + inside of cone	$2\pi rh+\pi r^2+\pi rl$
Hemisphere scooped out of each end of a cylinder	CSA of cylinder + 2 hollow hemisphere CSAs	$2\pi rh+4\pi r^2$

These eight shapes cover every question type in Exercise 12.1.

Exam Tip

A cavity or depression removes a flat face but adds a new curved face inside. So cutting material
out can make the surface area bigger. Draw a rough figure first and tick each face that is still
on the outside, so that no hidden face is added by mistake.

3 Volume of a Combination of Solids

Learn by heartRule 2: volume of a combination

The volume of a solid formed by joining two basic solids is the sum of the volumes of its
constituents (the parts it is made of), because no material is lost in joining them. If a part is cut out or hollowed out, its volume is subtracted.

3.1 Example 5: cuboid and half cylinder (Shanta’s shed)

Shanta runs an industry in a shed shaped like a cuboid surmounted by a half cylinder. The base is
7 m × 15 m and the cuboid part is 8 m high. Find the volume of air the shed can hold. If machinery takes
300 m³ and 20 workers take about 0.08 m³ each, how much air is left ( $\pi=\frac{22}{7}$ )?

Figure 6: The shed: volume = volume of cuboid + half the volume of a cylinder (diameter 7 m, length 15 m) — **Figure 6**: The shed: volume = volume of cuboid + half the volume of a cylinder (diameter 7 m, length 15 m)

Solved Example 5

Cuboid: $15$ m $\times 7$ m $\times 8$ m. Half cylinder: diameter $7$ m ( $r=3.5$ m), height (length) $15$ m.

Volume $=$ volume of cuboid $+\tfrac12$ volume of cylinder $=15\times7\times8+\tfrac12\times\dfrac{22}{7}\times\dfrac72\times\dfrac72\times15$
$=840+288.75=$ 1128.75 m³

Space taken by machinery and workers $=300+20\times0.08=300+1.60=301.60$ m³
Air in the shed $=1128.75-301.60=$ 827.15 m³

3.2 Example 6: a cylinder minus a hemisphere (the juice glass)

A juice seller’s glass is a cylinder of inner diameter 5 cm and height 10 cm, but its bottom has a
raised hemispherical portion that reduces how much juice it holds. Capacity means the volume of liquid a container can hold. Find its apparent capacity (what it seems to hold) and
its actual capacity (what it really holds) ( $\pi=3.14$ ).

Figure 7: Actual capacity = apparent capacity (whole cylinder) minus the raised hemisphere — **Figure 7**: Actual capacity = apparent capacity (whole cylinder) minus the raised hemisphere

Solved Example 6

$r=2.5$ cm, $h=10$ cm.
Apparent capacity $=\pi r^2h=3.14\times2.5\times2.5\times10=196.25$ cm³
Volume of hemisphere $=\tfrac23\pi r^3=\tfrac23\times3.14\times2.5\times2.5\times2.5=32.71$ cm³ (approx.)
Actual capacity $=196.25-32.71=$ 163.54 cm³

3.3 Example 7: hemisphere with a cone, inside a circumscribing cylinder

A solid toy is a hemisphere surmounted by a right circular cone (a cone whose tip is straight above the centre of its base). The cone is 2 cm high and the base
diameter is 4 cm. Find the volume of the toy. A right circular cylinder (an upright cylinder) circumscribes the toy (fits
exactly around it). Find the difference between the volumes of the cylinder and the toy ( $\pi=3.14$ ).

Figure 8: A is the tip of the cone, O the centre of the base, P the lowest point. Cylinder EFGH: radius HP = BO = 2 cm, height EH = AO + OP = 4 cm — **Figure 8**: A is the tip of the cone, O the centre of the base, P the lowest point. Cylinder EFGH: radius HP = BO = 2 cm, height EH = AO + OP = 4 cm

Solved Example 7

In Figure 8, O is the centre of the common base BC, A is the tip of the cone, P is the lowest point of the hemisphere, and EFGH is the cylinder.
Radius $BO=\tfrac12\times4=2$ cm (for both hemisphere and cone), cone height $h=2$ cm.

Volume of toy $=\tfrac23\pi r^3+\tfrac13\pi r^2h=\left[\tfrac23\times3.14\times2^3+\tfrac13\times3.14\times2^2\times2\right]=$ 25.12 cm³

Cylinder: radius $=2$ cm, height $EH=AO+OP=2+2=4$ cm, volume $=3.14\times2^2\times4=50.24$ cm³
Difference $=50.24-25.12=$ 25.12 cm³

3.4 Question types in Exercise 12.2

Type	What to do	Example in NCERT Exercise 12.2
Add the parts	sum the volumes of all pieces	cone on hemisphere; Rachel’s model; gulab jamun
Subtract a hollow	outer solid minus the removed part	pen stand; water left in a cylinder
Count objects	number $=$ total volume $\div$ volume of one object	lead shots (small lead balls) in a conical vessel
Find mass	mass $=$ volume $\times$ mass of 1 cm³	solid iron pole
Check a claim	calculate the true volume, then compare	spherical glass vessel

Exam Tip

When solid objects are dropped into a vessel that is full of water, the water that flows out has the same
volume as the objects. So in the lead-shot question: volume of water that overflows $=$ number of shots $\times$
volume of one shot. For a percentage (such as the syrup inside a sweet), find the full volume first, then take that
percentage of it.

4 How to Solve Any Combined-Solid Question

Step 1Name the basic solids and draw a rough figure

↓

Step 2Write

r

h

l

of each part; change all lengths to one unit; get missing heights from the total height

↓

Step 3Surface area: add only outside faces (watch for a ring or a cavity). Volume: add parts, subtract cut-outs

↓

Step 4Take out common factors (

\pi r

etc.) before multiplying, then write the unit (cm², cm³, m², m³)

✓ Do this

Halve the diameter before using $r$
Find $l=\sqrt{r^2+h^2}$ when only the cone’s height is given
For mass, multiply volume by the mass of 1 cm³

✗ Never do this

Use the cone’s height where its slant height is needed
Mix metres and centimetres in one formula
Forget the unit (cm² for area, cm³ for volume)

Quick Revision: read this the night before the exam

TSA of a combined solid = only the outside surfaces; never the sum of the parts’ TSAs
Volume of a combined solid = sum of the parts’ volumes; subtract any cavity or raised part
Different radii at a joint: add the ring $\pi(r^2-r'^2)$
A cavity removes a flat face but adds a curved face, so surface area can increase
Hemisphere: CSA $2\pi r^2$ , TSA $3\pi r^2$ , volume $\tfrac23\pi r^3$

Next, practise with this chapter’s important questions, MCQs and full NCERT solutions.

More for this chapter

Important Questions MCQ / Objective NCERT Solutions Short Notes