171
Mensuration
INTRODUCTION
In previous classes, you have learnt the concept of the surface area and the volume of
cubes and cuboids, measurement of surface area and volume of these solids by using basic
units. In this chapter, we shall extend our study to some other solids such as right circular
cylinders, right circular cones, spheres and combination of such solids. We shall also take
up the problems of melting and recasting i.e. conversion of solids from one shape to another
and finding the volume or the surface area of new solids.
17.1 Surface Area and Volume of a Solid
❑Solid. A figure which occupies a portion of space enclosed by plane or curved surfaces is called
a solid. A solid has three dimensions — length, breadth and height.
❑Surface area of a solid. The sum of the areas of the plane or curved surfaces (faces) of
a solid is called its total surface area. It is measured in square units.
By the words ‘Surface area of a solid’ we shall mean the total surface area of the
solid.
❑Lateral surface area of a solid. The surface area of the solid excluding the top and the
bottom, if any, is called lateral surface area of the solid.
❑Volume of a solid. The measurement of the space enclosed by a solid is called its
volume. It is measured in cubic units.
❑Cross-section of a solid. If a cut is made through a solid perpendicular to its length (breadth
or height), then the surface so obtained is called its cross-section.
If the surface made by the cut has the same shape and size at every point of its length
(breadth or height), then it is called a uniform cross-section.
Cross-section
perpendicular to length
Cross-section
perpendicular to height
(i)
(ii)
Cross-sections
perpendicular to height
(iii)
In figures (i) and (ii) above, the solids have uniform cross-sections.
In figure (iii) above, the solid has non-uniform cross-sections.
When a solid has uniform cross-section, then its
(i) volume = area of cross-section × length (breadth or height).
(ii) lateral surface area = perimeter of cross-section × length (breadth or height).
17.2 Surface Area and volume of a cylinder
Height
axis
Cylinder. A solid obtained by revolving a rectangular lamina about
one of its sides is called a right circular cylinder. In other words,
a solid whose cross-sections are circles congruent to each other is
called a right circular cylinder.
(i)The radius of any circular cross-section is called the
radius of the cylinder.
Radius
(ii)The line joining the centres of circular cross-sections
at the ends is called the axis of the cylinder.
(iii)The distance between the centres of circular cross-sections at the ends is called the
height (or length) of the cylinder.
In this book, we shall be dealing only with right circular cylinders. Therefore, whenever
we use the word ‘cylinder’, it will mean right circular cylinder.
❑ Surface area of a right circular cylinder
To find the surface area of a right circular cylinder, perform the following activity:
Take a hollow closed right circular cylinder made up of thick paper/cardboard and cut
it as shown in the figure i.e. first cut its two circular ends (top and base) and then cut
it along a line on its surface parallel to the axis of cylinder and prepare a net as shown
in the figure.
r
2πr
h
r
The net of the right circular cylinder consists of:
(i)a rectangular piece whose length is equal to the circumference/perimeter of the
circular base and its breadth is equal to the height of cylinder.
(ii)two circular pieces of same radius.
If r is the radius of the cylinder and h is its height, then curved surface area of
the cylinder = area of rectangular piece = length × breadth
= perimeter of base of cylinder × h = 2πr × h.
Hence, curved surface area of the cylinder = 2πrh.
Total surface area of cylinder = curved surface area + area of two circular pieces
= 2πrh + 2 × πr2
= 2πr (h + r).
Hence, total surface area of cylinder = 2πr (h + r).
Mensuration
4125
❑ Volume of a right circular cylinder
To find the volume of a right circular cylinder,
take few circular sheets of thick paper of same
size and stack them up in a vertical pile. We
obtain a right circular cylinder as shown in the
adjoining figure.
If A is the area of the circular sheet and h is the
height to which the circular sheets are stacked up
in a vertical pile, then
volume of the cylinder so formed = area of circular sheet × height = A × h.
Thus, volume of a cylinder = area of base × height.
It follows that if a cylinder has radius r and height h, then
volume of cylinder = πr2 × h = πr2h.
❑ Solid cylinder
Let r be the radius and h be the height of a solid cylinder, then
(i) Curved (lateral) surface area
= perimeter of cross-section × height = 2π r h.
(ii) Total surface area
= curved surface area + area of two circular ends
= 2π r h + 2π r2
= 2π r (h + r).
(iii) Volume= area of cross-section × height
= π r2h.
h
r
Remark.
When a rectangular lamina is rolled, we get a hollow right circular cylinder.
❑ Hollow cylinder
R
Let R and r be the external and internal radii of a hollow
r
h
cylinder, and h be its height, then
(i) Thickness of cylinder = R – r.
(ii) Area of a cross-section = π (R2 – r2).
(iii) External curved surface area = 2π Rh.
(iv) Internal curved surface area = 2π r h.
(v) Total surface area= external curved area + internal curved area + area of two ends
= 2π R h + 2π r h + 2π (R2 – r2)
= 2π (Rh + rh + R2 – r2).
(vi) Volume of the material = π R2 h – πr2h
= π (R2 – r2) h.
Illustrative Examples
4126
Example 1. The diameter of a cylinder is 7 cm and its height is 16 cm. Find :
(i) Lateral surface area of the cylinder.
(ii) Total surface area of the cylinder.
(iii) Volume of the cylinder.
Understanding ICSE mathematics – x
Solution. Let r be the radius and h be the height of the given cylinder, then
diameter = 2r = 7 cm (given)
⇒
r = 7 cm.
2
Also height of the cylinder = h = 16 cm (given).
(i) Lateral (curved) surface area of the cylinder = 2πrh
= 2 × 22 × 7 × 16 cm2 = 352 cm2.
7
2
(ii) Total surface area of cylinder = 2πr (h + r)
22
7
7
= 2 × 7 × 2  16 + 2  cm2
= 22 × 39 cm2 = 429 cm2.
2
(iii) Volume of the cylinder = πr2h
=
22  7  2
×   × 16 cm3 = 616 cm3.
 2
7
Example 2. 30 circular plates, each of radius 14 cm and thickness 3 cm are placed one above
the another to form a cylindrical solid. Find:
(i) the total surface area.
(ii) the volume of the cylinder so formed.
Solution. Height of the cylinder formed = number of plates × thickness of one plate
= (30 × 3) cm = 90 cm.
(i) Total surface area of cylinder= 2πr (h + r)
22
× 14 × (90 + 14) cm2
7
= 2×
= (88 × 104) cm2 = 9152 cm2.
(ii) Volume of the cylinder = πr2h =
22
× (14)2 × 90 cm3
7
= (22 × 28 × 90) cm3 = 55440 cm3.
Example 3. A cylindrical roller 2·5 m in length, 1·75 m in radius when rolled on a road was
found to cover an area of 5500 m2. How many revolutions did it make?
Solution. Length of cylindrical roller = 2·5 m =
5
7
m, radius = 1·75 m =
m.
2
4
Area covered by the roller in one revolution = curved surface area of roller
22
7
5

× ×  m2 = 55 m2.
= 2πrh =  2 ×
7
4
2
2
Let the number of revolutions made by the roller be n, then
area covered in one revolution × n = total area covered
⇒
55
× n = 5500 ⇒ n = 200.
2
Hence, the number of revolutions made by the roller = 200.
Example 4. If the lateral surface area of a cylinder is 94·2 cm2 and its height is 5 cm, then find
(i) radius of its base
(ii) its volume
(Use π = 3·14)
Solution. (i) Let r be the radius of the cylinder
Lateral surface area of a cylinder = 2πrh
Mensuration
4127
⇒ 2 × 3·14 × r × 5 = 94·2 ⇒ r = 3 cm
(ii) Volume = πr2h = (3·14 × 32 × 5) cm3 = 141·3 cm3.
Example 5. The inner diameter of a circular well is 3·5 m. It is 10 m deep. Find:
(i) its inner curved surface area.
(ii) the cost of plastering this curved surface area at the rate of ` 40 per m2.
7

1
m.
Solution. Radius of circular well =  × 3.5 m =
4

2
22
7


× × 10 m2 = 110 m2.
(i) Inner curved surface area = 2πrh =  2 ×


7
4
2
(ii) As the rate of plastering the surface = ` 40 per m ,
∴ the cost of plastering the surface = ` (110 × 40) = ` 4400.
Example 6. The total surface area of a cylinder is 6512 cm2 and the circumference of its base is
88 cm. Find the volume of the cylinder.
Solution. Let r be the radius and h be the height of the given cylinder.
Circumference = 2πr = 88 cm (given)
2 × 22 × r = 88
⇒ 7
⇒ r = 14 cm.
Total surface area = 2πr (h + r) = 6512 cm2 (given)
⇒ 88 (h + 14) = 6512
⇒ h + 14 = 74
⇒ h = 60 cm.
2
∴ Volume of the cylinder= πr h
( 2πr = 88 and r = 14)
= 22 × (14)2 × 60 cm3 = 36960 cm3.
7
Example 7. If radii of two circular cylinders are in the ratio 3 : 4 and their heights are in the
ratio 6 : 5, find the ratio of their curved surfaces.
Solution. Since the radii of two circular cylinder are in the ratio 3 : 4, let their radii be
3r and 4r respectively.
As their heights are in the ratio 6 : 5, let their heights be 6h and 5h respectively.
Curved surface area of first cylinder = S1 = 2π × 3r × 6h = 36πrh,
and the curved surface area of second cylinder = S2 = 2π × 4r × 5h = 40πrh.
S
36 rh
9
1
∴ S = 40πrh = 10 .
2
Hence, the ratio of their curved surface areas = 9 : 10.
Example 8. The radius and the height of a cylinder are in the ratio 2 : 7. If the volume of the
cylinder is 704 cm3, find the total surface area of the cylinder.
Solution. Since the radius and the height of a cylinder are in the ratio 2 : 7, let r (radius)
= 2k cm, then h (height) = 7k cm.
∴ Volume of the cylinder = πr2h
= 22 × (2k)2 × 7k = 704 (given)
7
⇒ 88k3 = 704 ⇒ k3 = 8 ⇒ k = 2.
∴Radius of cylinder = 2k cm = 4 cm and
height of cylinder = 7k cm = 14 cm.
4128
Understanding ICSE mathematics – x
∴ Total surface area of the cylinder= 2πr (h + r)
= 2 × 22 × 4 × (14 + 4) cm2
7
176
3168
4
=
× 18 cm 2 =
cm 2 = 452 cm2.
7
7
7
Example 9. The total surface area of a circular cylinder is 462 cm2. If its curved surface area
is one-third of its total surface area, find the volume of the cylinder.
Solution. Let the radius of the circular cylinder be r cm and its height be h cm.
Total surface area = 462 cm2 (given)
⇒ 2πrh + 2 × πr2 = 462
…(i)
Given, curved surface area =
⇒ 2πrh =
1
× 462
3
1
of total surface area
3
⇒ 2πrh = 154
Substituting this value of 2πrh in (i), we get
154 + 2πr2 = 462
…(ii)
⇒ 2 × 22 × r2 = 308 ⇒ r2 = 308 × 7
2 × 22
7
⇒ r2 = 49 ⇒ r = 7.
Substituting this value of r in (ii), we get
2 × 22 × 7 × h = 154 ⇒ 44h = 154 ⇒ h = 7 .
7
2
22
7
Volume of the cylinder = πr2h =
× 72 ×
cm3 = 539 cm3.
7
2
Example 10. The adjoining figure shows the frame of a lampshade
which is to be covered with a decorative cloth. The frame has a base diameter
20 cm and height 30 cm. A margin of 2·5 cm is to be given for folding it
over the top and the bottom of the frame. Find how much cloth is required
for covering the frame.
1

Solution. Radius of frame =  × 20  cm = 10 cm, height of frame = 30 cm.
2
Actual height of cloth= height of frame + 2 margins of 2·5 cm each
= (30 + 2 × 2·5) cm = 35 cm.
∴ The area of cloth required to cover the frame
22


=  2 × × 10 × 35 cm2 = 2200 cm2.
7
If
Example 11. It costs ` 2200 to paint the inner curved surface of a cylindrical tank 10 m deep.
the rate of painting is ` 20 per m2, find
(i) the inner curved surface area of the tank
(ii) the radius of the base
(iii) capacity of the tank in kilolitres.
Solution. (i) As the cost of painting the inner curved surface of the cylindrical tank at
the rate of ` 20 per m2 is ` 2200,
inner curved surface area =
2200
m2 = 110 m2.
20
(ii)Curved surface area = 2πrh
⇒
22
7
2 × 7 × r × 10 = 110 ⇒ r = 4 ⇒ r = 1·75 m
Mensuration
4129
(iii) Capacity of the tank= πr2h
 22  7  2

= 
×   × 10 m3 = 96·25 m3 = 96·25 kilolitres.
 4
7

( 1 m3 = 1000 litres = 1 kilolitre)
Example 12. A cylindrical container is to be made of tin sheet. The height of the container is
1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs ` 300 per m2,
find the cost of the tin sheet for making the container.
Solution. Height of circular cylinder = 1 m and
35
7
its radius = 1 of 70 cm = 35 cm =
m =
m.
100
2
20
The area of the tin sheet required for making the container
= curved surface area + area of base
= 2 πrh + πr2 = πr (2 h + r)
=
22
7
7
11
7
11 47 2
×  2 × 1 +  m 2 =  2 +  m 2 =
×
m .
7
20 
20 
10 
20 
10 20
∴ The cost of the tin sheet for making the container
11
47

× 300  = ` 775·50
= `  ×

10 20
Example 13. Find
(i)the lateral surface area of a closed cylindrical petrol storage tank that is 4·2 m in diameter
and 4· 5 m high.
1
(ii)how much area of steel was actually used, if
of the area of steel actually used was
12
wasted in making the tank?

1
Solution.(i) Radius =  × 4·2 m = 2·1 m, height = 4· 5 m.

2
22

Lateral surface of the tank =  2 ×
× 2·1 × 4·5 m2 = 59· 4 m2


7
(ii) Total surface of the tank = 2π(r + h)r
= 2 × 22 (2·1 + 4·5) × 2·1 m2 = 87·12 m2
7
Let x m2 be the actual area of steel used in making the tank. As
1
of the actual
12
1
11


× x m2 =
area is wasted in making, so remaining area =  x −
x m2.
12


12
∴
11
11
x = 87· 12 ⇒ x =
× 87· 12 = 95·04
12
12
Hence, the actual area of the steel used = 95· 04 m2.
Example 14. A cylindrical water tank of diameter 1· 4 m and height 2·1 m is being fed by a
pipe of diameter 3·5 cm through which water flows at the rate of 2 m/s. Calculate the time it takes
to fill the tank.
Solution. Radius of water tank = 1 × 1· 4 m = 70 cm.
2
Height of water tank = 2·1 m = 210 cm.
∴ Volume (capacity) of water tank = πr2h
= 22 × (70)2 × 210 cm3 = (22 × 700 × 210) cm3.
7
Radius of water pipe = 1 × 3·5 cm = 7 cm.
2
4130
4
Understanding ICSE mathematics – x
ength of water delivered by the pipe in one sec = 2 m = 200 cm.
L
∴ Volume of water delivered by the pipe in one sec = π r2h
=
22  7  2
×   × 200 cm3 = (77 × 25) cm3.
 4
7
∴ Time consumed by the water pipe in filling the water tank
=
22 × 700 × 210
sec = 1680 sec
77 × 25
=
1680
minutes = 28 minutes.
60
Example 15. A metal pipe has a bore (internal diameter) of 5 cm. The pipe is 2 mm thick all
round. Find the weight in kilogram, of 2 metres of the pipe if 1 cm3 of the metal weighs 7·7 g.
Solution. Internal radius (r) of the pipe = 1 × 5 cm = 5 cm.
2
2
2
1
he thickness of the pipe all round = 2 mm =
T
cm = cm .
10
5
5
1
27


∴External radius (R) of the pipe =  +  cm =
cm.
2 5
10 Length of the pipe = 2 m = 200 cm.
∴ The volume of the metal = π (R2 – r2) h
22   27  2  5  2 
   −    × 200 cm3
 2 
7   20 


22  27 5   27 5 
+   −  × 200 cm3
=

7  10 2   10 2 
=
=
22 52
2
88 × 52
× × × 200 cm 3 =
cm3.
7
10 10
7
Since 1 cm3 of metal weighs 7·7 g,
∴ the weight of the pipe = 88 × 52 × 7·7 g
7
=
88 × 52 × 77
= 5·0366 kg.
7 × 10 × 1000
Example 16. The difference between the outer and the inner curved surfaces of a cylinder
14 cm long is 88 cm2. Find the outer and the inner radii of the cylinder, given that the volume of
the metal is 176 cm3.
Solution. Let the outer and the inner radii of the cylinder be R cm and r cm respectively.
Length of the cylinder= 14 cm (given).
∴Outer curved surface = 2πRh = 2πR × 14 cm2 = 28 πR cm2.
Inner curved surface = 2πrh = 2πr × 14 cm2 = 28πr cm2.
According to the given data,
28πR – 28πr = 88
⇒
28 × 22 (R – r) = 88
7
⇒R – r
Volume of the metal
=
=
=
=
=
1
π (R2 – r2)h
π (R + r) (R – r) × 14 cm3
14π (R + r) × 1 cm3
14π (R + r) cm3.
…(i)
(using (i))
Mensuration
4131
According to the given data,
14π (R + r) = 176
⇒
14 × 22 (R + r) = 176
7
⇒R + r = 4
Adding (i) and (ii), we get
…(ii)
2R = 5 ⇒ R = 5 = 2·5
2
∴ From (ii), r = 4 – R = 4 – 2·5 = 1·5
∴ Outer radius = 2·5 cm and inner radius = 1·5 cm.
Example 17. The total surface area of a hollow metal cylinder, open at both ends, of external
radius 8 cm and height 10 cm is 338π cm2. Taking r cm to be inner radius, write down an equation
in r and use it to find the thickness of the metal in the cylinder.
Solution. External radius of hollow cylinder = R = 8 cm and its height = h = 10 cm.
Total surface area = (2π Rh + 2πrh + 2π (R2 – r2)) cm2
= 2π [h(R + r) + (R2 – r2)] cm2
= 2π (R + r) (h + R – r) cm2
= 2π (8 + r) (10 + 8 – r) cm2
= 2π (8 + r) (18 – r) cm2
According to given
2π (8 + r) (18 – r) = 338 π
⇒
(8 + r) (18 – r) = 169
⇒
144 + 10 r – r2 = 169
⇒
r2 – 10 r + 25 = 0 ⇒ (r – 5)2 = 0 ⇒ r = 5.
∴
Internal radius = 5 cm.
∴ Thickness of metal in the cylinder = (8 – 5) cm = 3 cm.
Exercise 17.1
Take π = 22 , unless stated otherwise.
7
1. Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave
your answer in terms of π.
2. An electric geyser is cylindrical in shape, having a diameter of 35 cm and height 1·2 m.
Neglecting the thickness of its walls, calculate
(i) its outer lateral surface area,
(ii) its capacity in litres.
3.A school provides milk to the students daily in cylindrical glasses of diameter 7 cm.
If the glass is filled with milk upto a height of 12 cm, find how many litres of milk
is needed to serve 1600 students.
4. In the given figure, a rectangular tin foil of size 22 cm
by 16 cm is wrapped around to form a cylinder of
height 16 cm. Find the volume of the cylinder.
16 cm
22 cm
5. (i) How many cubic metres of soil must be dug out to make a well 20 metres deep
and 2 metres in diameter?
(ii) If the inner curved surface of the well in part (i) above is to be plastered at the
rate of ` 50 per m2, find the cost of plastering.
4132
Understanding ICSE mathematics – x