Green C Math Class 10

= o 2ac + c² + a² – b² po 2ac – c² – a² + b² p
2a 2a

= (a + c)² – b² × b² – (a – c)²
2a 2a

= (a + c + b) (a + c – b) × (b + a – c) (b – a + c)
2a 2a

= (a + c + b) (a + c – b) (b + a – c) (b – a + c)
4a²

= (a + b + c) (a + b + c – 2b) (a + b + c – 2c) (a + b + c – 2a)
4a²

= 2s(2s – 2b) (2s – 2c) (2s – 2a)
4a²

= 16s (s – b) (s – c) (s – a)
4a²

= 4s(s – a) (s – b) (s – c)
a²

2 s(s – a) (s – b) (s – c)
∴ h =
a

Putting the value of h in equation (i), we get

Area of ∆ABC = 1 . base × height
2

= 1 2 s(s – a) (s – b) (s – c)
2 a.
a

= s(s – a) (s – b) (s – c)

Worked Out EXAMPLES

EXAMPLE 1 Find the area of triangle ABC. A
Solution : 10cm
In ∆ABC, a = 10cm, b = 8cm and c = 6cm

s = a + b +c 6cm 8cm
2

= 10 +8 + 6 = 12cm B C
2

Area of ∆ABC = s(s – a) (s – b) (s – c)

= 12(12 – 10) (12 – 8) (12 – 6)

= 12 × 2 × 4 × 6

= 24cm²

GREEN Mathematics Book-10 45

EXAMPLE 2 Find the area of the given kite: 16cm D 18cm
Solution : A 20cm C
In ∆ACD,
s = 20 + 18 + 16 = 27cm
By using formula, 2

Area of ∆ADC = s(s – a) (s – b) (s – c) B

= 27 (27 – 20) (27 – 18) (27 – 16)
= 27 × 7 × 9 × 11

= 136.79cm²
∴ Area of the kite = 2 × 136.79cm2 = 273.58cm2

Longer diagonal bisects the kite

EXAMPLE 3 Find the area of rhombus ABCD if diagonal AC = 6cm and BD = 8cm.
Solution :
Here,
AC = 6cm and diagonal B D = 8cm A D

Area of rhombus ABCD

= Area of ∆ABC + Area of ∆ADC O

= 1 AC . BO + 1 AC . DO B C
22

= 1 AC (BO + DO)
2

= 1 AC × BD
2

= 1 × 6 × 8 = 24cm²
2

Diagonals of a rhombus bisect
each other at right angle

EXAMPLE 4 Find the equal sides of an isosceles triangle if its area and base are
Solution : 12cm² and 6cm respectively.
Let a be the base and b and c be equal sides.

Thus a = 6cm and b = c
s = a + b +c = 6 + 2b = 3 + b

22
Area of an isosceles triangle = 12cm²

or, s(s – a) (s – b) (s – c) = 12

or, (b + 3) (b + 3 – 6) (b + 3 – b) (b + 3 – c) = 12

46 GREEN Mathematics Book-10

or, (b + 3) (b – 3)) (b + 3 – b) (b + 3 – b) = 12 [∴ b = c]

or, (b + 3) (b – 3) . 3. 3 = 12

or, (b² – 9) × 9 = 12

or b² – 9 = 4

or, b² – 9 = 16

or, b² = 25

or, b = 5

Also, b = c = 5cm

∴ Equal sides of the triangle are 5cm each.

EXAMPLE 5 The perimeter and area of a triangle are 12cm and 6cm² respectively. If
one of the sides is 5cm, find the length of remaining sides.

Solution : Let a, b, and c be the three sides of the triangle where a = 5cm,
P = 12cm, A = 6cm2
s = a + b +c = 5 + b + c

22
or, 6 = 5 + b + c [∵ P = 12 cm, s = P = 12 = 6cm]

2 22

or, b + c + 5 = 12

or b = 12 – 5 – c
∴ b = 7 – c ............... (i)

Area of the triangle = s(s – a) (s – b) (s – c)

6 = 6(6 – 5) [6 – (7 – c)] (6 – c)

36 = 6 × 1 (c – 1) (6 – c)
or, 6 = 6c – c² – 6 + c

or, c² – 7c + 12 = 0

or, c² – (4 + 3) c + 12 = 0

or, c² – 4c – 3c + 12 = 0 Putting c = 4 in equation
(i), we get
or, c(c – 4) – 3 (c – 4) = 0
b=7–c==7–4=3
or, (c – 4) (c – 3) = 0
Putting c = 3 in equation
either (c – 4) = 0 → c = 4 (i), we get

or, (c – 3) = 0 → c = 3 b=7–3=4

∴ the remaining sides are 4cm and 3cm

GREEN Mathematics Book-10 47

EXAMPLE 6 The perimeter and area of a right angled triangle is 30cm and 30cm²
Solution : respectively. Find the length of the sides and the hypotenues.

Now,

Area = 30
1 . p. b = 30
∴ 2 pb = 60 .......................... (i)

Also, p + b + h = 30

h = 30 – (p + b) ........... (ii)

In right-angled triangle,

h² = p² + b²

or, [30 – (p + b)]² = p² + b² [∵ from equation (ii)]

or, 30² – 2. 30 (p + b) + (p + b)² = p² + b²

or, 900 – 60 (p + b ) + p² + 2pb + b² = p ² + b²

or, 900 – 60 (p + b) + 2 × 60 = 0 [∵ pb = 60]

or, 1020 = 60 (p + b)

∴ p + b = 17 .......................... (iii)

We know that,

(p – b)² = (p + b)² – 4pb

= 17² – 4 × 60

= 49

p – b = 7 ............................ (iv)

Adding equation (iii) and (iv)

p – b = 7

p + b = 17

2p = 24

p = 12

Putting the value of p in equation (i), we get
pb = 60
12 × b = 60
b = 5
Putting the value of p and b in equation (ii)
h = 30 – (p + b)
= 30 – (12 + 5)

= 13

∴ lengths of the sides are 5cm, 12cm and 13cm

48 GREEN Mathematics Book-10

EXERCISE 4.1

A. Very Short Questions

1.a. i. A triangular plot of land has its three sides 40m, 30m and A h
50m, find its perimeter. b
A
ii. Find the perimeter and area of right angled triangle ABC. p
x cm
iii. If the perimeter of an equilateral triangle is 3xcm, find its B C
area. C

iv. What is the area of a triangle having its sides x, y and z
respectively and semi-perimeter is S?

v. What is the area of ∆ABC where BC = x cm and AB = AC = y cm 12cmy cm
ycm? 10cm
12cm
vi. If a = 8cm, b = 6cm and c = 4cm, find the value of B
6cm
s(s – a) (s – b)(s – c) . A

b. i. The perimeter of a triangle is 30cm. If its two
sides are 13cm and 12cm respectively, find the
length of the third side. C
S
ii. In the figure if the area of ∆ABM = 10.5cm2, B M
find the area of ∆AMC and ∆ABC. P

iii. The area of quadrilateral PQRS is 40cm2. If the Q R
area of ∆PQR = 3 of the area of quadrilateral A D
PQRS, find the a5rea of ∆PRS.

iv. The area of ∆ABC = 12cm2 and ∆ACD = 3
∆ABC. Find the area of the quadrilateral
ABCD.

2. Find the area of the given figures: B 16cm C

aA bA cP

M14cm 5cm Q R
6 cm C 6cm
B
13cm

B C 3cm D
A
dX eA Df

5cm 14cm
10cm
C
4cm
Y ZB B
6cm C

GREEN Mathematics Book-10 49

g A hP

R

PQ A

B D B 6cm C
5cm Q
6cm
RS

C

B. Short Questions

3. a. Find the area of a triangle if the sides are in the ratio of 3 : 4 : 5 and perimeter is
24cm.

b. Find the area of an equilateral triangle if its perimeter is 36cm.

c. In ∆ABC, AB + BC + CA = 18cm and AB : BC : CA = 2 : 3 : 4, find the area ∆ABC.

d. The length of unequal adjacent sides of a kite-shaped piece of land are 120m and

160 m respectively. If the length of longer diagonal is 200m, find the area of the

piece of land. 21cm D

A 1 cm
1 cm
4. a. Find the area of trapezium ABCD.
20cm
B CE 1 cm D
16cm

b. Find the area of the given polygon. C

A 1 cm B

A

6cm 6cm

c. Find the area of the kite : B 8cm D

4cm 4cm

C. Long Questions

C

5. a. The area and base of an isosceles triangle are 672cm² and 28cm respectively.

Find the length of its equal sides.

b. The area of an isosceles triangle is 48cm². If each of the equal sides is 10cm, find
the length of the third side.

c. Perimeter and area of a triangular field are 84m and 336 m2 respectively. If one
of the sides of the field is 30m, find the length of remaining sides.

d. The perimeter of a right-angled triangle is 24cm and its area is 24cm². Find the
sides of the triangle.

50 GREEN Mathematics Book-10

4.2 Prism

A solid object whose at least a pair of opposite faces are parallel and congruent polygons,
while the lateral faces are rectangles is called prism. The rectangular faces are called
lateral surface. The distance between the bases of prism is the height of the prism. The
lateral surface is formed by joining the corresponding vertices of the base.

If the bases are regular figures, the prism is called a regular prism and the prism is called
a right prism if its lateral surfaces are perpendicular to the base.

The prisms are named according to their base. The base figures. If the base of prism is a
triangle, the prism is called triangular prism.

Some prisms are given below:

When a prism is cut such that the new surface is parallel to its base, the new surface is

called cross-section. Area of cross section is equal to the area of a base.

base (polygons) base (circle)

height
height

height
base (triangle)

Rectanguler Base height Right pentagonal base
(rectangular prism) Triangular prism prism
Circular prism
(cuboid) (Cylinder)

Cross section

base

base
base

Cross section

base Cross section

Surface area and volume of a triangular prism

bc bb c A triangular prism has three rectangular
c faces (lateral surface) and two triangular
faces as shown in the figure.
a h a
h c In the triangular prism
h h
b a. Area of lateral surface
a
= Area of three rectangles.
a bc = bh + ah + ch
Triangular prism
Net of triangular prism

= (a + b + c) × h = p × h

∴ Lateral surface area (LSA) = perimeter of triangular base × height of prism.

b. Total surface area (TSA) = Lateral surface area + 2 × area of triangular base = Ph + 2∆

c. Volume of triangular prism (V) = Area of triangular base × height of prism.

GREEN Mathematics Book-10 51

Worked Out EXAMPLES A'
B'
EXAMPLE 1 Find : A C 20 cm
a. lateral surface area
6cm C'
b. total surface area and
c. volume of the given triangular prism.B 8cm

Solution : In base ABC,

AB = 6cm and BC = 8cm

Using pythagoras theorem, AC = AB² + BC²

= 6² + 8²
= 100
= 10cm

a. Lateral surface area (LSA) = perimeter of base × height

= (AB + BC + AC) × CC'
= (6 + 8 + 10) × 20
= 24 × 20

= 480 cm²

b. Total surface area (TSA) = LSA + 2 area of ∆ABC

= 480 + 2 1 ×6 × 8p
o2

= (480 + 48)cm2

= 528 cm²
c. Base area (A) = 1 × 6 × 8 cm2

2
= 24cm²

Height (h) = 20cm

∴ Volume (V) = Base area × height

= A × h

= 24 × 20

= 480 cm³.

52 GREEN Mathematics Book-10

EXAMPLE 2 The base of a right prism is an equilateral triangle of area 173 cm² and
Solution : volume of the prism is 10380 cm³. Find the height and lateral surface
area of the prism. (Use 3 = 1.73)

Volume of prism = 10380 cm³

or, Area of base × height = 10380

or, 173 × h = 10380

∴ h = 60 cm.

Now, Area of base = 173cm²


or, 3 × a² = 173
or, 4
or, a² = 173 × 4
3

a² = 173 × 4
1.73

or, a² = 400
∴ a = 20cm
Then, lateral surface area (LSA) = perimeter of base × height

= 3a × h
= (3 × 20) × 60
= 3600cm² .

EXAMPLE 3 In the given solid triangular prism, if YZ = 6cm, X M
Solution : R
ZR = 20cm, XY⊥YZ and volume of the prism is N
7200 cubic cm, find the measurement of XY. Y 6cm Z 20cm 53

Here,

YZ = 6cm

ZR = 20cm

Volume of prism (V) = 720cm3

We have, V = Base area × height

or, 720cm3 = 1 XY . YZ . 20cm
2

or, 720cm3 = 1 XY . 6cm . 20cm
2

or, 720cm3 = XY . 60cm2

or, XY = 12cm



GREEN Mathematics Book-10

EXERCISE 4.2

A. Very Short Questions D

1. a. In the given triangular prism AB + BC + CA = 12cm A E F
and CF = 20cm. Find the lateral surface area.

B PC

b. In the given right triangular prism, perimeter and

area of ∆PQR are 30cm and 30cm². If RR' = 20cm, find R

i. the lateral surface area. Q

ii. the total surface area. P'

iii. the volume. Q' R'
D

P

c. What is shaded part in the figure? Also find its area. A Q3cm E F
R A'
B 4cm C

d. If the area of base ∆ABC in the figure is 12cm2 and A
BB' = 20cm, find the volume of the given prism.
B'
C'

BC

A'
P

e. In the given figure area of cross section PQR = 7.5cm2 A B'
R
and volume of prism is 225cn3. Find the length of Q C'

AA'. BC

A

f. The area of cross section DEF is 7.5cm2. If the volume B E C

of the prism is 150cm3 and CF = 5cm, find the volume

of the prism below the cross section. DF

H

GI

2. a. If the sides of triangular base of prism are 13cm, 20cm and 21cm respectively,
find the area of cross section.

b. The perimeter of base and height of a triangular prism are 30cm and 15cm
respectively. Find the lateral surface area of the prism.

c. The area of a base of a triangular prism is 80cm² and lateral surface area is
600cm². Find the total surface area of the prism.

d. The volume and the cross section of a triangular prism are 480 cm³ and 24cm².
Find the height of the prism.

54 GREEN Mathematics Book-10

B. Short Questions
3. Find LSA, TSA and volume of the given triangular prism.

D 55
6

c
A 15cmb
13cm
aE

B 30cm F 8cm 20cm
12cm C

d e 10cm
10cm
6cm
9cm 17cm 25cm
8cm15cm

22cm
4. a. The three sides of the base of a triangular prism are 5cm, 6cm and 7cm

respectively. If the height is 14cm, find the total surface area and volume.

b. If the lateral surface area of the given prism is 440cm², find A
6.5cm 20cm
the length of BC. A B C

5cm C
B

c. The volume of given prism is 450 cm³. Find A'
the height of the prism.

B' C'
13cm

d. The volume and height of prism having isosceles right angled triangle base are
1000cm³ and 16cm respectively. Find the length of equal sides of the base.

5. a. In the given solid triangular prism, if PQ = 8cm, P S
T
QR = 6cm, PQ⊥QR and volume is 480 cubic cm, find Q R
the height of the prism. 6cm Z Q'
Y
b. In the given solid triangular prism, if XY = 12cm, 5cm
12cm
X

YZ = 5cm, XY⊥YZ and height is 22cm. Find the area of T

rectangular surfaces of the prism. SR

Project Work:

a. Measure the sides of a triangular prism in your school's science lab and then
calculate the cross section area, lateral surface area and its volume.

b. Prepare 3 triangular prism by using paper, wood or bamboo and calculate their
total surface area, volume and cross section area.

GREEN Mathematics Book-10 55

4.3 Area and volume of cylinder

A cylinder is a circular prism. When we roll a rectangular sheet of paper as shown in the
figure, we get a curved surface figure with two circular faces. This is circular prism or a
cylinder.

l = 2πr Area of circular face = πr2

A B AB

C bh h A=l×b Area curved surface = 2πrh
l = 2πrh Area of circular face = πr²
D

CD

l = 2πr

From the above figure,

Lateral surface area of circular face = 2 × Area of circular base = 2πr2

Surface area (L.S.A.) = 2πrh

Total surface (T.S.A.) = L.S.A. + 2. area of base

= 2πrh + 2.πr²

= 2πr ( h + r).

Volume (V) = base area × height

= πr²h

Surface area of sphere A
R
Sphere is round object like football, cricket ball, etc. Archimedes (6th
century BC) discovered that the surface area of spherical object like tennis dh
ball, football etc. is equal to the curved surface of circumscribing cylinder.
Q
The figure is the cylinder circumscribing a sphere. Here diameter of the P B
sphere AB is equal to the diameter PQ of the cylinder and diameter of d
sphere is equal to the height QR of cylinder.
2r

We know that, surface area of sphere = curved surface area of the
circumscribing sphere

= 2πrh

= 2πr × d [∵ h = d]

= 2πr × 2r [d = 2r]

= 4πr²

∴ Surface area of a sphere = 4πr².

56 GREEN Mathematics Book-10

Surface area of a hemisphere A BA πr² (circular surface)
B
Hemisphere is one of the half parts of 2πr² (curved surface)
a sphere when it is divided into two
equal parts. It has two surfaces, circular
face (great circle) and curved surface.

The radius of hemisphere is equal to the

radius of sphere.

Curved surface area of hemisphere = 1 . curved surface area of sphere
2

= 1 × 4πr²
2

= 2πr²

Total surface area of hemisphere = area of circular surface + area of curved surface

= πr² + 2πr²

= 3πr²

∴ Total surface area of himisphere = 3πr²

Volume of a hemisphere and sphere dh

The figure is the cylindrical vessel
circumscribing a spherical plastic ball. When
the spherical ball is cut into two equal parts
we get two hemispheres.

Now, fill the hemisphere with rice or sand and
pour it into the cylinder completely.

Thus, 3 hemispheres = 1 cylinder

= πr²h

= πr² × d

= πr² × 2r

= 2πr³

∴ 1 hemisphere = 2 πr³
Thus,v olume of a hemisphere = 3

2 πr³
3

Now, Volume of a spheres = Volume of two hemispheres

= 2× 2 πr³
3

= 4 πr³
3

GREEN Mathematics Book-10 57

Worked Out EXAMPLES

EXAMPLE 1 Find the curved surface area, total surface area and the volume of the
given cylinder.

Solution : Here, diameter of the cylinder (d) = 7cm 7cm

7cm 21cm
2
Radius of the cylinder (r) = = 3.5cm

Now,

Curved surface area (C.S.A.) = 2πrh

= 2 × 22 × 3.5 × 21
7

= 462cm²

Total surface area (T.S.A.) = 2πr(r + h)

= 2 × 22 × 3.5 (3.5 + 21)
7

= 539cm²

Volume (V) = πr²h

= 22 × (3.5)² × 21
7

= 808.50cm³

EXAMPLE 2 25 circular plates, each of radius 7cm and thickness 0.5cm, are placed
Solution : one above another to form a circular prism. Find the volume of the
prism.

Here,
Thickness of 1 plate = 0.5cm
∴ Thickness of 25 plates = 0.5 × 25
= 12.5cm
Thus the height of the circular prism (cylinder) (h) = 12.5 cm
Radius of the cylinder (r) = 7cm
Now,

Volume (V) = πr²h

= 22 × 7² × 12.5
7

= 1925cm³

58 GREEN Mathematics Book-10

EXAMPLE 3 The largest sphere is carved out of a cube of side 7cm. Find the volume
Solution : of the sphere.

Here, diameter of sphere (d) = length of cube

= 7cm 7cm
7cm
Radius of sphere (r) = d = 7
2 2

= 3.5 cm
Now,
Volume of sphere (V) = 4 πr³
3

= 4 × 22 × (3.5)³
37

= 179.67cm³

∴ The volume of sphere is 179.67cm3.

EXAMPLE 4 Three metallic spheres of radii are 1cm, 6cm and 8cm respectively. If
Solution : these three spheres are melted to form a single sphere, find the radius
of the new sphere.

Here, radius of first sphere (r1) = 1cm

radius of second sphere (r2) = 6cm

radius of third sphere (r3) = 8cm

Now, volume of new sphere (V) = sum of volume of three spheres

4 πr³ = 4 πr1³ + 4 πr2³ + 4 πr3³
3 3 3 3

or, 4 πr³ = 4 π (r13 + r23 + r33)
3 3

or, r³ = 1³ + 6³ + 8³

or, r³ = 1³ + 6³ + 8³

or, r³ = 1 + 216 + 512

or, r³ = 729

or, r³ = 9³

∴ r = 9cm

So, the radius of the new sphere = 9cm.

GREEN Mathematics Book-10 59

EXAMPLE 5 A sphere of diameter 5cm is dropped into a cylindrical vessel partly
filled with water. The diameter of the base of the vessel is 10cm. By
how much will the level of water rise if the sphere is completely
submerged?

Solution : Here, diameter of sphere (d1) = 5cm

∴ diameter of vessel (d2) = 10cm
Now,
radius of vessel (r) = 10 cm = 5cm
2

Volume of the water risen up in the vessel = volume of sphere

πr². h = 16 πd1³

or, 5² × h = 1 × 5³
6

or, h = 65

∴ h = 56 cm

So, the water level rises by 5 cm.
6

EXERCISE 4.3

A. Very Short Questions

1. a. What are the shapes of base and curved surface of a cylinder?

b. If a cylinder is made with a rectangular sheet of paper of 220cm × 10cm, find the
area of the curved surface.

c. Which of the objects football or coin, is sphere?
d. What is the half part of a sphere called?
e. What is the total surface area of half part formed when a sphere of radius x cm

is cut into two halves?

2. a. The circumference of the base of cylinder is 22cm and its height is 10cm. Find
the curved surface area of the cylinder.

b. If the radius of base of a cylinder is 7cm and height is 21cm, find (i) the curved
surface area (ii) the total surface area (iii) volume.

60 GREEN Mathematics Book-10

c. If the curved surface area of a cylinder is 440cm² and area of a base is 154cm²,
find its total surface area.

d. The circumference of the base of a cylinder is 88cm and the sum of its radius and
height is 30cm, find the total surface area.

3. a. If the base area and height of a cylinder are 154cm² and 20cm respectively, find

the volume of the cylinder. C

b. The given solid is a cylinder. If AB = 14cm and DB = 21cm, find the D

volume. A 21cm
B. Short Questions 14cm B

4. a. If the volume of a cylinder is 4620cm³ and its height is 3cm, find its radius.

b. The volume of a cylindrical can is 1.54 liters. If the area of its base is 77cm², find
its height.

c. The base radius and the height of a cylinder are in the ratio of 5 : 7. If the volume
of the cylinder is 550cm³, find its radius.

d. If the height and radius of a cylinder are equal and curved surface area is 308cm2,
find its height.

e. The volume of a cylinder is 539 cm³ and the circumference of the base is 22cm.
Find the height of the cylinder.

5. a. Find the total surface and the volume of the given cricket ball of 35mm
radius 35mm.

b. Find the total surface area and volume of the given hemisphere of diameter
21cm.

c. Find the total surface area and volume of a sphere if the circumference of the
great circle is π cm.

d. Find the total surface area of two hemispheres when a sphere of radius 7cm is
cut into two halves.

6. a. If the total surface area of a solid sphere is 616 cm², what will be its radius?

b. Find the total surface area of a sphere whose volume is 1372 π cm³.
3

7. a. The curved surface area of a cylinder with height 15cm is 1320cm². Find the
total surface area and the volume.

b. The total surface area of a cylindrical can is 2464cm² and its height is equal to the
radius of the base. Find the volume of the can.

c. By how many times does the volume of a cylinder increase when its diameter is
doubled?

GREEN Mathematics Book-10 61

d. Th external and internal radii of a hollow cylindrical vessel 70cm high are 3.5cm
and 3cm respectively. Find the volume of the material contained by the vessel.

8. a. The radii of two spheres are in the ratio of 2 : 3. Find the ratio of their surface
area.

b. If the ratio of total surface area of two hemisphere is 1:4, find the ratio of their
radii.

9. a. Three metallic spheres of diameters 1cm, 6cm and 8cm are melted to form a
single sphere. Find its radius.

b. Three metallic spheres each of radius 4cm, 7cm and 9cm are melted to recast a
solid cylinder of radius 7cm. Find the height of the cylinder so formed.

c. The diameter of a metallic sphere is 6cm. It is melted and recast into a wire
having diameter of the cross-section as 0.2cm. Find the length of the wire.

d. How many balls, each of radius 1cm, can be made from a solid sphere of lead of
radius 8cm?

C. Long Questions

10. The area of curved surface area of a solid cylinder is equal to 2 of the total surface
3
area. If the total surface area is 924cm2, find the volume.

11. The internal diameter and height of a cylindrical bucket are 14cm and 35cm
respectively and it is filled with water completely. If the water is poured into a
rectangular vessel with internal length 28cm and breadth 17.5cm, and it is completely
filled with water, find the height of the vessel.

12. Find the volume.

a. b.
7cm 14cm
20cm 10cm 8cm 35cm

13. Find the total surface area.
a. b.

19cm

14cm

14. a. The total surface area of the given solid is 902cm2 and the sum of the radius of
hemisphere and the height of cylinder is 17cm. Find the height of the cylinder.



62 GREEN Mathematics Book-10

b. A combined solid made up of a cylinder of radius 3cm and length h cm and
hemisphere with the same radius as the cylinder has volume of 792cm3. Find
the value of h.

3cm

h

Project Work:

a. Collect 5 cylindrical objects at your home and measure their size and find the
volume, total surface area and cross section area.

b. A sheet of paper 42cm long, 21cm breadth is used making a cylinder. Find the total
surface area and volume of so formed figure.

4.4 Pyramid : Introduction

Pyramid is a three dimensional figure like a prism. It is a solid with polygonal base and
triangular lateral faces meeting at a common point called vertex or apex. This figure has
fascinated human beings from the ancient times. Pyramids of Egypt are one of the seven
wonders of the world. These pyramids were built during the period 3000 – 2000 B.C.

Pyramids other than tetrahedron are named according to the name of their bases such

as a square pyramid, pentagonal pyramid, etc. Pyramid with triangular base is called

tetrahedron. A

In the figure (i), A is the vertex and AO is the height of tetrahedron. P
Its base is ∆PQR and lateral surfaces are ∆APQ, ∆APR and ∆AQR. It Q OR
has 3 triangular faces altogether.
(I) Tetrahedron

A

Vertex

In the figure (ii), A is the vertex, AO is height, AM is the slant height

which is also the height of triangular faces. All the four triangular S

P O
R
faces are equal in area and PQRS is its base. ∆ARS, ∆APS, ∆APQ

M

and ∆AQR are lateral (triangular) faces of the pyramid. Q

(II) Square pyramid

If the foot of perpendicular (height) drawn from the vertex to the base is the
centre of the base, the pyramid is called a right pyramid.

GREEN Mathematics Book-10 63

Surface area and volume of a pyramid

In the given square pyramid PQRS is base whose one side is QR = a, AO is the height (h)
and AM is slant height (l).

Now,

Lateral surface area (L.S.A) = sum of area of four triangular faces

= 4 o 1 . a . l.p A
2
P S
= 2 a l h S
O
Area of total surface (T.S.A) = Area of square base + L.S.A M
Q R
= a² + 2 a l

Volume (V) = 1 base area × height of pyramid
3

= 1 A²h
3

Worked Out EXAMPLES

EXAMPLE 1 Find the lateral surface area and total surface area of the given square
Solution : - based pyramid.

Here, P D
length of square base (a) = 12cm
10cm
A

slant height of the pyramid (l) = 10cm BC
How, 12cm

lateral surface area (L.S.A.) = 2 a l

= 2 × 12 × 10

= 240 cm²

total surface area (T.S.A) = a² + 2 a l

= 12² + 2 × 12 × 10

= 144 + 240

= 384 cm²

64 GREEN Mathematics Book-10

EXAMPLE 2 Find the volume of the given square based pyramid if vertical height
Solution :
and edge of triangular face are 24cm and 25cm respectively.

A 25cm

In the pyramid, 24cm S
vertical height (AO) = 24cm P

Edge (AS) = 25cm O
QR

In right angled triangle AOS,

OS = AS² – AO²

= 25² – 24²

= 625 – 576

= 49

= 7cm

Thus, d iagonal of square base (d) = QS

= 2 × OS
= 2 × 7
= 14cm

Now, Area of square base (A) = 1 × d²
2

= 1 × 14²
2

= 98 cm²

Then, volume of the pyramid (v) = 1 A × h
3

= 1 × 98 × 24
3

= 784cm³

GREEN Mathematics Book-10 65

EXAMPLE 3 Find the total surface and the volume of the given tetrahedron.

Solution : Here, 8cm 10cm

length of base (a) = 12cm 12cm

slant height (l) = 10cm

and height of the pyramid (h) = 8cm

Now, Total surface area (T.S.A.) = Area of equilateral triangular base
+ three triangular lateral faces

= 3 a² + 3 1 . a . l
4 2

= 43 . 12² + 32 . 12 . 10

= (36 3 + 180)cm2

= 242.35 cm2

Volume of the pyramid (V) = 1 . base area × height
3

1 o 3 × a² × hp
= 3 4

= 3 . 12² × 8
12

= 96 3 cm³

EXAMPLE 4 In the given figure, the total surface area of the square based pyramid
Solution :
is 96cm² and the side of the square base is 6cm. Find the volume of the

pyramid. A

Here, PS

Side of square base (a) = 6cm Q R
6cm
Now,

total surface area of pyramid (T.S.A.) = 96cm²

or, a² + 2al = 96

or, 36 + 2.6.l = 96

or, 12l = 96 – 36
or, l = 60

12
or, l = 5cm

66 GREEN Mathematics Book-10

∴ l = 5cm

Then, height (h) = l² – o a 2

2 p

= 5² – o 6 2

2 p

= 16

= 4cm

Again,

Volume of the pyramid (V) = 1 × a² × h
3

= 1 × 6² × 4
3

= 48 cm³

EXERCISE 4.4

A. Very Short Questions P

1. a. Name lateral faces and base of the given pyramid.

b. Write the name of the following in the given figure. A D
h M
i. Point P ii. PO iii. PM C
O
iv. PD v. BC B

c. What is the relation between OM and BC?

d. Write the relation of PM, PO and OM.

2. a. If the area of base of a pyramid 64cm² and height is 6cm, find its volume.

b. One of the triangular faces of a square based pyramid is is 40cm². Find the area
of lateral surface of the pyramid.

c. The base area of a square based pyramid is 100cm2 and one of the triangular

faces is 65cm2. Find the total surface area of the pyramid. A

d. Answer the following questions from the given figure. P S
i. If OM = 5cm, find the area of base PQRS. M
R
ii. If OM = 5cm and OA = 12cm, find AM. O

Q

iii. If OM = 5cm and OA = 12cm, find he volume of the pyramid.

iv. If OM = 5cm, AM = 13cm, find the lateral surface area of the pyramid.

GREEN Mathematics Book-10 67

B. Short Questions

3. a. Find the volume of a pyramid whose base is an equilateral triangle of side 7cm
and height 15cm.

b. Find the lateral surface area of the24cm A 13cm
given pyramid. S

7cm

c. Find the area of triangular faces of the given pyramid. P

P

d. Find the volume of the pyramid A D QR
10cm
if PO = 8cm and OM = 6cm. OM P

BC AD

e. Find the volume of the given pyramid if AC = 12 2 and O

PO = 8cm. BC
10cm

4. a. The volume and base area of a square-based pyramid are 300cm³ and 100cm²

respectively. Find the height of the pyramid.

b. The length of the base of a square-based pyramid is 8cm and its total surface
area is 224cm². Find the slant height of the pyramid.

C. Long Questions A

5. a. Find the volume of the following pyramids.

17cm P

ab

AB
12cm
13cmc 12cmS
12cm
P
13cm
M
15cm
16cm B C QR

P 17cm e
8cm f
dA D 8cm

BC 3cm
13cm
6. Find the total surface area of the square based pyramids.

P 10cm A A 13cm
S
aA D b P Sc P M

M

BC QR QR
16cm 12cm 10cm

68 GREEN Mathematics Book-10

A 26cm e 18cm f
d PS 20cm
13cm

Q 24cm R 18cm

18cm

7. a. In the given figure, if the length of a side of the base

of the pyramid having square base is 12cm and

the volume of the pyramid is 384cm³, find the total

surface area of the pyramid. 12cm

b. The volume of a square based pyramid is 750 cubic cm and height 10cm. Find

the total surface area of the pyramid. E

8. a. In the given figure, the total surface area of the square based D C

pyramid is 96cm² and the side of the square base is 6cm. Find the

volume of the pyramid. PA B

6cm

b. The adjoining figure is a square based pyramid. If A D
its slant height is 13cm and its total surface area 360 O 13cm
square cm, find the volume of the pyramid.

BC

4.5 Cone : Introduction

When we cut off the sector POQ from the circular O
O
piece of paper and roll them such that OP and OQ
Q
join together then a cone is formed as shown in the PQ

figure. The curved surface of the cone is formed A Vertex

from the sector of the circle and circumference of Sector P
base of the cone is formed from the arc PQ. Thus a

cone is a pyramid with circular base.

The given solid figure is a cone. A is the vertex, O is the centre h l (slant height)
of circular base, OB is radius and AB is the slant height. As AOB
is a right angled triangle, l² = h² + r². OB
r
radius of base

GREEN Mathematics Book-10 69

O

Surface area and volume of the cone

The curved surface area of a cone = area of sector of a circle

= πrl PQ

Total surface area of a cone (T.S.A.) C.S.A. = πrl

= curved surface area + area of circular base

= πrl + πr²

= πr(l + r) Area of circular base

Cone is a circular base pyramid. Thus in generalised sense, the volume of cone can be
calculated by using the same formula as the volume of the pyramid.

∴ Volume of cone = Volume of circular base pyramid

= 31 A × h

= 1 πr²h.
3

Combined solids L.S.A. = 2πr² + 2πrh

r T.S.A. = 2πr² + 2πrh + πr²

a Volume (V) = volume of hemisphere + volume of cylinder

h = 2 πr³ + πr²h
3
br
Total surface area (T.S.A.) = Curved area of two hemispherical
h ends + curved surface area of cylinder

= 2πr² + 2πrh + 2πr2 = 4πr² + 2πrh

Volume (v) = volume of two hemispheres + volume of cylinders
= 2 πr³ + 2 πr³ + πr²h

33
= 4 πr³ + πr²h

3

cr Total surface area (T.S.A)

l

h1 = area of circle + C.S.A. of cylinder + C.S.A. of cone
h = πr² + 2πrh + πrl

Volume (V) = volume of cylinder + volume of cone

= πr²h + 1 πr²h1
3
70 GREEN Mathematics Book-10

hr Total surface (T.S.A.) = curved surface area of cone + curved
surface area of hemisphere
d
= πrl + 2πr²
Heighth' l(slant height)
Volume (V) = volume of cone + volume of hemisphere
= 1 πr²h + 2 πr³

33
= 1 πr² (h + 2r)

3
Lateral surface area (L.S.A.) = lateral surface of cuboid +

L.S.A. of pyramid = 4ah + 2al

Total surface area (T.S.A) = Area of square base + lateral
surface area of cuboid + lateral surface of pyramid

h = a² + 4ah + 2al

a Volume (V) = Volume of cuboid + volume of pyramid

= a² × h + 1 . a² h
3

Worked Out EXAMPLES

EXAMPLE 1 Find the curved surface area, total surface area and 5cm
Solution : volume of the given cone.

Here, 3cm
71
slant height of the cone (l) = 5cm

radius of the circular base (r) = 3cm
Now, height of the cone (h) = l² – r²

= 5² – 3²
= 16cm2

= 4cm

Curved surface area of the cone (C.S.A.) = πrl
= 22 × 3 × 5

7
= 47.14cm²

Total surface area of the cone (T.S.A.) = πr(l + r)

= 22 × 3 (5 + 3)
7

= 22 × 3 × 8
7

GREEN Mathematics Book-10

= 75.43cm²

Volume of the cone (V) = 1 πr²h
3

= 1 × 22 × 3² × 4
3 7

= 37.71cm³

EXAMPLE 2 Find the lateral surface area, total surface 6cm 20cm
area and the volume of the given solid. h'

Solution : Here, 12cm

Radius of the base (r) = 6cm

Height of the cylinder (h) = 12cm

Height of the cone (h1) = 20 – 12 = 8cm
Now, Slant height of the cone (l) = h1² + r²

= 8² + 6²

= 10cm

∴ Lateral surface area (L.S.A) = C.S.A of cylinder + C.S.A. of cone

= 2πrh + πrl

= 2 × 22 × 6 × 12 + 22 × 6 × 10
77

= 452.57 + 188.57

= 641.14cm²

∴ Total surface area (T.S.A.) = πr² + 2πrh + πrl

= 22 ×6²+2× 22 ×6× 12 + 22 × 6 × 10
77 7

= 113.14 + 452.57 + 188.57

= 754.28cm²

∴ Volume (V) = Volume of cylinder + volume of cone

= πr²h + 1 πr²h1
3

= 22 × 6² × 12 + 1 × 22 × 6² × 8
7 37

= 1357.71 + 301.71

= 1659.42cm³

72 GREEN Mathematics Book-10

EXAMPLE 3 Find the volume of the given solid:

Solution : Here, 45cm

Let the height of the upper part be h1 14cm
and lower part be h2
∴ h1 + h2 = 45
Now, Volume of the given solid (V)

= Volume of upper cone + volume of lower cone

= 1 πr²h1 + 1 πr²h2
3 3

= 1 πr²(h1 + h2)
3

= 1 × 22 × 72 × 45 [∵ r = d = 14 = 7]
37 22

= 2310cm³

EXAMPLE 4 A tent in the form of a cylinder of diameter 20m and height 2.5m is
Solution : surmounted by a cone of equal base and slant height 12.52m. Find the
cost of canvas at Rs. 100 per square meter.

Here,

diameter of base (d) = 20cm

radius of base (r) = d = 20 = 10m
22

height of cylinder (h) = 2.5 m

slant height of cone (l) = 12.52m

Now,

Surface area of canvas required to make tent (A)

= C.S.A. of cylinder + C.S.A. of cone

= 2πrh + πrl

= 2 × 22 × 10 × 2.5 + 22 × 10 × 12.52
77

= 157.14 + 393.49
= 550.63m²
Unit cost of canvas (R) = Rs 100 per m²
Thus, total cost of canvas (T) = A × R
= 550.63 × 100
= Rs. 55063

GREEN Mathematics Book-10 73

EXERCISE 4.5

A. Very Short Questions A
1. a. Is cone a prism or pyramid?

b. Name the following from the given solid figure. h
i. Point A ii. AO iii. OB iv. AB .
c. Write the relation of AB, AO and OB. OB
r

2. a. The area of circular base of a cone is 154cm2 and height is 6cm. Find the volume
of the cone.

b. The total surface area and curved surface area of cone are 301.71cm² and
188.57cm² respectively. Find the area of the base.

c. Slant height and vertical height of a cone are 8cm and 6cm respectively. Find the
volume of the cone.

d. Find the lateral surface area of the cone if its radius of base and slant height are
5cm and 13cm respectively.

B. Short Questions

3. Find the volume of the following cones:

A b 6cm A

a c

4cm

OB 30° 5cmOB
10cm 33
6cm13cm

4. Find the total surface area of the the following cones:

a b 10cm c

10cm 8cm 2cm

5. a. The total surface area of a cone whose base radius 5m is 282 6 m². Find the slant
7
height of the cone.

b. The total surface area of a cone is 814 cm². If the sum of its slant height and the
radius of its base is 37cm, find the slant height of the cone.

c. The total surface area and the curved surface area of a cone are 704cm² and 550
cm² respectively. Find the radius of the base of the cone.

74 GREEN Mathematics Book-10

d. The radius and slant height of a cone are in the ratio 7:25. If its curved surface
area is 2200cm2, find the radius.

6. Find the total surface of the given solids.

a b 12cm c 18cm 12cm

12cm 5cm

80cm 10cm 10 cm

18cm

d 15cm e 13cm 15cm

12cm

7cm

7. Find the volume of each of the following solids:

a 14cm 13cm b 30cm c

10 cm 25cm 13cm

54cm

25cm 14cm

d e f 26cm 4cm 14cm

10cm 24cm

8. a. In the adjoining solid, the area of the circular base is 100cm² 3cm
and the height of the cylinder is 3cm. If the volume of the
whole solid is 500cm³, find the height of the solid.

b. Given figure is a solid composed of a cylinder 30cm14cm
with hemisphere at one end. If the total surface 14cm
area and the height of solid are 770 cm² and
14cm respectively, find the height of the
cylinder. It radius of cylinder is 7cm.

c. The total height of the given solid is 30cm and radius of the base is
14cm. If the height of cylinder and cone are in the ratio of 3 : 2, find
the volume of the solid.
11cm
9. a. Calculate the cost of painting the inner 3cm
side of the tent at Rs. 25 per m².

12cm 75
GREEN Mathematics Book-10

b. A circus tent in the form of cylinder of diameter 20m and height 2.5m is
surmounted by a cone of equal base and height 7.5m. Find the cost of canvas at
Rs. 100 per m².

c. A container made up of metal sheet is in the form of cylinder. If radius of base
is 14cm and height is 21cm, find the cost of milk which can completely fill the
container at the rate of Rs 40 per liter.

d. A solid is composed of a cylinder with hemispherical ends. If the whole length
of the solid is 104cm and the radius of each of the hemispherical ends is 7cm,
find the cost of polishing its surface at the rate of Re 1 per m².

e. A cylindrical road-roller made of iron is 1m long. Its internal diameter is 54cm

and the thickness of the iron sheet used in making the roller is 9cm. If 1cm³ of

iron has 7.8g mass, find the mass of the roller. 5cm

10. a. Find the total surface area and the 14cm
volume of the given solid.

5 cm

6cm

b. In the adjoining figure, the total volume is 512cm³.
Find the total height of the solid figure.

8cm

8cm

Project Work:

a. Collect 5 different shapes of objects at your home and combine any two object and
measure their size. Also find the volume, total surface area and cross section area
and fill in the following table.

S.N. Name of objects Base area Total surface area Volume

b. Prepare the model of pyramids, prism and cone by using paper, mud, bamboo,
wood, etc. Prepare the combined solid object from the local materials. Also calculate
their total surface area and volume.

76 GREEN Mathematics Book-10

5

H.C.F. and L.C.M.

Estimated Teaching Periods : 10

H.C.F. and L.C.M., the world-famous Euclid's Algorithm that was invented
around 300 B.C., in order to find the HCF of two numbers. So, I take this
opportunity to honour the intellectual giant for providing us with this
great mathematical tool.

Contents

5.1 Highest Common Factor ( H.C.F )
5.2 Lowest Common Multiple ( L.C.M. )

Objectives

At the end of this unit, students will be able to:
meaning of H.C.F. and L.C.M.
know factorization of algebraic expression.

Materials

chart of formula and chart of rule of factorization.

GREEN Mathematics Book-10 77

5.1 Highest Common Factor ( H.C.F )

Highest Common Factor (H.C.F) is the common factor in all the 3x2y 6xy2
given algebraic expressions. So H.C.F. can exactly divide each of the x 3xy 2y
given expressions. For example, 3x2y and 6xy2 are two expressions.
In these expressions 3xy is common factor. So, H.C.F. = 3xy.

While we find H.C.F., we should use the following steps:

Steps: H.C.F.

i. Find the factors of the given expressions.

ii. Take common factors of the given expressions.

iii. Express the common factors in the product form which is the required H.C.F.

iv. If there is no any common factor in the given expressions, then H.C.F. is 1 since 1 is
a factor of any expression.

Look at the following worked out examples:

Worked Out EXAMPLES

EXAMPLE 1 Find H.C.F. of 9a³b² and 15b³c²d.
The 1st expression = 9a³b² = 3 × 3 × a × a × a × b × b
Solution : The 2 nd expression = 15b³c²d = 3 × 5 × b × b × b × c × c × d
Common factors = 3 × b × b = 3b²
∴ H.C.F = 3b²

Alternate method
1st expression = 9a³b²
2nd expression = 15b³c²d
H.C.F of coefficients 9 and 15 = 3
V ariable with the least power = b²
∴ H.C.F = 3b²

EXAMPLE 2 Find H.C.F. of 3(a² – b²) and 12a²b(a² + 2ab + b²)
Solution : T he 1st expression = 3(a² – b²)
= 3 (a – b) (a + b)
The 2nd expression = 12a²b (a² + 2ab + b²)
= 2 × 2 × 3 × a × a × b (a + b)(a + b)
∴ H.C.F = 3 (a + b)

78 GREEN Mathematics Book-10

EXAMPLE 3 Find H.C.F. of (a³ – b³) (a + b)², 3a4 + 2a³b – 5a²b² and a4 – b4.
The 1st expression = (a³ – b³) (a + b)²
Solution : = (a – b) (a² + ab + b²) (a + b)²
The 2nd expression = 3a4 + 2a³b – 5a²b²
= a² (3a² +2ab – 5b²)
= a² (3a² + 5ab – 3ab – 5b²)
= a² { a (3a + 5b) – b (3a + 5b) }
= a² (3a + 5b) (a – b)
= a² (a– b) (3a + 5b)
The 3rd expression = a4 – b4
= (a2 + b2) (a2 – b2)
= (a2 + b2) (a + b) (a – b)
∴ H.C.F. = (a – b)

EXAMPLE 4 Find H.C.F. of x2 – 5x + 6, x² + 4x – 12 and x2 – 2x
The 1st expression = x2 – 5x + 6
Solution : = x2 – 3x – 2x + 6
= x(x – 3) – 2(x – 3)
= x(x – 3) (x – 2)
The 2nd expression = x² + 4x – 12
= x² + 6x – 2x – 12
= x(x + 6) – 2(x + 6)
= (x + 6) (x – 2)
The 3rd expression = x² – 2x
= x(x – 2)
∴ H.C.F. = (x – 2)

GREEN Mathematics Book-10 79

EXAMPLE 5 Find H.C.F. of a2 – b2 – 2bc – c2, b2 – c2 – 2ca – a2 and c2 – a2 – 2ab – b2
The 1st expression = a2 – b2 – 2bc – c2
Solution : = a2 – (b2 + 2bc + c2)
= a2 – (b + c)2
= (a + b + c) (a – b – c)
T he 2nd expression = b2 – c2 – 2ca – a2
= b2 – (c2 + 2ca + a2)
= b2 – (c + a)2
= (b + c + a) (b – c – a)
= (a + b + c) (b – c – a)
The 3rd expression = c2 – a2 – 2ab – b2
= c2 – (a2 + 2ab + b2)
= c2 – (a + b)2
= (c + a + b) (c – a – b)
= (a + b + c) (c – a – b)
∴ H.C.F. = (a + b + c)

EXAMPLE 6 Find H.C.F. of a3 – 2a2b + 2ab2 – b3, a4 + b4 + a2b2, 4a4b + 4ab4

Solution : The 1st expression = a3 – 2a2b + 2ab2 – b3
= a3 – b3 – 2a2b + 2ab2
= (a – b) (a2 + ab + b2) – 2ab (a – b)
= (a – b) (a2 + ab + b2 – 2ab)
= (a – b) (a2 – ab + b2)
The 2nd expression = a4 + b4 + a2b2
= a4 + b4 + a2b2
= (a2)2 + 2a2 b2 + (b2)2 – a2b2
= (a2 + b2)2 – (ab)2
= (a2 + b2 + ab) (a2 + b2 – ab)
= (a2 + ab + b2) (a2 – ab + b2)
The 3rd expression = 4a4b + 4ab4
= 4ab(a3 + b3)
= 4ab (a + b) (a2 – ab + b2)
∴ H.C.F. = (a2 – ab + b2)

80 GREEN Mathematics Book-10

EXAMPLE 7 Find H.C.F. of a2 – 18a – 19 + 20b – b2, a2 + a – b2 + b, 4a2– 4b2 + 8b – 4

Solution : The 1st expression = a2 – 18a – 19 + 20b – b2
= a2 – 2.a.9 + 92 – 100 + 20b – b2
= (a – 9)2 – (102 – 2 . 10 . b + b2)
= (a – 9)2 – (10 – b)2
= {(a – 9) + (10 – b)} {(a – 9) – (10 – b)}
= (a – 9 + 10 – b) (a – 9 – 10 + b)
= (a – b + 1) (a + b – 19)
T he 2nd expression = a2 + a – b2 + b
= a2 – b2 + a + b
= (a + b) (a – b) + 1(a + b)
= (a + b) (a – b + 1)
The 3rd expression = 4a2– 4b2 + 8b – 4
= 4(a2 – b2 + 2b – 1)
= 4{a2 – (b2 – 2b + 1)}
= 4{a2 – (b – 1)2}
= 4(a + b – 1) (a – b + 1)
∴ H.C.F. = (a – b + 1)

EXAMPLE 8 Find H.C.F. of a(1 + ab) + a(a – b2) – b(1 + b), 5a4b – 5ab4 and 2ab3 – 2a3b

Solution : The 1st expression = a(1 + ab) + a(a – b2) – b(1 + b)

= a + a2b + a2 – ab2 – b – b2
= a2 – b2 + a – b + a2b – ab2
= (a + b) (a – b) + 1 (a – b) + ab (a – b)
= (a – b) (a + b + 1 + ab)

The 2nd expression = 5a4b – 5ab4
= 5ab (a3 – b3)
= 5ab (a – b) (a2 + ab + b2)

The 3rd expression = 2ab3 – 2a3b
= 2ab (b2 – a2)
= – 2ab (a2 – b2)
= – 2ab (a + b) (a – b)
∴ H.C.F. = (a – b)

GREEN Mathematics Book-10 81

EXERCISE 5.1

A. Very Short Questions b. 12x² – 27 and 4x² + 6x

1. Find the H.C.F. of the following:

a. 16x³y² and 24x²y³
c. a2 and b2

2. Find the H.C.F. of: d. (x² – y²) and (x + y)²
e. (c² – d²) and c² + 2cd + d²
a. 18p²q² and 12p³q²r
b. 15c²d² and 18cde
c. 45a4x5y², 54a³x6y³ and 72a5x3y

3. Find the H.C.F. of: b. 8x² –18y² and (2x + 3y)²
a. 24 (x³ – a³) and 10(x² – a²)

B. Short Questions

4. Find the H.C.F. of:
a. 6p² – 13p + 6 and 12p² + 7p – 10 b. a³ + b³ and a4 + a²b² + b4
c. 6x² – x – 1 and 54x4 + 2x

5. Find the H.C.F. of:

a. 8m³ + n³ and 16m4 + 4m²n² + n4
b. 2x³ – x² – x, 4x³ – x and 8x4 + x

6. Find the H.C.F. of:

a. a² – 13a + 12 and a² + 4a – 5 d. 4x² – 9 and 8x³ + 27
b. ax² + 7ax + 12a and a²x² + 6a²x + 9a² e. 8a³ + b³ and 8a³ – b³
c. x³ – 64y³ and x² – 16y²

C. Long Questions

7. Find the H.C.F. of:

a. 3a² + 9ab, a³ + 6a²b + 9ab² and 2a³ – 18ab²
b. a³ + b³, a³ – a²b + ab² and a4 + a²b² + b4
c­ . x² – 6ax + 9a², 2x2 – ax – 3a2 and 8x3 – 27a3
d. 2c³ – c² – c, 4c³ – c and 8c4 + c
e. 4p4 + 19p²q² + 49q4, 4p³ + 14pq² – 6p²q and 2p3 – 3p2q + 7pq²
f. x³ – 1, x4 + x² + 1 and x³ + 1 + 2x² + 2x
g. 1 – 2x – 4x2 + 8x3, 16x4 – 8x3 + 4x2 – 2x

82 GREEN Mathematics Book-10

8. Find the H.C.F. of:
a. (a + b)² – 4ab, (a – b)³ + 3ab (a – b) and a² – b²
b. 2x² – 8, x³ + 8 and x² – 2x – 8
c. (a + b)² – c², (b + c)² – a² and (c + a)2 – b²
d. (3x – 2y) (3x + 2y) – 8yz – 4z², (2z – 2y) (2z + 2y) – 9x² – 12xy and (3x + 2y)² – 4z²
e. 9m² – 4n² – 4nr – r², r² – 4n² – 9m² –12mn and 9m² + 6mr + r² – 4n²
f. 3x² – 8x + 4, 2x2 – 5x + 2 and x4 – 8x
g. 5x³ – 20x, x³ – 3x² – 10x and x³ – x² – 2x + 8
h. 8m³ + 27n³, 16m4 + 36m²n² + 81n4 and 4m³ – 6m²n + 9mn²
i. a³b + b4, a4 + a²b² + b4 , 2ma³ – 2ma2b + 2mab²
j. 8m4 + 27mn³, 8m³n +2m²n² – 15mn³ and 4m³n – 9mn³
k. a³ – 8, a4 + 4a² + 16 and a³ + 2a² + 4a
l. 4x4 + 16x³ – 20x², 3x³ + 14x² – 5x and x4 + 125x
m. 4z4– 9z², 6z³ – 5z² – 6z and 4z³ – 12z² + 9z

D. Very Long Questions

9. Find the H.C.F. of:
a. (a – b)² + 4ab, (a + b)³ – 3ab (a + b) and a² + 2ab + b²
b. 1 – 4x2 – 8x3 + 16x4 and 1 – 4x + 16x³ – 4x²
c. 4x2 – 9y2 + 6yz – z2, 9y2 – 4x2 + 4xz – z2 and z2 – 9y2 + 12xy – 4x2
d. a2 – 2bc – b2 – c2, b2 – 2ca – a2 – c2 and c2 – 2ab – a2 – b2
e. a(a + c) – b(b + c), b(a + b) – c(c + a) and c(b + c) – a (a + b)
f. a2 – b2 + a + 3b – 2 and a2 – 2ab + b2 + 5a – 5b + 6
g. x2 – 2x + 1 – 2xy + 2y and x2 – 4xy + 4y2 + 2x – 4y – 3

GREEN Mathematics Book-10 83

5.2 Lowest Common Multiple ( L.C.M. )

To calculate lowest common multiple of algebraic expressions, we x2+5x + 6 x2+3x+2
should convert them into simplest factors. Then taking common x+3 x+2 x+1
factors of the given expressions, the remaining factors should be
multiplied. L.C.M.

∴ L.C.M = Common factors × Remaining factors

For example: x2 + 5x + 6 and x2 + 3x + 2 are two algebraic expressions.

Worked Out EXAMPLES

EXAMPLE 1 Find the L.C.M. of 3x²yz, 4y²z and 5xz²

Solution : The first expression = 3x²yz = 3 × x × x × y × z

T he second expression = 4y²z = 2 × 2 × y × y × z

The third expression = 5xz² = 5 × x × z × z

Common factors = z

Remaining factors = 3 × 2 × 5 × 2 × x × x × y × y × z

= 60x2y2z

∴ L.C.M = Common Factors × Remaining factors

= z × 60x2y2z

= 60x²y²z²

EXAMPLE 2 Find the L.C.M. of m4 – 4m² and 3m² + 6m

Solution : The first expression = m4 – 4m²

= m² (m² – 4)

= m² (m² – 2²)

= m² (m – 2) (m + 2)

= m × m (m – 2) (m + 2)

The second expression = 3m² + 6m

= 3m(m + 2)

Common factors = m(m + 2)

Remaining factors = m(m – 2) × 3

∴ L.C.M = common factors × remaining factors

= m(m + 2) × 3m(m – 2)

= 3m² (m² – 4)

84 GREEN Mathematics Book-10

EXAMPLE 3 Find the L.C.M. of a² – 3a + 2, a4 + a³ –6a² and a³ + 2a² – 3a

Solution : The first expression = a² – 3a + 2

= a² – 2a – a + 2

= a(a – 2) – 1 (a –2)

= (a – 2) (a – 1)

The second expression = a4 + a³ – 6a²

= a² (a² + a – 6) Remember:
= a² (a² + 3a –2a – 6)
= a² {a(a + 3) – 2 (a + 3)} When we find L.C.M.
= a² (a + 3) (a – 2) we should take
= a × a (a + 3) (a – 2) common at least any
two terms.

The third expression = a³ + 2a² – 3a

= a( a² + 2a – 3)

= a (a² + 3a – a – 3)

= a { a(a + 3) – 1 (a + 3)

= a (a + 3) (a – 1)

Common factor = a

Common factors from any two ten = (a + 3) (a – 1)

Remaining factors = a (a – 2)

∴L.C.M = Common factors × remaining factors

= a(a – 2)(a – 1) (a + 3) × a

= a² (a – 1) (a – 2) (a + 3)

EXAMPLE 4 Find the L.C.M. of x2 – 5x + 6, x² + 4x – 12 and x4 – 2x

Solution : T he first expression = x2 – 5x + 6
= x2 – 3x – 2x + 6
= (x – 3) (x – 2)
T he second expression = x² + 4x – 12
= x² + 6x – 2x – 12
= x(x + 6) – 2(x + 6)
= (x + 6) (x – 2)
The third expression = x² – 2x
= x(x – 2)
∴ L.C.M. = Common factors × remaining factors
= (x – 2) × (x – 3) (x + 6) × (x)
= x (x – 2) (x – 3) (x + 6)

GREEN Mathematics Book-10 85

EXAMPLE 5 Find the L.C.M. of a2 – b2 – 2bc – c2, b2 – c2 – 2ca – a2 and c2 – a2 – 2ab – b2

Solution : T he first expression = a2 – b2 – 2bc – c2
= a2 – (b2 + 2bc + c2)
= a2 – (b + c)2
= (a + b + c) (a – b – c)
The second expression = b2 – c2 – 2ca – a2
= b2 – (c2 + 2ca + a2)
= b2 – (c + a)2
= (b + c + a) (b – c – a)
T he third expression = c2 – a2 – 2ab – b2
= c2 – (a2 + 2ab + b2)
= c2 – (a + b)2
= (c + a + b) (c – a – b)
∴ L.C.M. = Common factors × remaining factors
= (a + b + c) × (a – b – c) × (b – c – a) × (c – a – b)
= (a + b + c) (a – b – c) (b – c – a) (c – a – b)

EXAMPLE 6 Find the L.C.M. of a3 – 2a2b + 2ab2 – b3, a4 + b4 + a2b2, 4a4b + 4ab4

Solution : The first expression = a3 – 2a2b + 2ab2 – b3

= a3 – b3 – 2a2b + 2ab2

= (a – b) (a2 + ab + b2) – 2ab (a – b)

= (a – b) (a2 + ab + b2 – 2ab)

= (a – b) (a2 – ab + b2)

T he second expression = a4 + b4 + a2b2

= a4 + b4 + a2b2
= (a2)2 + 2a2 b2 + (b2)2 – a2b2
= (a2 + b2)2 – (ab)2
= (a2 + b2 + ab) (a2 + b2 – ab)
= (a2 + ab + b2) (a2 – ab + b2)

The third expression = 4a4b + 4ab4

= 4ab(a3 + b3)
= 4ab (a + b) (a2 – ab + b2)
∴ L.C.M. = (a2 – ab + b2) (a – b) (a + b) (4ab) (a2 + ab + b2)
= 4ab (a2 – b2) (a2 – ab + b2) (a2 + ab + b2)

86 GREEN Mathematics Book-10

EXAMPLE 7 Find the L.C.M. of a2 – 18a – 19 + 20b – b2, a2 + a – b2 + b, 4a2– 4b2 + 8b – 4

Solution : The first expression = a2 – 18a – 19 + 20b – b2
= a2 – 2.a.9 + 92 – 100 + 20b – b2
= (a – 9)2 – (102 – 2 . 10 . b + b2)
= (a – 9)2 – (10 – b)2
= {(a – 9) + (10 – b)} {(a – 9) – (10 – b)}
= (a – 9 + 10 – b) (a – 9 – 10 + b)
= (a – b + 1) (a + b – 19)
The second expression = a2 + a – b2 + b
= a2 – b2 + a + b
= (a + b) (a – b) + 1(a + b)
= (a + b) (a – b + 1)
The third expression = 4a2– 4b2 + 8b – 4
= 4(a2 – b2 + 2b – 1)
= 4{a2 – (b2 – 2b + 1)}
= 4{a2 – (b – 1)2}
= 4(a + b – 1) (a – b + 1)
∴ L.C.M. = (a – b + 1) 4(a + b – 1) (a + b) (a + b – 19)
= 4 (a – b + 1) (a + b – 1) (a + b) (a + b – 19)

EXAMPLE 8 Find the L.C.M. of 1 + 4a + 4a2 – 16a4, 1 + 2a – 8a3 – 16a4 and 1 – 8a3

Solution : T he first expression = 1 + 4a + 4a2 – 16a4

= (1)2 + 2.1.2a + (2a)2 – (4a2)2
= (1 + 2a)2 – (4a2)2

= (1 + 2a + 4a2) (1 + 2a – 4a2)

The second expression = 1 + 2a – 8a3 – 16a4

= 1(1 + 2a) – 8a3(1 + 2a)
= (1 + 2a) (1 – 8a3)
= (1 + 2a) (1 – 2a) (1 + 2a + 4a2)

T he third expression = 1 – 8a3

= (1)3 – (2a)3
= (1 – 2a) (1 + 2a + 4a2)

∴ L.C.M. = Common factors × remaining factors
= (1 + 2a + 4a2) (1 – 2a) × (1 – 2a + 4a2) (1 + 2a)
= (1 – 8a3) (1 + 8a­3)

GREEN Mathematics Book-10 87

EXAMPLE 9 If the L.C.M. and H.C.F. of algebraic expressions are (x2 – 5x + 6) and
(x – 3) respectively and one of expressions is (x2 – 3x + 2), find the
other expressions.

Solution : Here,

L.C.M. = x2 – 5x + 6
= x2 – 3x – 2x + 6
= x(x – 3) – 2(x – 6)
= (x – 3) (x – 2)
H.C.F. = (x – 3)
Given expression = x2 – 3x + 2
= x2 – 2x – x + 2
= x(x – 2) – 1(x – 2)

= (x – 2) (x – 1)

L.C.M. × H.C.F.
Other expression = Given expression

(x–3)(x–2)×(x–3)
=

(x–2)(x–1)
x2–6x+9
=

(x–1)

EXERCISE 5.2

A. Very Short Questions

1. Find the L.C.M. of the following expressions:

a. 4a³b², 10a²b² and 15a²b³ b. (a – b)2 and (a3 – b3)

c. 2x2 – 2y2 and x2 + 2xy + y2 d. 20x3 – 20y3 and 5x – 5y

e. 9x2 – 4y2, 6x – 4y and 3x – 2y f. 20x3y – 10xy and 2x2y – y2

g. 2x2 – 4xy + 2y2 and 3x2 – 3y2 h. 5x + 10y and x2 + 4xy + 4y2

i. (a – b)2 + 4ab and a2 – b2

B. Short Questions

2. Find the L.C.M. of:

a. x² – y² and x² + xy – 2y² b. p² – 4pq + 4q² and p² + 3pq – 10q²

c. 4x²(x² – a²), 9x²(x³ – a³) and 6x³(x – a)²

88 GREEN Mathematics Book-10

d. 40m²ab², 30ma²b² and 60m³a³b³ e. a² – 3a + 2, 2a³ – 16

f. 6x² – x – 1, 54x4 + 2x g. a2 – 4a + 4 and a2 – 5a + 6

h. x3y – xy3, 3x + 3y and (5x – 5y) i. a3 + 8b3, 2a2 – 4ab + 8b2 and 4a + 8b

C. Long Questions

3. Find the L.C.M. of:
a. mx² – 3mx + 2m, m²x4 + m²x³ – 6m2x² and m³x³ + 2m³x² – 3m³x

b. 8x³ – 27y³, 8x³ + 27y³ and 16x4 + 81y4 + 36x²y²

4. Find the L.C.M. of:
a. 3a(a² – b²), 2b(a² + b²) and 8a²b²(a4 – b­4)
b. x² – y² , x³ + y³ and x³ – y³
c. p² – 4, p³ – 8 and p³ + 8
d. a³ – 4a, a4 – a³ –2 a² and a³ – 8
e. x² + 2xy + y² – z², y² + 2yz + z² – x² and z² + 2zx + x² –y²
f. x² – 6ax + 9a², 6x² – 21ax +9a² and x² – 9a²
g. a³ – 1, a³ + 1 and a4 + a² + 1
h. x² + 2x – 15, x² + 9x + 20 and x² + 4x – 21
i. 6b² – b – 1, 2b² + 3b – 2 and 3b² + 7b + 2
j. 2x² – 8, x³ – 8 and x² – 2x – 8
k. a4 + a² + 1, a³ – 1 and a³ – a² + a
l. x4 + (2b² – a²) x² + b4 and x4 + 2ax³ + a²x² – b4
m. a² – b² + 2bc – c², (a + b)² – c² and (a + b – c)²

D. Very Long Questions

5. Find the L.C.M. of:

a. (a – b)² + 4ab, (a + b)³ – 3ab (a + b) and a² + 2ab + b²
b. 1 – 4x2 – 8x3 + 16x4 and 1 + x – x2 – 10x3
c. 4x2 – 9y2 + 6yz – z2, 9y2 – 4x2 + 4xz – z2 and z2 – 9y2 + 12xy – 4x2
d. a2 – 2bc – b2 – c2, b2 – 2ca – a2 – c2 and c2 – 2ab – a2 – b2
e. a(a + c) – b(b + c), b(a + b) – c(c + a) and c(b + c) – a (a + b)
f. a2 – b2 + a + 3b – 2 and a2 – 2ab + b2 + 5a – 5b + 6
g. x2 – 4y2 – 3x + 2y + 2 and x2 – 4xy + 4y2 + 2x – 4y – 3

6. Solve the following questions:

a. H.C.F. and L.C.M. of two expression MP (x + 2) and (x2 + 7x + 10), if one
expression is (x2 + 5x), find the other expression.

b. If the product of two expression is a3 – 8b3 and L.C.M. of two expressions
is (a2 + 2ab + 4b2), find its H.C.F.

GREEN Mathematics Book-10 89

6

Indices and Surds

Estimated Teaching Periods : 10

Body Mass Index, derived from a simple math formula, was devised in
the 1830s by Lambert Adolphe Jacques Quetelet (1796-1874), a Belgian
astronomer, mathematician, statistician and sociologist. BMI is said to
estimate how fat you are by dividing your weight in kilograms by your
height in meters square. However, as mentioned earlier, the measurement
is flawed, especially if the person carries a lot of muscle.

Contents

6.1 Introduction
6.2 Laws of indices
6.3 Exponential equations
6.4 Surds
6.5 Rationalization of surds
6.6 Equation involving surds

Objectives

At the ends of this unit, students will be able to:
introduc the laws of indices.
explaine the generalization of exponential equations.
conjugate of given surds.
solve equation involving surds.

Materials

Panel board, chart of major laws of indices.

90 GREEN Mathematics Book-10

6.1 Introduction

Review

When a number or a variable is multiplied by For example :
itself two or more times, then the product can
be written in the short form by a symbol which x × x × x can be written as x³. In this
is known as the index of the number or variable. case, x is multiplied itself three
times.
The index is also called the power or exponent.
In y5, 5 is called the index of y. The plural from So, in x³, x is called the base and 3
of index is called indices. If x is multiplied m is called the index of x.
times i.e. x × x × x × ......... m times, is equal to xm,
where m is called the index of x. Let's take an algebraic term 4x³.

In 4x³, 4 is the coefficient of x³, x is
the base and 3 is the index of x.

6.2 Laws of indices

There are certain rules which are used to solve the problems of indices. These rules are
called the laws of indices.

Multiplication law of indices with same base

The index in the For example :
multiplication of two or x³ × x4
more algebraic terms with = x × x × x × x × x × x × x = x7 = x3 + 4
the same base is the sum of ∴ xm × xn = (x×x×x .... m times) × (x×x×x .... n times)
the indices of each term. = x × x × x .............. (m + n) times
= xm + n
Thus, xm × x­n = xm+n, where m and n are positive
integers and x ≠ 0.

Law of negative index

An algebraic term with For example : x2 x × x 1 1
negative index is the x4 x × x x×x x²
reciprocal of the term with i. x–2 = x2– 4 = x2 ÷ x4 = = x × × x = =
the positive index.
ii. x– m = xm – 2m = xm ÷ x2m

= xm = xm = 1
x2m xm × xm xm

∴ xm = 1 , where x ≠ 0.
x–m

GREEN Mathematics Book-10 91

Division law of indices with same base

TheLientd'sexseienththaet,d ivision of For example :
two algebraic terms with the
same base is the difference of Let, x5 ÷ x²

v. the :indices of each term. = x5 = x × x × x × x × x = x³ = x5–2
x² x × x
Let's see that,
∴ xm ÷ xn

= (x × x × x ..... m times) ÷ (x × x × x ........... n times)

= (x × x × x ......... (m – n) times

= xm – n, where m > n .

Thus, xm ÷ xn = xm – n, where m and n are positive
integers and x ≠ 0 and m > n.

Law of index of indices

The index of indices of any For example :
algebraic term is the product
of two indices. Let, (x³)² = x³ × x³
If x³ = y, then (x³)² = y² = y × y
It can also be extended = x3 + 3 = x6 = x3×2
∴ (xm)n = xm × xm × xm ............n times
[{xmn}p]q = xm + m + m..... n times
= xmn
= [xmnp]q ∴ (xm)n = xmn, where m and n both are integers

= xmnpq and x ≠ 0.

Law of zero index

Any algebraic term having For example :
the index 0 is always equal
to 1. But 00 ≠ 1. x0 = x2 – 2 = x² ÷ x² = x² = 1, where x ≠ 0.
x²
1
y0 = ym–m = ym × y–m = ym ym = 1, where y ≠ 0.

92 GREEN Mathematics Book-10

Root law of indices

na = a 1 For example :
n
2
The square root of 32 = 32 = 3 2 = 3
3
The cube root of 33 = 3 33 = 3 3 = 3
4
The fourth root of 34 = 4 34 = 3 4 = 3
c
Similarly, a bc = b a
m
Thus, if is term with base a and power m, then,
an
mn
a n = n am

Worked Out EXAMPLES

EXAMPLE 1 Find the value of 3 3 512

Solution : 3 3 512

11

= {(83) 3 )} 3

1

= (8) 3

1

= (23) 3

=2

EXAMPLE 2 Simplify: o 64 – 1 × qo 36 p 1 ÷o 9 – 1 r
125 3 25 2 25 2
p p

Solution : = 64 – 1 × 36 1 ÷ 9 – 1
3 25 2 25 2
o125p qo p o p r

= o 4 3.o– 1 p × 6 2o 1 p÷ 3 2.o– 1 p
5 3 qo 5 p 2 5 2
p o p r

= 4 – 1 × 6 ÷ 3 –1
5 5 o5p r
o p q

= 5 × q 6 ÷ 5
4 5 3r

= 5 × 6 × 3
4 q5 5r

1 5 18 9
= 2 4 × 25 5

9
= 10

GREEN Mathematics Book-10 93

EXAMPLE 3 Simplify : 3 2x3y13 × 3 4x6y–1

Solution : 3 2x3y13 × 3 4x6y–1

= 3 2x3y13 × 4x6y–1

= 3 23 × x9 × y12

11 1

= (23) 3 . (x9) 3 . (y12) 3

= 2.x3.y4

= 2x3y4

EXAMPLE 4 310 + n × 93n – 5
Solution : Simplify: 34n + 1 × 33n – 1

310 + n × 93n – 5
34n + 1 × 33n – 1

310 + n × 32(3n – 5)
= 34n + 1 + 3n – 1

310 + n × 36n – 10
= 37n

310 + n + 6n – 10
= 37n

37n
= 37n
= 37n – 7n
= 3°
= 1

94 GREEN Mathematics Book-10