Math 10

4. (a) In the given solid, upper part is a square based pyramid of 5 cm
slant height 5 cm and lower part is a square based prism
having each side of base 8 cm. If the volume of whole figure
is 448 cm3, find the height of the prism.

8 cm

(b) In the given solid, the lower part is a square based prism 20cm
and the upper part is square based pyramid. The length of
the base of the prism is 10cm and its height is 20 cm. If the 10cm
volume of whole solid is 2400 cm3, find the slant height of the
pyramid.

Answers

1. (a) 300 cm3 (b) 448 cm3 (c) 832 cm3 (d) 1200 cm3 (e) 240 cm3 (f ) 1200 cm3

(g) 624 cm3 2. (i) 24 cm (ii) 25 cm (iii) 175 cm2 (iv) 700 cm2 (v) 56 cm

(vi) 1120 cm2 (vii) 196 cm2 (viii) 1820 cm2 (ix) 2016 cm2 3. (a) 860 cm2, 960 cm2

(b) 2,380 cm2, 2,576 cm2 (c) 480 cm2, 480 cm2 (d) 720 cm2, 763.30 cm2 4. (a) 6 cm (b) 13 cm

7.5 Geometrical Bodies

Pemba is planning to plaster the yard in front of his home. The cost of plastering per
square feet is Rs. 70. He has to calculate the cost of plastering his yard.

The structure of his yard is as shown in the figure.

C

A
B



E D

Let's find the area of the yard. The measurement of each side is given below. AB = 12 ft,

BC = 15 ft. CD = 17 ft, ED = 14 ft., AE = 10 ft.

A C
12 ft 15ft

B

10 ft

17ft

E D
14 ft



Let's take the measurement of BD and EB.

BD = 8 ft and BE = 16 ft.

Now, the area of the yard is the area of ∆ABE + Area of ∆BDE + Area of ∆BDC.

For ∆ABE,

Oasis School Mathematics-10 145

three sides, a = 10 ft, b = 12 ft, c = 16 ft.

s= a+b+c = 10+12+16 = 19 ft.
2 2

Area of ∆ ABE = s(s-a) (s-b) (s-c)

= 19(19–10) (19–12) (19–16)

= 19 × 9 × 7 × 3

= 3,591

= 59.92 square ft.

Again, for ∆ BDE,

three sides, a = 14 ft, b = 16ft, c = 8 ft.

s= a+b+c = 14+16+8 = 19 ft.
2 2

Area of ∆ ABE = s(s-a) (s-b) (s-c)

= 19(19–14) (19–16) (19–18)

= 19 × 5 × 3 × 11

= 3,135

= 55.99 square ft.

For ∆ BDC,

three sides, a = 17 ft, b= 15 ft, c = 8ft.

s= a+b+c = 17+15+8 = 20 ft.
2 2

Area of ∆ BDC = s(s-a) (s-b) (s-c)

= 20(20-17) (20-15) (20-8)

= 20 × 3 × 5 × 12

= 3,600

= 60 square ft.

Area of the yard = (59.92 + 55.99 + 60) sq. ft.

= 175.91 sq. ft.

Here, Cost of plastering 1 sq. ft. = Rs. 70

Cost of plastering 175.91 sq. ft. = Rs. 70 × 175.91 = Rs. 12313.70.

There are many such examples which come in our daily life like,

• Estimation of the cost of constructing the well.

• Estimation of the cost of constructing the pillar of a gate.

• Estimation of the cost of constructing a water tank, etc.

146 Oasis School Mathematics-10

Worked Out Examples



Example: 1
A water tank having dimensions 3m × 2.5m × 2m is to be constructed. Find
(i) how much water the tank can hold.
(ii) the area of its four walls
(iii) cost of plastering its four walls at the rate of Rs. 125 per square meter.

Solution:

Here, the length of the tank (l) = 3 m
breadth of the tank (b) = 2.5 m
height of the tank (h) = 2 m
(i) Volume of the tank (V) = l × b × h

= 3 m × 2.5 m × 2 m

= 15 m3

= 15 × 1000 liters [∴1m3 = 1,000 l]

= 15000 liters

(ii) Area of 4 walls = 2h(l+b)

= 2 × 2(3 + 2.5)m2

= 4 × 5.5 m2 = 22 m2

(iii) The cost of plastering 1m2 wall = Rs. 125

The cost of plastering 22m2 wall = Rs. 125 × 22

= Rs. 2,750

Example: 2
The diameter of a cemented ring of a well is 3 ft. Its height is 1.1 ft. Find
(i) the volume of the well if it contains 40 such rings.
(ii) cost of constructing a well if the cost per ring is Rs. 1,100.
(iii) volume of water, if the level of water is on the 15th ring from the top.

Solution:
Given, Diameter of the ring (d) = 3 ft.

∴ Radius (r) = 1.5 ft.

Height of a ring = 1.1 ft.

Height of 40 rings = 40 × 1.1 ft. = 44 ft.

We have,

Volume of the cylinder (well) = πr2h

= 22 × (1.5)2 × 44 = 311.14 ft3
7

Oasis School Mathematics-10 147

(ii) Cost per ring = Rs. 1100

Cost of 40 rings = Rs. 1100 × 40 = Rs. 44,000

(iii) Here, (40–15) rings are covered by water.

∴ Height of 25 rings = 25 × 1.1 ft.

= 27.5 ft.

Radius (r) = 1.5 ft.

We have, volume of water = πr2h

= 22 × (1.5)2 × 27.5 ft3
7

= 194.46 ft3

Example 3:

A house contains 12 pillars of dimension 30 cm × 30 cm × 3 m. Find the total cost of
constructing the pillars if the construction cost per cubic meter is Rs. 1,450. Also, find
the cost of plastering all the pillars at the rate of Rs. 90 per square meter.

Solution:

Dimension of a pillar = 30 cm × 30 cm × 3 m

∴ Volume of a pillar = 0.3 m × 0.3 m × 3 m

= 0.27 m3

Volume of 12 pillars = 12 × 0.27 m3 3m

= 32.4 m3 30 cm
30 cm
Construction cost per m3 = Rs. 450

∴ Total cost of constructing the pillars = 32.4 × 1450 = Rs. 46,980

Again, lateral surface area of a pillar,

= perimeter of base × height

= 2 (0.3 + 0.3) × 3 m

= 3.6 m2

LSA of 12 pillars = 12 × 3.6m2

= 43.2 m2

Cost of plastering per m2 = Rs. 90

Cost of plastering 43.2 m2 = Rs. 90 × 43.2

= Rs. 3,888

Example 4:

A cylindrical shaped tent is made for the accommodation of 77 people in such a way
that each person gets the area 2m2 on the ground and 10m3 of air to breathe. Find the
height of the tent and cost of canvas required to prepare the tent if the cost per m2 of

cloth is Rs. 80 per m2.

Solution:

148 Oasis School Mathematics-10

Here, total number of people = 77

Area to be occupied by each people = 2m2

Then, the area of base = 77 × 2 m2 = 154 m2

Again,

Since each person needs 10m3 of air to breathe

Volume of the tent = 77 × 10 cm3 = 770 cm3

We have,

Volume = Area of base × height

or, 770 = 154 × h

or, h= 770 = 5 cm.
154

Again,

Area of base = 154 m2

πr2 = 154

22 r2 = 154
7

r2 = 154 × 7= 154
22

∴ r = 7 cm.

Again,

Area of canvas required to prepare the tent

= CSA of tent + area of top

= 2πrh + πr2

= 2× 22 ×7×5+ 22 × 72
7 7

= 220 m2 + 154 m2

= 374 m2

Since the cost per m2 cloths = Rs. 80

Total cost = Rs. 374 × 80

= Rs. 29920.
Example 5:

A square based Pyramid shaped tent is made for the accommodation of 100 people.
For each person 4 sq. meter ground and 12m3 of air to breathe. Find,

(i) height of the ten (ii) slant height of the tent
(iii) LSA of the tent (iv) total cost to prepare the tent at the rate of Rs. 120 per m2.

Solution:
Total number of people = 100.
Since each person needs 4 square meter area on the grounds.
Area of the base (A) = 100 × 4 = 400 m2
Again, each person needs 12m3 of air to breathe

Oasis School Mathematics-10 149

Then, volume of tent (V) = 100 × 12 = 1200 m3

(i) We have,

Volume of the Pyramid = 1 Area of base × height
3
1
12000 = 3 × 400 × h

h= 1200 × 3 =3m
400

(ii) Again,

Area of base = 400 m2

a2 = 400

a = 20 m.

In square based Pyramid,

l2 = h2 + 92
4
202
l2 = 92 + 4

l2 = 81 + 100

l2 = 181

l = 181

= 13.45 m.

(iii) Again, LSA of Pyramid = 2al

= 2 × 20 × 13.45

= 538 m2

(iv) Cost per m2 of cloths = Rs. 120

Total cost = 538 × 120

= Rs. 64,560.

Example 6:

A semi cylindrical tunnel is made on a rectangular field of dimension 50 ft. × 28 ft. The
tunnel is covered by plastic. Find

(i) the cost of plastic to cover the tunnel at the rate of Rs. 20 per per ft2
(ii) the cost to harvest the vegetable on the field at the rate of Rs. 50 per square feet.

Solution:

In the given figure,

ABCD is a rectangular field of dimension 50 ft. × 25 ft., A 28 ft. D
50 ft.
A semi cylindrical shaped tunnel is made on it.

The area to be covered by plastic B C

= Its CSA + Area of its 2 semi circular face.

= πrh + 2 × 1 πr2
2

= πrh + πr2

150 Oasis School Mathematics-10

= 22 × 14 × 50 + 22 × (14)2
7 7

= 2200 + 616

= 2816 square feet.

(i) Since the cost to cover the tunnel by plastic at the rate of Rs. 20 per ft2

Total cost = Rs. 2816 × 20

= Rs. 56,320.

(ii) Again, area of the field = (50 × 28) square feet.

= 1400 square feet.

Hence, cost to harvest the vegetable at the rate of Rs. 50 per m2 is
Rs. 1400 × 50 = Rs. 70,000.

Example 7:

The adjoining figure is the figure of two pillars mounting a square pyramid on the
top. Find the cost of tiling both pillars at the rate of Rs. 20 per square feet.

Solution:

In the given figure, the lower part is a square based Prism and the upper part is squared
based pyramid.

Taking lower part,

Perimeter of the base = 4a 1.5ft. 4 ft.

= 4 × 6 ft

= 24 ft.

We have, LSA of the prism (A1) 10 ft. 10 ft.
= Perimeter of the base × height

= 24 × 6 6 ft. 6 ft.
= 144 ft2

For Pyramid, a = 6ft, h = 4 ft.

We have, = l2 + a2
4

= 42 + 62
4

= 16 + 9 = 25.

∴ l = 5 ft.

We have, LSA of pyramid (A2) = 2al

=2×6×5

= 60 ft2

TSA of a pillar = A1 + A2

= 144 ft2 + 60 ft2 = 204 ft2

Oasis School Mathematics-10 151

Cost of tiling a pillar at the rate of Rs. 20 per square feet is Rs. 204 × 20

= Rs. 4080

Cost of tiling two pillars = 2 × Rs. 4080 = Rs. 8160.

Exercise 7.4

1. (a) Convert 1 m3 into liter.

(b) Convert 1 cm3 into liter.

(c) Convert 1 m3 into cm3.

2. Find the area of the given geometrical shapes.

(a) B (b) P (c) (d) A

A 4 cm 12cm E HB
6cm
8cm C16 ft.
8cm 6cm

10cm 6 ft.
3cm 7cm
6cm
13cm
7cm

7cm
E 5cm C Q T I 13cm 9cm E
12cm 13cm 5cm
F 6cm J D
10cm
6cm
8cm R 9cm S
D
G

3. The structure of a yard in front of Zenith's house is as shown in the A 22 ft. F
figure. The measurement of each side is as shown in the figure.

(i) If the measurement of AE, AD and BD are 20 ft, 6 ft and 14 ft D 16 ft. E

respectively, find the area of the yard. 18 ft.

(ii) How many tiles of 1 ft2 area are required for the yard? B

(iii) Find the total cost, if the cost of paving per tile is Rs. 150. 6 ft. C

4. (a) The inner length and height of a tank having a square base is
4.5m and 3m. Calculate the volume of the tank in liter.

(b) Calculate the area of the 4 walls of a cubical tank having each side 3.5 m.
(c) The dimensions of a water tank are 5m × 4.5m × 3m.
Calculate - (i) volume of the tank in liters, (ii) area of 4 walls.
5. (a) The diameter of a cemented ring of a well is 1.6m. The height of each ring is

30cm. Find
(i) volume of the well in litre if it has 30 such rings.

(ii) total cost of constructing the well if the cost per ring is Rs. 1,200.

(iii) volume of water in the well if the level of water is on the 20th ring from the top.

(b) Find the volume of water in the well having 50 rings if the diameter and height of
each ring are 2m and 0.3m respectively. Also find the volume of water in it if the
level of water is on the 2nd ring from the top.

152 Oasis School Mathematics-10

6. (a) A cylindrical water tank has height 3m and diameter 7m. If it has a hemispherical
top, find how much water can be held by the tank.

(b) A community has a cylindrical underground water tank having diameter 14m and
depth 5m. It has a hemispherical top above the ground.

(i) Find the cost of plastering its inner wall, base and inner surface of the hemisphere
at the rate of Rs. 105 per square meter.

(ii) Find the capacity of the tank.

7. (a) At a wedding party, a man managed a conical tent which contains 23100 m3 air. If
the height of the tent is 50 cm, find the cost of the cloth required to prepare the tent.
Given that the cost per square meter of cloth is Rs. 250.

(b) In a wedding ceremony, a man has to make arrangements for the accommodation of
150 people. For this purpose, he plans to build a conical tent in such a way that each
person has 4 square meters of space on the ground and 20 cubic meter of air. Find
the height of the tent.

(c) A conical tent is made for the accommodation of 96 people. For each person there is
6π sq. meter ground and 44 cubic meter space. Find,

(i) height of the tent
(ii) cost to prepare the tent at the rate of Rs. 42 per square meter.

8. (a) A cylindrical shaped tent is made for the accommodation of 88 people in such a way
that each person has 3 m2 space on the ground and 15m3 of air to breathe, find,

(i) area of cloths required to make the tent

(ii) cost to prepare the tent at the rate of Rs. 50 per square meter.

(b) A cylindrical shaped ten of height 10 m is made in such a way that it contains 360 m3

of air, find

(i) area of the base (ii) radius of the base

(iii) total surface area of the tent

(iv) cost to prepare to tent at the rate of Rs. 50 per m2

9. (a) A square based Pyramid shaped tent is made in such a way that it contains 192 m3
of air. If the height of the tent is 9m, find

(i) the area of the base
(ii) a side of the base.
(iii) slant eight of the tent
(iv) area of cloths required to make the tent
(v) cost to prepare the tent at the rate of Rs. 70 per m2.

(b) A square based Pyramid shaped tent is made for the accommodation of 90 people in
such a that each people gets 10 ft2 area on the ground and 30 ft3 of air to breathe find

(i) height of the tent

(ii) a side of the base (iii) slant height of the tent

(iv) surface area of the tent

(v) cost to prepare the tent at the rate of Rs. 40 per ft2

Oasis School Mathematics-10 153

10. (a) A school compound gate has two pillars having a

square base. Each pillar has a square based pyramid on 1.5ft. 1.5ft.

the top as shown in the figure. Find

(i) total surface area of each pillar.

(ii) cost of paving tiles on its wall at the rate of Rs. 70 7 ft
per square foot. 7 ft

1 ft 1 ft 1 ft 1 ft

(b) A shopping complex gate has two pillars. Each pillar is

square based with dimension 2 ft. × 2ft. On the top of each pillar there is a square

based Pyramid of height 2 ft. If the total height of each pillar is 10 ft. Find the cost of

tiling both pillars at the rate of Rs. 120 per square feet.

(c) An apartment compound gate has two pillars

cylindrical in shape. The diameter of the base of

each pillar is 2.8 ft. Each pillar has a hemispherical 13.4 ft 13.4 ft
top. The height of both pillars is 13.4 ft. Find,

(i) total surface area of each pillar. 2.8 ft 2.8 ft

(ii) cost of paving tiles at the rate of Rs. 85 per
square foot.

11. (a) A half cylinder tunnel is made on the rectangular field of 14 ft.
dimension 20ft × 14 ft. as shown in the figure. Find

(i) thecosttocoverthetunnelbyplasticattherateofRs.20 20 ft.
per square feet.

(ii) the cost to harvest the vegetable on the field at the rate of Rs. 15 per square feet.

(b) A half cylinder shaped tunnel is made on the rectangular field of dimension
50 ft. × 28 ft. along its length. An aluminium gate of dimension 3 ft × 2 ft is made on
one of its semi-circular face. Find,

(i) the cost to prepare the aluminium door at the rate of Rs. 200 per square feet.

(ii) cost to covert the tunnel by plastics except door at the rate of Rs. 80 per

(iii) cost to harvest vegetable at the rate of Rs. 30 per square feet.

Answers
1. (a) Consult your teacher 2. (a) 35.01 cm2 (b) 125.3 cm2 (c) 110.2 cm2 (d) 87.11 cm2
3. (i) 133.61 ft2 (ii) 134 (appox.) (iii) Rs. 20, 100
4. (a) 60,750 liters (b) 49m2 (c) 67,500 liters, 57m2
5. (a) (i) 18,102.85 litre (ii) Rs. 36000 (iii) 6,034.29 liters (b) (i) 47,142.85 liters
(ii) 45,257.14 liters 6. (a) 2,05.333m3 or, 2,05,333 liters (b) (i) Rs. 71610 (ii) 1488666.67 liters
7. (a) Rs. 8,94,811.29 (b) 15m (c) 7m, Rs. 79,200 8. (a) (i) 5m (ii) 552 m2 (iii) Rs. 27,600
(b) (i) 36π m2 (ii) 6m (iii) 156π m2 (iv) Rs. 24514.28
9. (a) (i) 64 m2 (ii) 8m (iii) 9.85 m (iv) 157.58 m2 (v) Rs. 11030.72
(b) (i) 9 ft. (ii) 30 ft (iii) 17.49 ft (iv) 1049.40 ft2 (v) Rs. 41976
10. (a) 31.16 ft2 (ii) Rs. 443 (approx.) (b) Rs. 17506.56 (c) (i) 117.92 ft2 (ii) Rs. 20,046.40
11. (a) Rs. 11880 (ii) Rs. 4200 (b) (i) Rs. 1200 (ii) Rs. 2,24,800 (iii) Rs. 42,000

154 Oasis School Mathematics-10

Miscellaneous Exercise

Area of plane figure D (c) B C (d) 21 cm
1. Find the area of the following figures. A 13 cm
D
(a) P (b) A 16 cm 5 cm
F 10 cm E (Ans: 156cm2)
10 cm 25 cm10 cm(Ans: 259.8cm2)
11 cm 12 cm 9 cmC

QS R B 13 cm

12 cm

(Ans: 48cm2) (Ans: 121.4 cm2)

2. Find the area of a right-angled isosceles triangle, if the perimeter and longest side of the

triangle are 24 cm and 10 cm respectively. (Ans: 24.5 cm2)

3. (a) The length of the sides of a triangular field are 20 m, 13 m and 21 m. Find

(i) the area of the field.

(ii) the cost of fencing at Rs. 250 per meter.

(iii) the cost of levelling at Rs. 125 per square meter.

[Ans:(i) 126 cm2, (ii) Rs. 13,500, (iii) Rs. 15,750]

(b) The ratio of the sides of triangular field is 12 : 17 : 25 and the perimeter is 540 ft.
find,

(i) the sides of the field

(ii) the area of the field

(iii) cost of paving stone at the rate of Rs. 250 per square foot.

[Ans:(i) 120 ft, 170 ft, 250 ft (ii) 9000 sq.ft. (iii) Rs. 2250000]

4. Find the area of the following polygons: E
D
(a) A 10 cm B (b) P 60 cm Q (c) 4 cm

8 cm 26 cm A

D CS R 4 cm 4 cm

15 cm 75 cm B 4 cm C
(Ans: 30.64cm2)
(Ans: 99.5cm2) (Ans: 1641.45cm2)

5. (a) An umbrella is formed by stitching 12 pieces of triangular pieces of cloth having

three sides 12 cm, 12 cm and 8 cm. Find the area of the cloths required to form the

umbrella. [Ans: 543.06 cm2]

(b) A tent is formed by stitching 10 pieces of cloth having 3 sides 8 ft, 6 ft and 10 ft.

Find the surface area of the tent so formed. [Ans: 240 sq.ft.]

Oasis School Mathematics-10 155

Cylinder, Sphere and Hemisphere 20 cm

1. Find the volume of the 14 cm
given solid object. (Ans: 1,540 cm3)

2. A cylinder of height 60 cm and diameter 14 cm is split vertically into two equal halves.

Find the volume of each half. (Ans: 4,620 cm3)

212c1mcm
3. Find the total surface area
of the given solid. (Ans: 3,046.5cm2) 50 cm

4. A cylinder of height 40 cm and radius 3.5 cm is cut vertically into two equal halves. Find

the total surface area of each part. (Ans: 758.5 cm3)

5. The volume of a cylinder is half its total surface area. If the radius of its base is 7 cm, find

the volume. (Ans: 179.67cm3)

6. A solid sphere of radius 6 cm, is melted down and drawn out into a long cylindrical wire

of uniform thickness of 0.2 cm. Find the length of the wire. (Ans: 288m)

7. A sphere of radius 21 cm is made of brass and is melted to form a cylinder of radius 7 cm

and height 14 cm. How many cylinders can be formed? (Ans: 18)

8. The surface area of a sphere is 18 times the area of the curved surface of a cylinder. If the
radius and height of the cylinder are 21 cm each, find the volume of sphere. ( 10,47,816 cm3)

9. Surface area of a sphere of radius 14 cm is equal to the surface area of a cylinder of radius

7 cm. Find the height and volume of the cylinder. (Ans: 149cm, 7,546cm3)

10. A cylindrical bucket 56 cm in diameter and 24 cm. high is full of water. If the water is

poured into a rectangular tank 77 cm long and 35 cm wide, find the height of water level

in the tank. (Ans: 21.94cm)

11. A cylindrical jar of radius 7 cm contains water. Some iron spheres each of radius 1.4 cm

are immersed in the water. How many spheres are necessary to raise the level of water by

3.5 cm.? (Ans: 47)

Cone

1. The volume of a right circular cone is 392π cm3. If its height is 24 cm, find its

(i) radius (ii) slant height (iii) total surface area (Ans: 7 cm, 25cm, 704cm2)

2. Curved surface area of the cone is 154 10cm². If its slant height is7 10cm, find its

(i) radius. (ii) height (iii) volume (Ans: 7cm, 21cm, 1078cm3)

3. The radius of the base and the height of the cone are in the ratio 7:12, and the volume is

616cm3. Find its total surface area. (Ans: 459.58cm2)

156 Oasis School Mathematics-10

Combined Solids

1. A circus tent is cylindrical in shape up to a height of 22 m and conical above it. If the
diameter of the tent is 120 m and height of the conical portion is 11 m, find

(i) the surface area of the tent. (ii) the amount of air in it. 20 cm
2. Find the volume of the given solid.
10 cm

18 cm 37 cm

14 cm (Ans: 2147.76cm3)

3. Find the TSA and CSA of the given solid. 38 cm

(Ans: 1304.82cm2, 1304.82cm2)

4. The given figure is a solid composed of a cylinder with hemisphere at one 14 cm

end. If the total surface area and the height of the solid are 770 sq. cm and 14

cm respectively, find the height of the cylinder. (Ans: 7cm)

20 cm

5. The curved surface area of the combined solid as shown in the figure

is 19,494cm2. Calculate the radius. [Ans: 21cm]

6. A cone is filled with water. If this water is poured into a cylindrical tank, upto what height

does the surface of water rise in the cylinder? [Ans: 2.22cm]

Triangular Prism 14 cm

1. The height of a prism having its base as a right-angled 20cm
triangle is 25 cm. If the lengths of the sides of the right- 15cm 14 cm
angled triangle containing the right angle are 6 cm and 8
cm, calculate the volume of the prism. (Ans: 600 cm3)

2. If a prism is 15 cm high with a triangular base having sides 6 cm, 8 cm and 10 cm, find

its volume. (Ans: 360 cm2)

3. The volume of a prism having a base right-angle is 864 cm3. If the length of the sides

of the right angled triangle containing the right angle are 8 cm and 9 cm, calculate the
A (Ans: 24 cm)
height of the prism.

B C
A' 30 cm
4. The lateral surface area of the given triangular prism is

900cm².IfA’B’=8cm,B’C’=12cmandCC’=30cm,findthe

length of A’C’. (Ans: 10 cm) 8cm

B' 12 cm C'

5. Total surface area of the given prism is

198 3 .cam² ². Find the side of the base. (Ans: 6cm) 10 3 .cam²
4 4

Oasis School Mathematics-10 157

Pyramid

1. Find the volume of a pyramid whose base is an equilateral triangle having perimeter 36

cm and height of the pyramid is 15cm. [Ans: 180 3cm3]

O

2. In the given figure, DC = OP and if the lateral surface area is A B
Q
64 5cm3, find its volume. [Ans: 170.67cm3] P C

OD

3. Find the TSA and volume of the given pyramid, 10cm

if OB = 10cm and DE = 6 2cm. [Ans: 132cm2, 63.44cm3] A B

4. Find the volume of the following figures. 6 2cm E C
D

(a) (b) (c) (d)

12 cm 6 cm 15 cm 6 cm 5 cm 18 cm 8 cm
8 cm
12 cm

6 cm 6 cm 3 cm 4 cm 8 cm
6 cm (Ans: 300cm3­ ) (Ans: 256cm3­ )
(Ans: 288cm3­ ) (Ans: 68cm3­ )

5. Find the TSA of the given solid.
(a) (b)

15 cm 12 cm 14 cm 10 cm

8 cm 6 cm
8 cm 6 cm
(Ans: 528cm2­ ) (Ans: 336cm2­ ) 2.5 ft. 2.5 ft.

Geometrical Bodies

1. A school compound has two pillars in its gate. Each pillar has 6 ft. 6 ft.
conical top as shown in the figure. Find

(i) the cost of plastering its wall at the rate of Rs. 50 per square foot. 1.4ft 1.4ft

(ii) the number of tiles of size 1 ft × 1 ft. needed to pave its surface. B'
C'
(iii) the cost of paving tiles at the rate of Rs. 90 per tile. NB
C
(i) (Ans: Rs. 3,190) (ii) (Ans: 64) (iii) (Ans: Rs. 5,760) A'

2. In the given figure, ABCD is the base of the tank. A'B'C'D' is

its top. Here the diagonal of the base BD = 15 ft, AM = 5ft, CN D'

= 6ft. Height of the tank is 9ft. Its 4 walls are rectangular in A
shape. If A'B' = 12ft, B'C' = 9 ft, C'D' = 13 ft. and A'D' = 8ft, find

DM

158 Oasis School Mathematics-10

(i) area of the base of the tank. (iv) (Ans: Rs. 4,125)

(ii) area of its 4 walls.

(iii) cost of plastering its walls at Rs. 45 per square foot.
(iv) cost of plastering its base at the rate of Rs. 50 square foot.
(i) (Ans: 82.5 ft2) (ii) (Ans: 378ft.2) (iii) (Ans: Rs. 17010)

3. An umbrella is made by stitching 8 triangular pieces of cloth, which are equal in area. If

three sides of a triangular piece are 20 cm, 20 cm and 10cm (i) Find the area of the cloths

required to make an umbrella. (ii) Find the total cost required to make the umbrella if the

cost per square cm of cloth is Rs. 2. [Ans: (i) 776cm2 (ii) Rs. 1552]

4. A tent is made using 6 pieces of triangular shaped cloth. Three sides of each piece of cloth
are 5 ft, 7 ft and 8 ft. (i) Find the area of cloth required to make the tent. (ii) Find the total
cost of the cloth required to make the tent if the cost per square foot is Rs. 60.

[Ans: (i) 103.92 square feet (ii) Rs. 6235.20]





Attempt all the questions: G roup – A [5 × 1 = 5] Full Marks: 30

1. (a) If 'a', 'b', 'c' be the three sides of a triangle and 2S be its perimeter, write the
formula to calculate the area of the triangle.

(b) Find the volume of the cylinder whose area of the base is 154 cm2 and the height
is 20cm.

(c) Find the surface area of a sphere if its radius is 1cm.

(d) Write the formula to calculate the lateral surface area of a prism.

(e) In a cone, what are l, r and h? Write the relation among them.

Group – B [6 × 2 = 12]

A

2. (a) Find the area of the given triangle.
5 cm
6 cm

BC
7 cm

A 50 cm A'

(b) Find the total surface area of the given triangular 5 cm

prism. 12 cm B B'
C C'

3. (a) The circumference of the base of a cylinder is 33
cm. If the area of the curved surface is 528 cm², find its volume.

(b) If the total surface area of a sphere is 48πcm², find its volume.

Oasis School Mathematics-10 159

14 cm

4. (a) Find the TSA of the given cone:

25 cm

14 cm
(b) Find the volume of the given hemisphere.

Group – C [2 × 4 = 8]

5. If the curved surface area of the combined solid is
858 cm², find its common radius.

25 cm

25 cm

6. Find the volume of the given square-based pyramid.

14 cm

Group – D [1 × 5 = 5]

7. A man has to construct a water tank of dimensions 4.5m × 3m × 2m.
(i) find the capacity of the tank.
(ii) find the area of the four walls of the tank.
(iii) find the cost of plastering its walls at the rate of Rs. 90 per m2.
(iv) find the area of the base.
(v) find the cost of plastering its base and ceiling at the rate of Rs. 70 per square

meter.

Project Work

I. Take the measurement of the dimensions of the yard of your home. Find its
area. Take the size and rate of tiles from a tile house. Calculate how many
tiles you need to cover it. Calculate the total cost of paving the yard.

II. Visit a contractor. Ask him or her about the rate of plastering, coloring the
wall, ceiling and floor. Measure the dimensions of the rooms of your home
and estimate the cost of -

• plastering the wall, ceiling and floor.
• coloring the four walls.

160 Oasis School Mathematics-10

Algebra

36Estimated Teaching Hours

Contents

• H.C.F. and L.C.M.
• Simplification of rational expressions
• Indices
• Exponential equation
• Radical equation
• Verbal problems on simultaneous equation
• Verbal problems on quadratic equation

Expected Learning Outcomes

At the end of this unit, students will be able to develop the following
competencies:
• To find the H.C.F. of at most three algebraic expressions by

factorisation method
• To find the L.C.M. of at most three algebraic expressions by

factorisation method
• To solve radical equations
• To simplify an expression containing a surd
• To simplify simple rational expressions
• To solve word problems of linear simultaneous equation and

quadratic equation

Materials Required

• Graph sheet, Flash cards, A4 size paper.

Oasis School Mathematics-10 161

Unit

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

8.1 Warm-up Activities

Discuss the following in your class and draw a conclusion.
• What is the formula of a2–b2?
• What is the formula of a3–b3?
• What is the formula of a3+b3?
• What are the factors of (a+b)3?
• How to factorise the following expressions?
• 4a2 – b2
• 2a3 – 16
• 3a3 + 81

• x2 – 5x + 6
• x4 + x2 + 1

8.2 Highest Common Factor (H.C.F.)

Let’s take any two algebraic expressions.

x2 – 4, x2 – 5x + 6

Let’s find the factors of x2 – 4

x2 – 4 = (x)2 – (2)2

= (x + 2) (x – 2)

Again, x2 – 5x + 6

= x2–(3 + 2)x + 6

= x2 – 3x – 2x + 6

= x(x – 3) –2 (x – 3)

= (x – 3) (x – 2)

Is there any common factor between the two expressions?

Yes their common factor is (x–2).

∴ (x–2) is the H.C.F. of these two expressions.

Let’s represent this relation in a Venn diagram.

162 Oasis School Mathematics-10

x+2 x–2 x–3 If the factors are the
elements of a given set
of algebraic expressions
then the H.C.F. is their

intersection.

x2 – 4 x2–5x+6

Hence, the common factor or the product of all common factors of given
algebraic expressions is called the H.C.F. It is known as the highest common

factor. It is also called the greatest common divisor.

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

Let’s take any two algebraic expressions .

2x3–16, x3–3x2+2x

Here, 2x3 – 16

= 2(x3 – 8)

= (x3 – 23)

= 2(x–2) (x2+2x+4)

Again, x3 – 3x2 + 2x

= x (x2–3x+2)

= x[x2–(2+1)x+2]

= x[x2–2x–x+2]

= x[x(x-2)-1(x-2)]

= x(x–2) (x-1)

Is there any common factor between these two expressions?
Let’s represent them in a Venn diagram.

2, Union of two sets
represents L.C.M.
(x2+2x+4) x–2 x, (x-1)

2x3–16 x3–3x2+2x

What is the union of these sets?
{(x–2), 2, (x2+2x+4), x, (x–1)}
What is their product?

Oasis School Mathematics-10 163

(x–2) × 2 (x2+2x+4) × x (x-1)


Common Remaining factors
factor

Hence, the product of all the common factors and the remaining factors of the
given expressions is their L.C.M.

Note:

In three algebraic expressions
L.C.M. = common factor from all × common factors from two expressions × remaining factors.



Worked Out Examples



Example: 1

Find the H.C.F. of: 2x3 – 16, x2 – 4x + 4, x2 – 3x + 2
Solution: Here,

1st expression = 2x3 – 16
= 2(x3 – 8)
= 2(x3 – 23)
= 2(x – 2)(x2 + 2x + 4)
2nd expression = x2 – 4x + 4
= x2 – (2 + 2) x + 4
= x2 – 2x – 2x + 4
= x(x – 2) – 2(x – 2)
= (x – 2)(x – 2)
3rd expression = x2 – 3x + 2
= x2 – (2 + 1) x + 2
= x2 – 2x – x + 2
= x(x – 2) – 1(x – 2)
= (x – 2)(x – 1)
∴ H. C. F. = (x – 2).

Example: 2

Find the H. C. F. of: x2 – y2 + 2yz – z2, (x + y)2 – z2
Solution: Here,

164 Oasis School Mathematics-10

1st expression = x2 – y2 + 2yz – z2
= x2 – (y2 – 2yz + z2)
= x2 – (y – z)2
= (x + y – z)(x – y + z)
2nd expression = (x + y)2 – z2
= (x + y – z)(x + y + z)
∴ H. C. F. = (x + y – z)

Example: 3

Find the H. C. F. of: 8x3 + y3, 16x4 + 4x2­y2 + y4, 4x2y – 2xy2 + y3

Solution: Here,

1st expression = 8x3 + y3 Alternative way of factorisation
= (2x)³ + (y)3 Second expression
= (2x + y)(4x2 – 2xy + y2) = 16x4 + 4x2y2 + y4
2nd expression = 16x4 + 4x2y2 + y4
= (4x2)2 + 2.4x2.y2 + (y2)2 – 4x2y2 = (4x2)2+(y2)2 + 4x2y2
= (4x2 + y2)2 – (2xy)2 = (4x2+y2)2–2.4x2.y2+4x2y2

= (4x2+y2)–8x2y2+4x2y2
= (4x2+y2)2–(2xy)2

= (4x2+y2+2xy) (4x2+y2–2xy)

= (4x2 + y2 + 2xy)(4x2 + y2 – 2xy)

= (4x2 + 2xy + y2) (4x2 – 2xy + y2)

3rd expression = 4x2y – 2xy2 + y3

= y(4x2 – 2xy + y2)

∴ H. C. F. = (4x2 – 2xy + y2)

Example: 4

Find the H.C.F. of: x4–10x2+169, x3+x(x+13)–7x2, 3x2–18x+39.

Solution:
1st expression = x4–10x2+169

= (x2)2+(13)2–10x2

= (x2+13)2–2.x2.13–10x2

= (x2+13)2–26x2–10x2

= (x2+13)2 – 36x2

= (x2+13)2 – (6x)2

= (x2+13+6x) (x2+13–6x)

= (x2+6x+13) (x2–6x+13)

2nd expression = x3+x(x+13)–7x2

= x3+x2+13x–7x2

Oasis School Mathematics-10 165

= x3–6x2+13x
= x(x2–6x+13)
3rd expression = 3x2–18x+39
= 3(x2–6x+13)
∴ H.C.F. = (x2–6x+13)
Example: 5
Find the L. C. M. of: 6x2 – x – 1, 54x4 + 2x
Solution:
1st expression = 6x2 – x – 1

= 6x2– (3–2) x – 1

= 6x2 – 3x + 2x – 1

= 3x(2x – 1) + 1(2x – 1)

= (2x – 1) (3x + 1)
2nd expression = 54x4 + 2x
= 2x(27x3 + 1)
= 2x{(3x)3 + (1)3}
= 2x(3x + 1)(9x2 – 3x + 1)
∴ L. C. M. = (3x + 1) (2x – 1). 2x (9x² – 3x + 1)
= 2x(3x + 1)(2x – 1)(9x2 – 3x + 1)

Example: 6

Find the L. C. M. of : 1 + 4x + 4x2 – 16x4, 1 + 2x – 8x3 – 16x4, 1 – 4x2

Solution:

1st expression = 1 + 4x + 4x2 – 16x4

= (1 + 2x)2 – (4x2)2

= (1 + 2x + 4x2) (1 + 2x – 4x2) Common from all = none
common from two expressions =
2nd expression = 1 + 2x – 8x3 – 16x4
= 1(1 + 2x) – 8x3(1 + 2x) (1 + 2x + 4x²) ( 1 + 2x), (1 – 2x)
= (1 + 2x)(1 – 8x3) Remaining factor = (1 – 2x + 4x²)

= (1 + 2x)(1 – 2x)(1 + 2x + 4x2)

3rd expression = 1 – 4x2

= (1 + 2x)(1 – 2x)

∴ L.C.M. = (1 + 2x)(1 – 2x) (1 + 2x + 4x2)(1 + 2x – 4x2)

166 Oasis School Mathematics-10

Example: 7

Find the H.C.F. and L.C.M. of: (x – 1)3, x4 + x2 + 1, x3 – 1
Solution: Here,

1st expression = (x – 1)3

= (x – 1)(x – 1)(x – 1)

2nd expression = x4 + x2 + 1

= (x2)2 + 2.x2.1 + 12 – x2

= (x2 + 1)2 – (x)2 Common from all = 1
= (x2 + 1 + x)(x2 + 1 – x) Common from two expression
= (x2 + x + 1)(x2 – x + 1) = (x–1) (x2+x+1)
3rd expression = x3 – 1 Remaining terms
= (x – 1)(x2 + x + 1) = (x–1), (x–1) and (x2–x+1)
∴ H. C. F. = 1 ∴ LCM =
and L. C. M. = (x-1) (x2+x+1) (x-1)2 (x2-x+1) (x–1) (x2+x+1) (x–1)2 (x2–x+1)

= (x – 1)3 (x2 – x + 1) ( x² + x + 1)

= (x–1)3 (x4+x2+1)

Example: 8

Find the H.C.F. and L.C.M. of: (a2 – 1) (b2 – 1) – 4ab, a2b2 + 2b (a2 + 1) + a2 – b2 –1
Solution: Here,
1st expression = (a2 – 1) (b2 – 1) – 4ab
= a2b2 – a2 – b2 + 1 - 4ab
= (a2b2 – 2ab + 1) – (a2+b2+2ab)
= (ab – 1)2 – (a+b)2
= [(ab – 1) + (a+b)] [(ab–1) – (a+b)]
= (ab–1+a+b) (ab–a–b–1)
2nd expression = a2b2 + 2b (a2 + 1) + a2 – b2 – 1
= a2b2 + 2ba2 + 2b + a2 – b2 – 1
= a2b2 + 2ba2 + a2 – b2 + 2b – 1
= [(ab)2 + 2.ab.a+a2] – (b2–2b+1)]
= (ab+a2)–(b–1)2
= [(ab+a) + (b–1)] [(ab+a) – (b–1)
= (ab + a + b–1) (ab + a – b + 1)
∴ H.C.F. = (ab + a + b – 1)
L. C. M = (ab + a + b – 1) (ab – a – b – 1) (ab + a – b + 1)

Oasis School Mathematics-10 167

Exercise 8.1

1. (a) What is the H.C.F. of two expressions if there is no common factor?
(b) What is the L.C.M. of two expressions if there is no common factor?
(c) What is the relation of two expressions with their H.C.F. and L.C.M.?
(d) What is the H.C.F. of two expressions?
(e) What is the L.C.M. of two expressions?

(f) What is the L.C.M. of three expressions?

2. Find the H.C.F. and L.C.M. of:

(a) (a–b) (a–b) and (a + b) (a – b) (b) a(x–y) and ab (x+y)

(c) (x–2y) (x2 + 2xy + 4y2) and (x+2y) (x–2y) (d) 2x2 – 5x and 4x2 – 10

3. Find the H.C.F. of: (b) x2–4y2 and x3 – 8y3
(a) (x–y)2 and (x–y)3 (d) x2 + 2x + 4 and x3 – 8
(c) a2–16 and 4a + 16

(e) x2 – xy and x3y – xy3

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

(a) (a–b)2 and (a–b)3 (b) (x–2y)2 and 2x3y – 8xy3
(c) x3 + x2 + x and x4 – x (d) (a+b)2–4ab and 2a2b – 4ab2 + 2b3

(e) x2 – 3x + 3y – y2 and y2 + xy – 3y

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

(a) 9(4x2 – 25) and 15(2x2 + x – 15) (b) x² – 4, x2 + 10x + 16 and x2 – 7x – 18

(c) 16(a2 + 2a – 3), 24(a2 + 2a – 3) and 48(a2 – a – 12)

(d) ab2 – b(a – c) – c, b2 – 3b + 2, b3 – 1 (e) 2(x2 – y2), 4(x3 – y3) and 6(x4 – y4)

(f) 2x3 + 16, x2 + 4x + 4, x2 + 3x + 2 (g) m4 + m2n2 + n4 and m2n3 + m5

(h) 8y3 – z3 and 16y4 + 4y2z2 + z4

(i) (x – y)2 + 4xy, 2x3 + 4x2y + 2xy2 and x3 + 2x2y + xy2
(j) ab(x2 + y2) – xy(a2 + b2) and b2x2 – a2y2

(k) x2 + 2xy + y2 – z2, y2 + 2yz + z2 – x2 and z2 + 2zx + x2 – y2

(l) x² + y² + 2xy – 1, y² – x² + 2y + 1 and x² – y² + 2x + 1

(m) 9x2– 4y2 – 8yz– 4z2, 4z2– 4y2–9x2–12xy and 9x2 + 12xz + 4z2–4y2

(n) x2 + 4x – 21 and x3 + 7x2 – 9x – 63 (o) x² – 5x + 6, x3 – 2x² – x + 2, x3 – 8

(p) x3 + 3x2 – 6x – 8 and x2 + 5x + 4 (q) x3 – 1, x4 + x2 + 1 and x3 + 2x2 + 2x + 1
(r) a3–a2+a–1 and 2a3 – a2 + a – 2 2
(t) a3 – 1, a4 – a² – 2a – 1, a² – 3a + 2 (s) a3 – 1 – 2a + a and a3 + 1
a3 a3

(u) 16x4 – 4x2 – 4x – 1and 8x3–1

(v) a4+8a2+144, a3+(a+12)a–5a2, 2a2–8a+24 (w) a4+16a2+256, a3+64, a3+(16+a)a–5a2

(x) x2–y2+2yz–z2, x3–3xy(x-y)–y3+z3 (y) (a+b+c)2, a3+3ab(a+b)+b3+c3, a2–b2–c2–2bc

168 Oasis School Mathematics-10

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

(a) (x – 3)(x2 + x – 2) and (x2 – 5x + 6) (b) (a + 1)3(a – 1) and (a + 1)2(a2 – 1)

(c) a2 + b2 + 2ab – c2, a2 – b2 + ac – bc

(d) a3 + b3, a4 + a2b2 + b4 and a5 – a4b + a3b2 (e) 8x3 + y3, 8x3 – y3 and 16x4 + 4x2y2 + y4

(f) 2x2 – 3x – 9, 4x2 – 5x – 21 and x3 – 9x (g) 8x3 + 27y3, 8x3 – 27y3 and 4x2 – 9y2

(h) 1 + 4x + 4x2 – 16x4 and 1 + 2x – 8x3 – 16x4 (i) x4+x3–x–1, x3–1, 1+2x+x2–x4

(j) x3 – 3x2 – x + 3, x3 – x2 – 9x + 9 (k) a3 – b3, a6 – b6, a4 + a2b2 + b4

(l) (a – 1)³, a4 + a² + 1, a³ – 1

7. Find the H. C. F. and L. C. M. of :
(a) 2x2 – 8, 3x2 – 9x + 6, 6x2 + 18x + 12
(b) 10(x3 + 2x2 – 3x), 15(x3 – 9x) and 20(x4 + 27x)
(c) x4 + x² + 1 and x4 – x3 + x – 1
(d) a³ + b³, a4 + a²b² + b4, a³ – a²b + ab²
(e) (a² – b²) (c² – d²) + 4abcd, a²c + abd + a²d – abc.
(f) a² – 12a – 28 + 16 b – b², a² + 2a – b² + 2b
(g) (x+y+1)2, x2–y2–4x–6y–5, x2+xy+x
(h) a2– 6a – 40 + 14b – b2, 5a2–10ab + 5b2 – 80
(i) x6 – 1, x5 + x2 + x, x3 + 2x2 + 2x + 1
(j) x4 + 4y4 + (x2 + y2) 5y2 and x4 + 2x3y + x2y2 – 9y4

Answer

1. Consult your teacher 2. Consult your teacher 3. (a) (x–y)2 (b) x – 2y (c) a + 4
(d) x2+2x+4 (e) x(x–y) 4. (a) (a-b)3 (b) 2xy(x-2y)2 (x+2y) (c) x(x3-1) (d) 2b (a-b)2 (e) y(x-y) (x+y-3)
5. (a) 3(2x-5) (b) (x+2) (c) 8(a+3) (d) (b–1) (e) 2(x-y) (f) (x+2) (g) m2-mn+n2
(h) 4y2+2yz+z2 (i) (x+y)2 (j) (bx-ay) (k) (x+y+z) (l) (x+y+1) (m) 3x+2y+2z
(n) (x+7) (x-3) (o) (x-2) (p) (x2+5x+4) (q) (x2+x+1) (r) a–1 (s) a2-1+1/a2) (t) 1 (u) 4x2+2x+1
(v) a2–4a+12 (w) a2–4a+16 (x) (x–y+z) (y) (a+b+c) 6. (a) (x-3) (x+2) (x-1) (x-2) (b) (a+1)3

(a-1) (c) (a+b+c) (a+b-c) (a-b) (d) a3 (a3+b3) (a2+ab+b2) (e) (8x3+y3) (8x3-y3) (f) x(x2-9) (2x+3) (4x+7)
(g) (8x3+27y3) (8x3-27y3) (h) (1-4x2) (1+2x+4x2) (1+2x-4x2) (i) (x2+x+1) (1+x–x2) (x2–1) (j) (x2-1) (x2-9)
(k) (a6–b6) (a2 + b2) (l) (a-1)3 (a2+a+1) (a2-a+1)
7. (a) H.C.F = 1, L.C.M = 6(x2-4) (x2-1) (b) H.C.F = 5x(x+3), L.C.M = 60x(x2-9) (x-1) (x2-3x+9)
(c) H.C.F = x2-x+1, L.C.M = (x3+1) (x3-1) (d) H.C.F = a2-ab+b2, L.C.M = a(a3+b3) (a2+ab+b2)
(e) H.C.F = (ac+bd+ad-bc), L.C.M = a(ac+bd+ad-bc) (ac+bd+bc-ad)
(f) H.C.F = (a-b+2), L.C.M = (a+b) (a-b+2) (a+b-14) (g) H.C.F = (x+y+1), L.C.M = x(x+y+1)2 (x-y-5)
(h) H.C.F = (a–b+4), L.C.M = 5(a–b+4) (a–b–4) (a+b–10)
(i) H.C.F = x2+x+1, L.C.M = x(x–1) (x+1) (x2+x+1) (x2–x+1)
(j) H.C.F. = (x2+xy+3y2), L.C.M.= (x2+xy+3y2) (x2+xy–3y2) (x2–xy+3y2)

Project Work

Take any two algebraic expressions. Factorise them. Find their H.C.F. and L.C.M. Find the
product of the H.C.F. and L.C.M. Find the product of two expressions. Draw a conclusion.

Oasis School Mathematics-10 169

Unit

9 Radicals and Surds

9.1 Warm-up Activities

Discuss the following in your class and draw a conclusion.

• What are natural numbers?

• What are whole numbers?
• Is –2 an integer?

• Is '0' a whole number?

• What are integers?
• Are all integers a whole number?

9.2 Surds p
q
A rational number is a number which can be expressed in the form of , where p and q

both are integers and q ≠ 0.

1, –2, 1 , 1 , 1 , 111, etc. are examples of rational numbers. When the rational number is
2 4 3

expressed in a decimal,

1 = 0.5 (terminating decimals)
2
1
4 = 0.25 (terminating decimals)

1 = 0.3333 ............ (non-terminating recurring decimals)
3
3
11 = 0.272727 ........ (non-terminating recurring block of digits)

∴ Therefore, a rational number is either terminating or non-terminating but a recurring

decimal.

There are other sets of nbue mexbperressswedhiicnh thheavfoermnopqn-wtehrmerienaptianngd and non-recurring
decimals, i.e., they cannot q both are integers

and q ≠ 0. Such numbers are irrational numbers.

2 = 1.4142135 ........ (non-terminating, non-recurring decimals)
Let us take some irrational numbers like 2, 3, 3 5 , etc.

Here, 2 is the square root of a positive rational number 2, but it is not a square of any
rational number.

Similarly, 3 5 is a cube root of a positive rational number, but it is not a cube of any
rational number. Such irrational numbers are surds or radicals.

170 Oasis School Mathematics-10

Therefore, n m is a surd if m is a positive rational number and n m is an irrational number.
Hence, an "irrational root of a rational number" is a surd.
In a surd n m, n is the order of the surd and m the radicand.
Now, discuss the following questions in your class.

• Is 5 an irrational number?

• Is 4 an irrational number?

• Is 1 an irrational number?
3

• Why is 3 a rational number?

• Is 2 a surd?

Addition and subtraction of surds

Two or more than two surds can be added or subtracted if they have the same irrational
factors. i.e. if they are like surds.

Example: A surd 3 2 can be added to a surd 2 2, because both the surds have the same
irrational factor 2.

These operations can be done by adding or subtracting the co–efficient and keeping the
irrational factor the same.

i.e. 3 2 + 2 2 = (3 + 2) 2 [Adding the co–efficient and keeping irrational factor the same]

= 5 2.

Multiplication and division of surds

Two or more than two surds can be multiplied to each other if they are of the same order.

n a . n b = n ab
Similarly, we can divide two surds if they are of the same order.

i.e. n a = n a
n b b

If the surds are of a different order, first we have to make them of the same order. Then

only can we multiply or divide the surds.

Let us see an example, 2 × 3 3

First make both of them of the same order

L.C.M. of the orders 2 and 3 is 6.

2 × 3 3 = 6 23 × 6 32 = 6 8 × 6 9 = 6 72

Similarly, 2 = 6 23 = 68 = 6 8
33 6 32 69 9

Remember !

• Order of the surd 2 is 2 and the order of the surd 5 2 is 5.
• Two surds can multiply or divide if their order is the same.
• If their order is different, to multiply or divide make their order the same.

Oasis School Mathematics-10 171

Worked Out Examples

Example: 1 Example: 2
Simplify: 3 2 × 3 3 × 4 5
Simplify: 5 27 – 6 12 – 2 48
Solution: Solution:

5 27 – 6 12 – 2 48 = 5 32 × 3 – 6 22 × 3 – 2 42 × 3 3 2 × 3 3 × 4 5 = 3×4 2×5 × 3 3
= 12 10 × 3 3
= 5×3 3–6×2 3–2×4 3

= 15 3 – 12 3 – 8 3 = 12 6 103 × 6 32

= (15 – 12 – 8) 3 = 12 6 1000 × 6 9

= –5 3 = 12 6 1000 × 9

Example: 3 = 12 6 9000

Simplify: (5 3 – 2 2) (3 2 – 2 3)
Solution:

(5 3 − 2 2) (3 2 − 2 3) = 5 3×3 2 −5 3×2 3 −2 2×3 2 +2 2×2 3
= 15 6 − 30 − 12 + 4 6
= 19 6 − 42

Example: 4

Simplify: 3 3 81 − 3 3 24 + 2 3 375 = 33 33 × 3 − 33 23 × 3 + 33 53 × 3
Solution: 13 3 192 13 3 43 × 3

3×3 3 3 −3×2 3 3 +2×5 3 3
= 13 × 4 3 3

3 3 81 − 3 3 24 + 2 3 375 = 33 33 × 3= − 393 23 33×−36+3 333+513 0×333
13 3 192 13 3 43 × 352 33

3 × 3 3 3=−133××1523243333 3+ 2 × 5 3 3
33
=

93 3− 6=3 3 +1310 3 3
13 33 3 3512
= 52
= = 4

52 3 3

= 13
52

= 1
4

172 Oasis School Mathematics-10

Exercise 9.1

1. (a) What is the order of the surd 5?

(b) Is 4 a surd? Justify your answer.

(c) On which condition is m a rational number?
n

(d) On which condition is m an irrational number?
n

(e) What do ‘n’ and ‘m’ represent in the surd n m?

2. Simplify: (d) 4 2 (e) nm
(a) 2 2 + 3 2 (b) 3 × 2 (c) 3 4 × 3 5 4 3 np
3. (a) Convert 2 and 3 3 in the same order.

4. ((Scbi))m((ca((FP)p)cail)r)niofd(5vy2e(5:t2ht2h3e2(3av(−ta−:1l18u62889e+2)+2)o=2f333322323))× 5 2 . ((bb)) ((22 33 −− 66)) ((22 33 ++ 66))

((dd)) ((33 22 −− 33))22 −−((33 22 ++ 33))22

(a()e()e)2 (34(35 x+x−3−222022)+)(5(45 8xx0++33 22)) (b)((ff)) 2 12xx522+−−329 45 + 3 5

(c) g( )g23)3333x953xx294x2−+−+−+25332555104244+−54 3 250 (d) 2 4 16428 −−3 +42147225(5i00) 0− +3xxxx6242−−−82−2244 44−−xx22
(e() 11 (f )((hh))3 2 +4 634 22−−xx
−−

(g() j()j 5) 5 8080−−66 2200++77 118800 (h)((kk)) 2233 16 + 5 3 54 − 6 33 22
55 2200++33 112255 3 250

5. Simplify:

(a) 5 2( 1(8a+) 352)2( 18 + 32) (b) (2 3 − 6(b))(2 (32+ 36−) 6 ) (2 3 + 6 )

(c) (2 3 − (c2))2 (2 3 − 2 )2 (d) (3 2 − 3(d)2) − (3(3 22+− 3 )32)2 − (3 2 + 3 )2

(e) (3 x − 2(e)2) ((53 xx+−32 22)) (5 x +(3f) 2) x2 − 9 (f) x2 − 9
x+3 x+3

(g) 3 x2 − 25 (g) 3 x2 − 25 (h) 48 − 2(7hh))(i) x42 8− −4 −274(−i)x2 x2 − 4 − 4 − x2
3 x+5 3 x+5 3 x − 23 2− x x − 2 2 − x

(j) 5 80 A−2n60s+(w2j30)e5r+172855018−2060 20 + 7 180(k) 23 16 +3523550(4k)− 6233 216 + 5 3 54 −63 2
5 +3 125 3 250

1. Consult your teacher 2. Consult your teacher 3. (a) 6 8 and 6 9 (c) 15 1944

2.4. (aa)) 2288 55 (b) (b7) 75 5(c) –33 2(c) −(d3)3 –2124 2 ((ed)) −5121412 (f) 11(e)25 11(g) 2 (h) 13 (i) 1
5

3.5. (a) 70 (b) 6 (c) 14 − 4 6 (d) − 12 6 (e) 15x − 2x − 12

(f) x−3 (g) 3 x − 5 (h) 01 (i) 0 ( j) 2 (k) 13
x+3 5

Oasis School Mathematics-10 173

9.3 Rationalization of Surds

A surd is an irrational number. When a surd is multiplied by another surd, it may
be changed into a rational number. The process of converting a surd into a rational
expression is called the rationalizations of surds.

For example: 3 × 3 = ( 3)2

= 3 a rational number.
∴ 3 is the rationalizing factor of 3.

Again, 8 × 2 = 8 × 2

= 16
= 4 a rational number.

∴ 2 is the rationalizing factor of 8.

Conjugate of surds
A surd in the form of a + b can be rationalized if it is multiplied by a – b .

∴ a + b and a – b are said to be conjugate to each other.

If an expression involving surds both in the numerator and denominator, is to be simplified,
the denominator should be rationalized.

Worked Out Examples

Example: 1 5 3+2 2
3− 2
Rationalize the denominator of

Solution: Here,

5 3+2 2
3− 2

Multiplying both numerator and denominator by 3 + 2

5 3+2 2 5 3 + 2 2 3+ 2
3− 2 3 − 2 3+ 2
= ×

= 5 3× 3+5 3× 2+2 2× 3+2 2× 2
( 3 )2 −( 2 )2

= 15 + 5 6 + 2 6 + 4
3−2

= 19 + 7 6
1

= 19 + 7 6

174 Oasis School Mathematics-10

Example: 2

Simplify: 3+ 2 + 3− 2
Solution: 3− 2 3+ 2

3+ 2 + 3− 2
3− 2 3+ 2
Alternative method

+ 3− 2 = 3+ 2 × 3+ 2 + 3− 2 × 3− 2 3+ 2 + 3– 2
3+ 2 3− 2 3+ 2 3+ 2 3− 2 3– 2 3+ 2

= ( 3 + 2 )2 + ( 3 − 2 )2 = ( 3 + 2)2 + ( 3 – 2)2
3 )2 − ( 2 )2 3 )2 − ( 2 )2 ( 3 – 2) ( 3 + 2)
( (
( 3)2 + 2 3 2 + ( 2)2+( 3)2–2. 3 2+( 2)2
= ( 3 )2 + 2 3 × 2 + ( 2 )2 + ( 3 )2 − 2 3 × 2 + ( 2 )2 = ( 3)2 – ( 2)2
3−2 3−2
= 3+2+3+2
3 + 2 6 + 2 3 − 2 6 + 2 3–2
= 1 + 1
= 10

= 5+2 6 +5−2 6

= 10

Example: 3 1
3+ 2− 5
Rationalize the denominator :
Solution:

1 = 1 ( 3+ 2)+ 5
3+ 2− 5 ×
( 3 + 2 − 5) ( 3 + 2)+ 5

= 3+ 2+ 5
( 3 + 2 )2 − ( 5 )2

= 3+ 2+ 5
( 3 )2 + 2 × 3 × 2 + ( 2 )2 − ( 5 )2

= 3+ 2+ 5
3+2 6 +2−5

= 3+ 2+ 5
26

= 3+ 2+ 5× 6
26 6

= 6× 3+ 6× 2+ 6× 5
2 6× 6

= 18 + 12 + 30
2×6

= 3 2 + 2 3 + 30
12

Oasis School Mathematics-10 175

Example: 4

Simplify: 42 6 + 6 3− 3
2+ 2+ 6+ 3

Solution: H=ere, 4 2 × 6 − 2 + 6 × 3 − 2 − 3 × 6 − 3
+2466+2+( 2−×62+(−6 623−2)62−)−+2 23 3 + 2 3− 2 6+ 3 6− 3
4 4( 62)2 3) 6− 3
=2 6 +6( 33 − 2) − 3( 6− 33)2 ×
2 )2 3
= ( 3 )2 −6 ( × −( 62)2−− (
+
×264446612(2×+−262)(314222−−23−−4624328−−(32−−+88+−22++83)2)6213)++−83+2313−−28(3−22−2−−262312(−22−)2312233−−32−(−−(+33−(6123−2(−28−)2+3)2(6−13233(1−8−3−()−232333333)−6−−)()132))6−2(− 6+ 3 6− 3
= 3)
= 4 3 )2
(
= 4

= 4
=

==

== 4((22 432−−21))+ 3 2 −2 3 − 3( 2 − 1)
3

= 2 3 −2+3 2 −2 3 − 2 +1

= (2 2 − 1)

Example: 5

If 7 −1− 7 +1= a+b 7 , find the values of 'a' and 'b'.
7 +1 7 −1

Solution: Here, 7 -1 - 7 +1= a+b 7
7 +1 7 -1
or,
( 7 -1)( 7 -1) - ( 7 +1)( 7 +1) = a+b 7
or, ( 7 +1)( 7 -1) ( 7 -1)( 7 +1)

or, ( 7 )2 - 2 7 ×1+(1)2 - ( 7 )2 +2 7 ×1+(1)2 = a+b 7
or, ( 7 )2 - (1)2 ( 7 )2 - (1)2
or,
or, 7 - 2 7 +1 - 7+2 7 +1=a + b 7
or, 7 -1 7 -1
or,
or, 8-2 7 - 8+2 7 =a+b 7
∴ 6 6

8-2 7 - 8+2 7 =a+b 7
6 6

8-2 7 -8-2 7 = a+b 7
6

-4 7 =a +b 7
6

- 4 7=a+b 7
6

- 2 7 =a+b 7
3

a= 0, b = -2
3

176 Oasis School Mathematics-10

Exercise 9.2

1. (a) What is the rationalizing factor of a?
(b) What is the rationalizing factor of 3 a ?

(c) What is the rationalizing factor of 3+1?

(d) What is the conjugate of 5 – 2?

(e) If a + b 2 = 3 + 4 2, what are the values of ‘a’ and ‘b’?

2. Rationalize the denominator and simplify:

(a ) 63 −2 3 (b) 43 (c) 2+ 3
2 3− 2 2− 3

(d) 3−2 2 (e) a+b− a−b (f) 2x
3+2 2 a+b+ a−b 1+ 1−x

(g) 4 (h) 1 (i) 2a + 3 + 2a − 3
2+ 3 + 7 3 − 2 −1 2a + 3 − 2a − 3

3. Simplify:

(a) 1 1 − 1 (b) 6+ 2 + 6− 2
2+ 2 −1 6− 2 6+ 2

(c) 2− 3 − 2 + 3 (d) p+ q p− q
2+ 3 2 − 3 p− q − p+ q

(e) a2 − b2 − a + a2 − b2 + a (f) y − y2 −1 y + y2 −1
a2 − b2 + a a2 − b2 − a −
y + y2 −1 y − y2 −1

(g) x+y − x−y x+y + x−y
−

x+y + x−y x+y − x−y


4. Simplify:

(a) 1 3 + 1 1 3 + 1 2
2+ + 3+

(b) 32 3 − 43 2 + 23
6− 6− 6+ 2

(c) 63 2 + 42 3 + 25
5− 5+ 2− 3


5. (a) If 4− 15 = a + b 15 , find the values of a and b.
4+ 15

(b) If a = 5 + 2 6, find the value of a+ 1 .
a

(c) If a = 2+ 3 and b = 2– 33, find the value of (a+b)2 and (a-b)2.
2– 3 2+

(d) If a = 3+ 2 = and b = 3- 22, prove that a2 + b2 = 98.
3– 2 3+

Oasis School Mathematics-10 177

Answer
1. Consult your teacher

2. (a)(2 3 + 6) (b) 12 +2 6 (c) (7 + 4 3 ) (d) (17 − 12 2 ) (e) a − a2 − b2
5 b

(f) 2(1 − 1− x) (g) 2 3 − 3 21 + 3 (h) − 6 −2− 2 (i) 2a + 4a2 − 9
4 3

3. (a) − 2 (b) 4 (c) − 8 3 (d) 4 pq (e) 2(b2 − 2a2 )
p−q b2

(f ) − 4y y2 −1 (g) −2 x2 − y2 4. (a) 3 − 2 2+ 3 (b) 4 3 −3 2 − 6
y 2 2

(c) 0 5. a = 31, b = − 8 (b)10 (c)14996 and 192



9.4 Equations involving surds

An equation with surds is called an equation involving surds, i.e., in this type
of equation the variable is in the form of a surd. It is also called a radical equation.
x = 2 and 3 x − 1 = 3 are some examples of an equation involving surds.

Radical equations can be solved by raising the power by the index of the surds, on both
sides. Then every root of the original equation is also the root of the new equation so
formed. This new equation may have other roots also, which do not satisfy the given
equation.

Worked Out Examples

Example: 1 Checking

Solve: 2x + 5 = 3 2x + 5 = 3

Solution: or, 2 × 2 + 5 = 3

2x + 5 = 3 or, 4 + 5 = 3

Squaring both sides, we get- or, 9 = 3

2x + 5 = 9 or, 3 = 3 (true)

or, 2x = 9 – 5 ∴ x = 2

or, 2x = 4 Note: Sometime the value of x may not
or, 4 satisfy the given equation. In such a
∴ x = 2 situation the given equation has no solution.

x = 2

178 Oasis School Mathematics-10

Example: 2 Checking

Solve: x = 3 + x + 9 When, x = 0

Solution: x = 3+ x+9 or, x = 3 + x + 9

or, x+9 = x–3 or, 0 = 3 + x + 9

Squaring both sides, we get- or, 0 = 3 + 3
or, ( x + 9)2 = (x – 3)2 or, 0 = 6 (false)
∴ x = 0 is not a solution of the given
or, x + 9 = (x)2 – 2.x.3 + (3)2 radical equation.

or, x + 9 = x2 – 6x + 9 Again, when, x = 7

or, x2 – 6x + 9 – x –9 = 0 or, x = 3 + x + 9
or, 7 = 3 + 7 + 9
or, x2 – 7x = 0 or, 7 = 3 + 16
or, 7 = 3+4
or, x (x – 7) = 0 or, 7 = 7 (true)

Either, x = 0,

or, x – 7 = 0

∴ x = 7

∴ The only solution of the given radical
equation is x = 7.

Example: 3

Solve: 3x + 1 – x + 1 = 2
Solution: 3x + 1 – x + 1 = 2

or, 3x + 1 = 2 + x + 1

Squaring both sides, we get-

or, ( 3x + 1 )2 = (2 + x + 1 )2
(2)2 + 2.2 x + 1 + ( x + 1 )2
or, 3x + 1 =

or, 3x + 1 = 4+4 x+1+x+1 Checking
= x+5+4 x+1
or, 3x + 1 = 4 x+1 When, x = 0
= 4 x+1
or, 2x – 4 = 2 x + 1 3x + 1 – x + 1 = 2

or, 2(x – 2) = (2 x + 1 )2 3×0+1 – 0 + 1 = 2
= 4 (x + 1)
or, (x – 2) = 4x + 4 or, 1– 1 = 2
= 0
= 0 or, 0 = 2 (false)

Again, squaring both sides Again, when x = 8

(x – 2)2 3x + 1 – x + 1 = 2

or, (x)2 – 2.x.2 + (2)2 or, 3.8+1 – 8 + 1 = 2

or, x2 – 4x + 4 or, 25– 9 = 2

or, x2 – 4x + 4 – 4x – 4 or, 5–3 = 2

or, x2 – 8x or, 2 = 2 (true)

Oasis School Mathematics-10 179

or, x (x – 8) = 0

∴ Either, x = 0

or, x – 8 = 0

∴ x = 8

∴ The only solution of this radical equation is x = 8 .

Example: 4

Prove that the values of x in the equation 2x + 3 = 3x – 5 + 1 are 23 and 3. Can both the
values be the solution of this equation? Justify your answer.

Solution:

2x + 3 = 3x – 5 + 1

Squaring both sides

or, ( 2x + 3 )2 = ( 3x – 5 + 1)2

or, 2x + 3 = ( 3x – 5)2 + 2 3x – 5.1 + (1)2

or, 2x + 3 = 3x – 5 + 2 3x – 5 + 1

or, 2x + 3 = 3x – 4 + 2 3x – 5

or, 2x + 3 – 3x + 4 = 2 3x – 5

or, –x + 7 = 2 3x – 5

or, 7 – x = 2 3x – 5 Checking
or,
(7 –x)2 = (2 3x – 5)2 When, x = 23

or, (7)2 – 2.7.x + x2 = 4 (3x – 5) 2x + 3 = 3x – 5 +1

or, 49 – 14x + x2 = 12x – 20 3(23)+3 = 3(23)–5 +1

or, x2 – 14x + 49 – 12x + 20 = 0 or, 49 = 64 +1
or, 7 = 8 + 1
or, x2 – 26x + 69 = 0 or, 7 = 9 (false)

or, x2 – 23x – 3x + 69 = 0 When, x = 3

or, x (x – 23) –3(x – 23) = 0 2x + 3 = 3x – 5 +1

or, (x – 23) (x – 3) = 0 or, 2 ×3+3 = 3× 3 – 5 +1

Either, x – 23 = 0 or, 9 = 4 +1

or, x – 3 = 0 or, 3 = 3 (true)

or, x = 23 Since only one value satisfies the radical
or, x = 3 equation, the only solution is x = 3

Hence the values of x are 23 and 3.

180 Oasis School Mathematics-10

Example: 5 Checking

Solve: x2+3x–3 – x2 – x – 3 = 2 x2+3x-3 – x2–x–3 = 2
Solution:
42+3.4–3 – 42– 4 – 3 = 2

Here, or, 28–3– 16–7 = 2

x2+3x–3 – x2 – x – 3 = 2 or, 25 – 9 = 2

or, x2+3x–3 = 2 + x2 – x – 3 or, 5 – 3 = 2

Squaring both sides, we get- 2 = 2 (true)

or, (x2 + 3x – 3)2 = (2 + x2 – x – 3 )2

or, x2 + 3x – 3 = 4 + 4 x2 – x – 3 + x2 – x – 3

or, x2 + 3x – 3 = 4 + 4 x2 – x – 3 + x2 – x – 3

or, x2 + 3x – 3 – x2 + x – 1 = 4 x2 – x – 3

or, 4x – 4 = 4 x2 – x – 3

or, 4 (x – 1) = 4 x2 – x – 3

Again, squaring both sides, we get-

or, (x – 1)2 = ( x2 – x – 3 )2

or, x2 – 2x + 1 = x2 – x – 3

or, –2x + x = –3 – 1
or, –x = –4
or, x = 4

∴ Solution of this radical equation is x = 4

Example: 6

Solve: x−1 = 4+ x −1 Checking
x +1 2
x−1 x −1
Solution: x +1 = 4+ 2

xx− +11 = 4 + x2− 1 or, 81 − 1 = 4+ 81 − 1
81 + 1 2

or, ( x )2 − (1)2 = 4+ x −1 or, 80 = 4 + 9 − 1
( x + 1) 2 9+1 2

( x + 1)( x − 1) or, 8 = 4+4
( x + 1)
or, = 4+ x −1 or, 8 = 8 (true)
2

or, x − 1 = 8+ x −1
2

or, 2 x − 2 = 7+ x
or, 2 x − x
or, x = 7+2

=9 Oasis School Mathematics-10 181

or, ( x + 1) = 4+ x −1
2

or, x −1 = 8+ x −1
2

or, 2 x − 2 = 7 + x

or, 2 x − x = 7 + 2

or, x =9

Squaring both sides x = 81

Solution of this radical equation is x = 81.

Exercise 9.3

1. (a) If x = 5, find the value of x.

(b) If 3 x = 3, find the value of x.

(c) If x – a = b , find the value of x.

(d) If 3 x+a = 3 b , find the value of x.

2. Solve:

(a) x – 4 = 3 (b) 3 2x–1 = 2 (c) 4 3x+4 = 4

3. Solve: (b) 2x–3 = x + 4 (c) x + 1 = –3
(a) 3x+1 = 2x+5
(d) 2x–3 = x + 7 (e) 4 1 = 1 (f) 2 = 1
x–3 y–1 3
4. Solve:
(a) x + 7 = 1 + x (b) y – 5 = y – 1
(c) x + 3 = x + 1 (d) x – x = 6
(e) x + 9 = x + 1 (f) 2x–1 = x + 1

(g) x + 5 = x + 1 (h) 9x2-20 = 3x – 2
(i) 2x+3 = 2x+33

5. (a) (i) Find two values of ‘x’ of the radical equation 3x+4 = x + 2 + 2.

(ii) Among these two values, which value does not satisfy the given radical
equation?

(iii) What is the solution of this equation?

(b) Prove that the values of ‘x’ in the radical equation x + 6 + 2x+5 = 7x+11 are
-3
10 and 2 . Which one of these values does not satisfy the given equation? Find

the solution of this equation.

6. Solve:

(a) x2 – 3x + 5 = 1 + x2 – x + 1 (b) x2 – 2x + 1 = 2 + x2 – 5x + 3

15 (d) x+ x+13 = 91
(c) x + 5 + x = x + 5 x+13

182 Oasis School Mathematics-10

7. Solve:

((aa)) 55yy −− 44 == 22 ++ 55yy −− 22 ((bb)) yy −− 2255 == 44 ++ yy −− 55
55yy ++ 22 22 55 ++ yy 55

((cc)) xx −− 11 == 2121 ++ xx22−− 22 ((dd)) xx ++ 55 xx ++ 66 == xx33++ 22 ++ 77
xx ++ 11 xx ++ 22
55yy −− 33
((ee)) 55yy −− 44 == 22 ++ 22 ((ff)) 33xx −− 44 −− 33xx −− 22 == 22
55yy ++ 22 22 ++ 33xx 22


8. So((gglv))e: yy −− 2255 ==xxxx44xx−−−−++++333333==yy+−553131−− 55
((aa))
5xx5xx(+++++a33)33 yy−− x −3 = 1 ((bb)) 2222xxxx(++++b5555) −− 22222xxxxx ++−−−− 5151111111+−== 141422xx − 11 = 1
++ x −3 3 ++ − 11 4
((cc)) xxxx(+−+−c)2222 ++ xxxxxx+−+−+−222222 ==+ 33
x −2 = 3 ((dd)) yy ++(d)yy −− y11+−− yyy==−11 1 − y =1
x +2
((aaS))olve33: xx22(a++) 33 3 xx3 x==244+ ((bb)) yy22 (−−b55) yy3 y++266−==5003
9. 3 33 x = 4 3 3 y + 6 = 0
3 3

Answer

1. Consult your teacher

2. (a) 13 (b) 9/2 (c) 84 3. (a) 4 (b) 7 (c) No solution (d) 10 (e) 4 (f) 49

4. (a) 9 (b) 9 (c) 1 (d) 9 (e) 16 (f) 8, (g) 4 (h) 2 (i) 8

5. (a) (i) 7, -1, (ii) -1 (iii) 7 (b) (i) -3 does not satisfy, 10 is the solution.
2

6. (a) 5 (b) 6 (c) 4 (d) 36 7. (a) 36 (b) 100 (c) 1 (d) 49 (e) 5 (f) 12
8 5

8. (a) 5 (b) 10 (c) 20 (d) 16/25 9. (a) 1, -64 (b) 8, 27

Project Work

I. Find the square root and cube root of the numbers from 1 to 10. Identify whether they
are a rational number or an irrational number. Make a chart and present it in your class.

II. Collect five radical equations which have no solution. Explain the reason why they
have no solution.

Oasis School Mathematics-10 183

Unit

10 Indices

10.1 Warm-up Activities

Discuss the following in your class and draw a conclusion.

Simplify the following,

(i) 2 × 2 × 2 × 2 (ii) 35 × 32

(iii) 56 (iv) 40
54

(v) ( 2 )-2 (vi) 5 210

3

• If ax = a2, what is the value of x?

• If 3x = 1, what is the value of x?
• What value of x satisfies the equation 5x+1 = 125?

• If 53 = x3, what is the value of x?

• Is there any solution to the equation 2x = -1? Give reason.

10.2 Indices

Let x be any real number and n any positive number such that the product of x, n is

given by

x × x × x× x ... (n. factors)

= xn.

Here, xn is called an exponential expression, where x is the base and n the power or
index or exponent of x of the expression.

Further, 9x5 Power
Take an example of 9x5.
Here, 9 is called the coefficient of x5. Co-efficient Base
5 is called the power or index of x.

Some Important Laws of Indices

Consider 'a' as any real number and 'm', 'n' as any positive or negative numbers. Then
the following laws are found to be established.

(i) am × an = am + n, am × an × ap = am+n+p (v) a0 = 1
(vi)
(ii) am ÷ an = am – n (vii) a-m = 1
am
(iii) (am)n = amn = (an)m
1m = 1
m
(viii) ( a )-m = ( b )m
(iv) n am = a n
ba

184 Oasis School Mathematics-10

Worked Out Examples

Example: 1 Example: 2
Simplify: 3 35566xx7 y7 y1111√÷√3 377xx4 y4 y5 5
Simplify: 5m+2 − 5m
5m+1 + 5m
Solution: Solution: 3 35566xx7 y7 y1111√÷√3 377xx4 y4 y5 5
56x7 y11
5m+2 − 5m 5m (5² – 1)
5m+1 + 5m = 5m (51 + 1) = 3 7x4 y5
= 3 8x7-4 .y11-5

= 25 − 1 = 3 8x3.y6
5+1

= 24 = 4 = 3 23.x3.y6
6
= (23.x3.y6)1/3

Example: 3 = 2 .x .y3 × 1 3 × 1 6 × 1
3 3 3

Simplify: x . x . x1 1/b 1 1/c 1 1/a = 2xy2

bc ca ab
x1/c x1/a x1/ b

Solution: x . x . x1 1/b 1 1/c 1 1/a Alternative method

bc ca ab
x1/c x1/a x1/ b
111
1 1/bc 1 1/ca 1 1/ab
1 11  x b  ×  x c  ×  x a 
= x . x . xbc 1/b−1/c ca 1/c −1/a ab 1/a −1/b    x 1   x 1 
 1   a   b 
x1 x1 1   
c−b a−c b−a xc
bc bc ca ca . ab
= . x ab x 1 ×bc x 1 ×ca x 1 ×ab
b c a
= × ×
c−b 11 1 1 ×bc 1 ×ca 1 ×ab
 1/ bc a−c  1/ca b−a  1/ab c a b
=  x bc  .  x ac  .   x x x
   x ab
    xc xa xb
= xb × xc × xa
c−b bc a−c ca b−a ab
=  x bc  .  x ac  .  x ab  =1
  
   

= xc−b . xa−c . xb−a

= xc−b+a−c+b−a

= x0

=1

Example: 4
Simplify:
a+ 1 a+b b 1 a+b
b a
( ) ( ) × −

 b2 − 1 b × a2 − 1 a
a2  b2 
a+b a+b

×
( ) ( ) a+ 1 b − 1
b a
Solution:
 1 b × a2 1 a
b2 − a2  − b2 

Oasis School Mathematics-10 185

aa++1b1b aa++bb.. bb--1a1a aa++bb
( () ( ) ) (( )) ( )== bb..
bb++1a1a bb bb--1a1a aa-- 11 aa.. aa++1b1b aa
bb
aa++1b1b ..aa++bb--aa bb--1a1a aa++bb--bb
bb++1a1a bb .. aa--1b1b aa
( ( ) ) (( ))==

aa++1b1b bb.. bb--1a1a aa
bb++1a1a bb.. aa--1b1b aa
(( )) (( ))==

== abab++++1b1a1b1a bb ..abab----1b1a1b1a aa
bb.. aa
==  aabb++11  aabb--11
aabbb+b+11 aabbaa--11

aa bb
bb.. bb aa
( ) ( )==aa aa
bb
bb.. aa --aa
( ) ( )==aa bb
bb
aa bb--aa
( )==bb

Example: 5 1 1 1
1+ xa + x−b xb + xc +
If a + b + c = 0, prove that: + 1 + x−c + 1 + x−a =1

Solution: Given a + b + c = 0, a + b = –c, b + c = –a and c + a = –b.

LHS = 1 + 1 + 1 =1
1+ xa + x−b b+ xc +
1 + x x−c 1 + x−a

= 1 + x−b + x−c ) + xa + x−a )
1+ xa + x−b x−b (1 + xb xa (1 + xc

= 1 + x−b x−b + xa xa + x0
1+ xa + x−b + x0 + x−b−c + xa+c

= 1 + x−b x−b + xa xa +1 [a+ b+ c = 0]
1+ xa + x−b + 1+ xa + x−b

1+ x−b + xa
= 1+ xa + x−b

=1


186 Oasis School Mathematics-10

Example: 6

2 -2

If x² + 2 = 53 + 5 3 , prove that 5x³ + 15x = 24.

Solution: 2 -2
2 -2
53 – 2 + 5 3
Here, x² + 2 = 53 + 5 3
( ) ( )1²– 2 . 1 . -1 + -1 ²
or, x² =
53 53 53 53
or, x² =

1 -1

53 – 53
( ) ²
or, x² =

1 -1

or, x = 5 3 – 5 3

Cubing both sides, we get-

1 -1

53 – 53
( ) x³ = ³

( ) ( ) ( – ) = 1 ³– -1 ³ – 3. 1 . -1 2 -1
or, x³
53 53 53 53 53 53

or, x³ = 5 – 5–1 – 3. 50. x [ (a–b)3 = a3 – b3 – 3ab (a–b)]

or, x³ = 5 – 1 – 3x
5

or, x³ = 24 – 3x
5

or, 5x³ + 15x = 24
Hence proved.

Exercise 10.1

1. (a) If am × an = ax, find the value of x in terms of ‘m’ and ‘n’.

(b) If ax = am ÷ an, find the value of x in terms of ‘m’ and ‘n’.

(c) If xa = xm × xn , write ‘a’ in terms of ‘m’, ‘n’ and ‘r’.
xr

(d) If ya = (yp)q × yr , find the value of ‘a’ in terms of ‘p’, ‘q’, ‘r’ and ‘t.
yt

(e) If n x5 = x5/4, find the value of n.

( ) (f) If a m b2, find the value of m.
b a2
=

2. Find the value of : ( )(c) 8–2 1/6

( ) (a) 8 2/3 ( )(b) 64 -2/3
27 27

( ) (d) 216–2 2/3 ( )(e) 625 -3/4 (f) 3 43
81

Oasis School Mathematics-10 187

3. Simplify:

(((((((((((Sddggaajjdga))i)))m)))))) 111115522p5255555552255xxxl×××xxx+++i333xxx×××××f22233××3×+++99xxxy55522233xx++x++++mm33322xx2x--22-:---++333mm---22----2-11555555×××332232--++××××××xxx55xx5x33××--×-55599--3311-31xxx11221555xxxmmxx××x+++xx--x-11++221233+++1111122mm -1 (b) 11883383933xxxxxx88x8++++++xx1122xx.12++x11++3322+××3×--3-22233xxx11--111x+++xx–××665500×501××9955988××8--×–-44××××77×33733111188xx8××x×x--33-xx11x1888xxxx--x11 (c) 111199114993311xx31xxxxx22x××+2++xx++111x+11199++22669×+211111××166--××2xx-×x++88+55115××--1×+5555--665599116199××99××xx6×119×1--xxxx11116622x116--×116611××61××99xx29x22---8811xx21x44x1133+1
-1 (b) 3123x+x1+ 2 3- 6x-52 –× 133xx-1 (c) 49xx+1+1××38 –- 92x-1××8821x/3
9x × 66 - 9x-2 × 243
(e) . (f)
((eb)) 13x × 41 ((fc))
2. (h) (i)
((he)) ((if))
(k) (l)
(h) (i)

542233...... ((((((SSSSaaeeaejiii)iSS)))))mmmmiimm188ppp8p333211884481444pp4lllliiii666------××××llfff44f88489ii×××yyyyff2232myy33377:::7332:3+m::///3331333444×××××-×××+×99939999---2552252m×222222+----3--66336312m-1(((fff))) (b) -2218855 33××××114466--2233 ××××-6612244--55--22 (c) 112222117777 × 112266886666 (d) 113344556666××××1166555555
(b) (c) ×× (d)
×
 11(2b227)55  --3312185÷÷3××122469955-23 ××-22-61124-5-2 ( ) ( ) ( )(g)(c1)861 12−423177×××126826965 −3 146 ×155
27 −3 ÷(d52) 356 × 65
2 

 125  3 ÷  25  2
27 9

3. Simplify : 1 1
6 444.... ((((((SSSaaccaciiiS))))))mmmi(ma((((((ppp557757)555plll777xxxaaaiiixxxl--ff-f33))3)+++111xiaayyayxxx111f))()-------11y1aaxa1121:::+++:+y11÷÷1÷)××y×(((-2555-2233232777xxx554454---aaa1112233xx23x)))+++aaxxax111(+++b111 ) 1-ax-y 1-ay-x
(p+q)-1 (p-1+q-1) (c) + (d) (1-xm-n)-1 + (1-xn-m)-1

(b) (((((22222882aa22a2aaaaa)))++--+222aa11a11aa1a---))++11+1aa111++÷÷÷11 (((×××22244aa224a2111---aaaa11aa1666++22))+2)11aa--1aaa-a333+++111÷÷÷ 1
(b) (8a-1 )a+1 --1111××
((44aa++11 ))aa 2222aa +5
(b) +5
(d)
(d) (4a+1 )a-1 × 22a+5
(d)

7. Simplify: (b) 3  125a3 -2 (c) 4 125x13y13 × 4 5x-5y19
(a) (64x3 ÷27a-3 )-2/3  27b-3  (f) 3 25a5b-1c-3 × 3 40ab-5c9
(d) 3 9a-2b7 × 3 3a5b2
(e) 3 8x6y-9 × 4 81x-8y12

(g) 3 27x3 × 4 625x4y4 (h) 4 32x7y11 ÷ 4 2x3y7 (i) 4 16x8y4 ÷ 3 8x6y3

(j) 4 8-1 x5y-2 ÷ 4 2x-3y6 (k) 18x3y-1 × 2x-1y

(l) (p+q)-1 × (p - q)(p2 - q2 ) 1
(m) (a - b)-1 × (a2 - b2 )-1(a+b)
(n) 3 (a+b)-7 ×(a+b)3

188 Oasis School Mathematics-10

8. Simplify:
2 2
( )( ( ) )( ( ) ) ( )(a) 22 (b) (xbxm)l l2 +xlxmmmll+m2ll22×++llmm++xxmmmn22 ×m2x+xmmmnnn+nmm222×++mmnn++xxnnnl22 ×n2+xxnlnnll+l2nn22 ++nnll++ll22
xa+b .xaax++bbb+c22 . 2.x xbb++ccc+a 22. x cc++aa
(a)
(xa .xb .(xxcaa).4xbb .xcc )44
1 xx bbac11cc 1 x cca11aa 1  aa11bb
 xb  bc ×xxbbcc ×xxaacc x b 
(c) (cx)c   ×ca ×ab xaa (d) (x(da-)b )c (.(xxaa--bbb-c))cca..((xxbbc---cca))aab.(xcc--aa )bb
x bb
xaxaa-22b--++aa.bbbb22 xxbaa2---bbb+cc.2 xbxbb-22c--bb++.cccc22 xxcbb2---+cccaa.2 xcxcc-22a--++ccaaaa22
(e) ((xea)+b )a(-bx.aa(++xbbb)+aac--bb)b.(-cx.bb(++xccc)+bba--)ccc.-(axcc++aa )cc--aa (f) (xf)xa2-+abb2 cc--aa

xy(haa)yx . xxyyyz aaaayyxxyz .. yyzzzx aaaayyzzxz
(g) (g)x .x x . . x x . xa+b a2 -aab++2bb b+aac22 --bb22b2 -bbc++2cc c+bba22 --ccc222 -acc++2 aa cc22 --aa22 (h) . zzxx a zz
a xx
b+c-xxa bbcc×bb++xxccac--aa ×c+a -xxb aacc×cc++xxaa --abbb a +b-c aa++bb--cc y2 /z2 z2 /x2 x2 /y2
(i)  xb ×  (j) (j) aa ×aa aa× aa× aa× aay2z2z2yy/22yzz222yyz222//xzz222 x2zz/22zxx222 zz22x2// xxy222 y2xx/22xyy222 xx22 //yy22
(ix) c  xaa zz22 //yy22 xx22 //zz22 yy22 //xx22
 xbb
1 1 1
(ka)x1-y x -z xx.11--yy yxx111---zzz .y -x yyz111---xxx  zz11--yy
(k) a yy11.--zz a .z-ay zz11--xx  (l) yz(laa) y/z yy.zz zaaxzzyyaa////yyzzzx//xz. zz.xx xaayzzxx//aa//xxzzyx /y xxyy a xx//yy
a a   z/y a yy //xx
/.x

9. Prove the following:

( ) ( )(a)� 1 + y a 1 -y a (( )) (( )) ( )(b)b+ 1 3 1 -b 3
x x a 3 a -a 3
= y 2� a � 1 = b 6
1 a 1 a x 1 b a
y  y  b �

 + x  - x a +
 

ab ( )(d� )  1 a+b 1 a+b
(( )) (( ))(c)� a a-b . b a-b  y  x
1+ b 1- a b a x + . y - = y a-b
a a-b b x 
= � b a
a-b . 
b +1 a -1  y2 - 1 . x2 - 1 
a b x2  y2

 p2 - 1 x  p - 1 y-x = p x+y
 - q2   q  q 
 
(e)� 1 y
p2  x - y
   1 
 q 2  q + p 



10. Prove the following:

(a)� 1 + 1 + a y 1 + a z-x + 1 + 1 + ax-z =1
1 + ax-y + az-y ay-z
-x

(b)� 1 + 1 + xb 1 + xb -a + 1 + 1 xc-b =1
1 + xa-b + xa-c xc-a +
-c

(c)� p2 - (p 2p -1 + (p - 1 = y2
(p - y)y - y)y y)y-2 (p - y)y

(d)� a3 + (x 3a2 + (x 3a + (x - 1 = x3
(x - a)x - a)x-1 - a)x-2 a)x-3 (x - a)x

(e)� a2 - (b 2a + (b - 1 = (2a - b)2
(b - a)b - a)b-1 a)b-2 (b - a)b

Oasis School Mathematics-10 189

11. Prove the following.

11 y1/ 3
x + (xy2 )3 -(x2 y)3 x1/ 3
(a)� x+y ×  
1 +  = 1

11  1 
 
(b) m + (mn²)3 + (m²n)3  1 - n3  = 1

m-n  1

m3
12.
(a(a)�)� IfIfaa++bb++cc==00, ,pprroovveetthhaatt::
I1fI1+fa+abxbxca1ca1=+=+1x1,x-p,b-pbr+ro+o1v1v+e+etxhtxbh1ba1+at+tx:x:-c-c ++11++xxc1c1++xx-a-a ==11
�� 11++aa1+1+bb-1-1++11++bb11++cc-1-1++11++cc11++aa-1-1 ==11
(b(b)� )� IfIfaabbcc++11==00, p, prroovveetthhaatt::

��
(c()c� )�
� � 11--aa1-1-bb-1-1++11--bb11--cc-1-1++11--cc1-1-aa-1-1==11

13. 1 -1

(a) If x = 33 + 3 3 , prove that :3x3 - 9x = 10

(b) If x = a1/3 - 1 , prove that : x3 + 3x = a - 1
a1/ 3 a

2 -2

(c) If x2 - 2 = 2 3 + 2 3 , prove that : 2x3 - 6x - 5 = 0


(d) If x2 + 2 = m2/3 + m-2/3, prove that x3 + 3x = m – 1
m



Answers

1. Consult your teacher 2. (a) 4 (b) 9 (c) 1 (d) 1 (e) 27 (f) 2
9 16 2 1296 125

3. (a) 3 (b) 1 (c) 1 (d) 10 (e) 1 (f) 2 (g) 6 (h) 4 (i) 1 (j) 5 (k) 1 (l) 2
2 2

4. (a) 1 (b) 8 (c) 4 (d) 2 (e) 1 (f) 1 (g) 1 5. (a) x2+y2 (b) 1
7 5 pq

(c) 1 (d) 1 6. (a) 1 (b) 1 (c) 1 (d) 16 7. (a) 9 (b) 9 (c) 5x2y8
25 4 16x2a2 25a2b2

(d) 3ab3 (e) 6 (f) 10a2c2 (g) 15x2y (h) 2xy (i) 1 (j) x2 (k) 6x (l) p-q (m) 1
b2 2y2 a-b
1
(n) (a+b)2 8. (a) 1 (b) 1 (c) 1 (d) 1 (e) 1 (f) 1 (g) 1 (h) 1 (i) 1 (j) 1 (k) 1 (l) 1

190 Oasis School Mathematics-10

10.3 Exponential Equations

Let's consider the terms 5x, 10x, ax , etc.

In each of the above terms, the base is constant and power is variable. Such terms are experiential
terms. Equations involving exponential terms are called exponential equation like.

2x+5 = 42x + 1
3x+2 + 3x = 30, etc.

Note:

• ax = a2 ⇒ x = 2

i.e. to equate the indices on both sides of the exponential equation, the bases should be
maintained the same, here 'a' is any number other than zero.

• 24 = x4 ⇒ x = 2

i.e. to equate the bases on both sides of the exponential equation, the powers should be
maintained the same.

• This condition may not always hold, for example (2)4 = (–2)4 but 2 ≠ –2

Worked Out Examples

Example: 1 Example: 2

Solve : 27x = 9(x + 4) Solve: 2x – 2 + 23 + x = 33

Solution: Here, Solution: Here,

27x = 9(x + 4) 2x – 2 + 23 + x = 33
or, (33)x = 32(x + 4)
or, Substitute x = 8 in the or, 2x.2–2 + 23.2x = 33
33x = 32x + 8 given equation and check
whether it satisfies the (( ))or, 22xx 4141++88 = 33
3x = 2x + 8 given equation or not. (( ))or,
or, 3x – 2x = 8 11++3322
∴ x = 8 22xx 44 = 33

or, 2x ( 33 ) = 33
4

or, 2x = 22
∴ x = 2

Oasis School Mathematics-10 191

Example: 3

Solve: 3x + 1 = 3 1
3x 3
1 1
Solution: Here, 3x + 3x = 3 3

Let, 3x = a

∴ Above equation reduces into

a + 1a = 10
3
or, a2a+1 = 10
or, 3a2 + 3 = 3
10a

or, 3a2 – 10a + 3 = 0

or, 3a2 – 9a – a + 3 = 0

or, 3a (a – 3) – 1 (a – 3) = 0

or, (a – 3) (3a – 1) = 0

Either, a – 3 = 0 or, 3a – 1 = 0

or, a = 3 3a = 11
a = 3
or, 3x = 31 or,

∴ x = 1 or, 3x = 3–1

∴ x = –1

Hence, x = ± 1

Example: 4

Solve: 9x – 10 × 3x + 9 = 0
Solution: Here,

9x – 10×3x + 9 = 0

or, (32)x – 10.3x + 9 = 0 Check whether both values satisfy the
or, (3x)2 – 10.3x + 9 = 0 given equation or not.
Let, = a, then
∴ 3x = 0 or, a – 1 = 0
or, a2 – 10a + 9 = 0 or, a = 1
or, a2 – 9a – a + 9 = 0 or, 3x = 30
or, a(a – 9) – 1 (a – 9) = 0 ∴ x = 0
Either (a – 9) (a – 1) = 0 Hence, x = 2, 0
or, = 9
or, a – 9 = 32
a
3x

∴ x = 2



192 Oasis School Mathematics-10

Example: 5

If a = 10x, b = 10y and aybx = 100, prove that xy = 1.

Solution: Here,

Given, a = 10x, b = 10y and aybx = 100

Here, ay.bx = 100

or, (10x)y. (10y)x = 102

or, 10xy. 10xy = 102

(10)2xy = 102

∴ 2xy = 2

or, xy = 1 proved.

Exercise 10.2

1. (a) If x2a+1 = (x3)a, find the value of ‘a’.

(b) If ax = b2x, what is the relation of ‘a’ and ‘b’?

(c) If 32x–1 = 1, find the value of x.

(d) If 5x+1 = 1, find the value of x.

(e) If 25x+1 = 26x–1, find the value of x.

2. Solve: (b) 4x = 614 (c) 2x–3 = 16
1 (f) 27 x/3 = 3
(a) 2x = 64 (e) 8-x/3 = 2
(d) x-2 = 116

3. Solve the following equations:

x+5 2x+5

=
(a) 32x+1 = 92x−1 ( ) ( )(b) 2 32

1 2x+1 x+1
22
=
2x+1 23x ( ) ( )(d)
(c) = × 3 16 5 64

(e) 5x ×(25)x2 = 125 (f) (10)3 y − 3 = 1
0.001

(g) (0.4)x/2 = 0.16 (h) (25)x+3 = 1
0.04

( ) ( )(i)2x 27 −2 (j) 2x2 = (16)x+3 (k) 1 = (25)−2
3 8 5x2
=

4. Solve the following:

(a) 23x – 5. ax – 2 = 2x – 2.a1 – x (b) 75x–4. a4x–3 = 72x – 3.ax –2
(c) 2x – 2 + 2x = 5
(e) 5x + 5x + 1 + 5x + 2 = 155 (d) 3x + 2 + 3x+3 =2
3

(f) 3x+3 + 3x = 28

Oasis School Mathematics-10 193

(g) 3x–2+ 3x = 10 (h) 32x + 3 – 2×9x+1 = 1
9 (j) 3

(i) 2x–5 × 5x–4 = 5 2x+3 × 3x+4 = 18.

5. Solve: 1 1
2x 4
(a) 4x – 3×2x + 2 = 0 (c) 2x + =4

(b) 4x – 6×2x + 1 + 32 = 0 (d) 4x + 1 = 16 1
4x 16

(e) 52x + 2 – 126×5x + 5 = 0 (f) 4×3x+1 – 9x = 27

(g) 2x+3 + 1 –9= 0 (h) 5×4x+1 – 16x = 64
2x

(i) 7x + 343 = 56 (j) 3x+2 + 1 = 30
7x 3x-2

(k) 5x–1 + 52–x = 6 (l) 4x–1 + 42–x = 5

(m) 16x – 3×22x + 2 = 0 (n) 5x+1 + 52–x = 126

6. If a = bx, b = cy and c = az, prove that xyz = 1.

7. If ax = by and bx = ay, show that x = y.

1 11

8. If a=x b=y c z and abc = 1, prove that x + y + z = 0 .
2xz
9. If ax = by = cz and b2 = ac, prove that y = x+z .

10. If 2x = 3y = (12)z, prove that, 1 – 1 = 2 .
z y x

11. If 8712 = 2x × 3y × 11z, find x, y and z.

12. If 2400 = 2x.3y.5z, find x, y and z.

Answer

1. Consult your teacher 2. (a) 6 (b) –3 (c) 7 (d) 4 (e) 1 (f) 1
3 3 -1 -3
3. (a) 2 (b) 5 (c) 2 (d) 11 (e) 1, 2 (f) 2 (g) 4 (h) -2 (i) 6 (j) 6, -2 (k) ± 2

4. (a) 3 (b) 1 (c) 2 (d) -2 (e) 1 (f) 0 (g) 0 (h) -3 (i) 5 (j) -2
2 3 2

5. (a) 0, 1 (b) 2, 3 (c) ± 2 (d) ± 2 (e) 1, -2 (f) 1, 2 (g) 0, -3 (h) 1, 2 (i) 1, 2 (j) ± 1 (k) 1, 2
1
(l) 1, 2 (m) 2 ,0 (n) -1, 2 11. 3, 2, 2 12. 5, 1, 2

Project Work

Collect some problems related to the indices in our daily life and present them in your class room.

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194 Oasis School Mathematics-10