Oasis Math 9

(b) The cost of fencing a circular ground at the rate of Rs. 20 per meter is Rs. 3080.
Find the circumference of the ground.

(c) The total cost of fencing a square field having perimeter 64m is Rs. 2560. Find
the cost per meter of fencing material.

3. (a) Given rectangular field is to be fenced. Find: 28 m

(i) the length of fencing material. 20 m

(ii) cost to fence at the rate of Rs. 40 per meter.

(b) Given circular field is to be fenced 3 times by a fencing mate- 28 m.
rial. Find-

(i) the length of fencing material required to fence the field.

(ii) total cost to fence at the rate of Rs. 50 per meter.

4. (a) The length and breadth of a rectangular field are 60 m and 40 m respectively.
Find the cost of fencing it at the rate of Rs. 50 per meter.

(b) The radius of a circular field is 35 m. Find the length of fencing material to
surround it 5 times. Find the total cost of fencing at the rate of Rs. 80 per meter.

(c) The cost of fencing a square field at the rate of Rs.150 per meter is Rs. 30,000. Find

(i) the perimeter of the field. (ii) the area of the field.

5. (a) Find the cost of carpeting a room having area 80m2 at the rate of Rs. 150 per
square meter.

(b) Find the cost of carpeting a room 4 m by 3 m at the rate of Rs. 150 per square meter.

(c) Total cost of carpeting a square room at the rate of Rs. 200 per square meter is
Rs. 3200. Find (i) area of the room. (ii) length of the room.

(d) The structure of the floor of a room is as shown in the 10 m
given figure, find, (i) area of the room.

18 m

(ii) total cost of carpeting at the rate of Rs. 120 per
square meter.

6. (a) A room is three times as long as it is broad. The cost 15 m
of carpeting the room at the rate of Rs. 175 per square
meter is Rs. 8400.

Find, (i) the area of the room (ii) the perimeter of the room.

(b) A room is twice as long as it is broad. Find the cost of carpeting its floor at the
rate of Rs. 250 per square meter if its perimeter is 42 m.

(c) A square window of each side 7 ft. has semicircular top. Find the cost of fitting
the glass on it at the rate of Rs. 1225 per square feet.

7. (a) Find the cost of carpeting a room with a carpet having length 20 m at the rate
of Rs. 80 per meter.

Oasis School Mathematics-9 95

(b) 120m carpet is required to carpet a hall at the rate of Rs. 240 per meter. Find the
total cost of carpeting.

(c) The cost of carpeting a room at the rate of Rs. 150 per meter is Rs. 6000. Find the
length of the carpet.

(d) The cost of carpeting a room with carpet of length 15m is Rs. 4500. Find the cost
per meter of the carpet.

8. (a) The length of a square room is 8 m. What will be the cost of carpeting the room
with carpet of width 128 cm at the rate of Rs. 120 per meter?

(b) A room is 6 m by 4 m. Find the cost of carpeting its floor with the carpet of
width 80 cm at the rate of Rs. 200 per meter.

(c) A room is 3 times as long as it is broad and its perimeter is 32m. Find the cost
of carpeting its floor with carpet 80 cm wide at Rs. 280 per meter.

(d) The cost of carpeting a room 10 m by 6 m at Rs .80 per meter is Rs. 1200. Find
the breadth of the carpet.

(e) A hall is 18 m long and 15 m wide. Find the breadth of the carpet, if the cost of
carpeting the hall with a carpet is Rs. 1800 at the rate of Rs. 50 per meter.

9. (a) Find the number of stones having area 0.3m2 required to pave the ground
having area 30m2.

(b) How many bricks of area 0.5m2 each are required to pave the ground of area
250m2?

(c) 400 stones of same size are required to pave the ground having area 260m2.
Find the area of each stone.

(d) 200 stones having area 0.25m2 each are required to pave the ground. Find the
area of the ground.

(e) A rectangular ground is 80m×60m. Find the number of bricks of size 20cm ×
10cm to pave the ground.

10. (a) A rectangular ground is 140 m by 80 m. Find the cost of paving the ground with
the stones of size 20 cm by 10 cm at the rate of Rs. 5 per stone.

(b) A rectangular field is 50 m longer than its breadth and its perimeter is 500 m. Find:

(i) the number of stones of size 50 cm × 30 cm to pave the field.

(ii) the cost of paving the field with stones, if cost per stone is Rs. 8.

(c) Length of a rectangular field is twice its breadth. If the perimeter of the field
is 120m, find

(i) the number of stones of size 25cm × 10cm required to pave the field.
(ii) the cost of paving if the cost per stone is Rs. 50.

96 Oasis School Mathematics-9



Answer

1. Consult your teacher 2. (a) Rs 1200 (b) 154m (c) Rs. 40 3. (a) (i) 96 m (ii) Rs. 3840

(b) i) 264m (ii) Rs. 13200 4. (a) Rs. 10,000 (b) 1100m, Rs. 88,000 (c) (i) 200m (ii) 2500m²

5. (a) Rs. 12000 (b) Rs.1800 (c) (i) 16m² (ii) 4m (d) (i) 210m² (ii) Rs. 25,200

6. (a) (i) 48m2 (ii) 32 m (b) Rs. 24,500 (c) Rs. 83606.25

7. (a) Rs. 1600 (b) Rs. 28,800 (c) 40m (d) Rs. 300 8. (a) Rs. 6000 (b) Rs. 6000

(c) Rs.16,800 (d) 4m (e) 7.5m 9. (a) 100 (b) 500 (c) 0.65m² (d) 50m² (e) 2,40,000

10. (a) Rs. 28,00,000 (b) (i) 100,000 (ii) Rs. 8,00,000 (c) (i) 32,000 (ii) Rs.16,00,000

5.4 Area of Pathways

Particularly, we need to make the paths around the field, park, garden etc.
According to the shape and size, the paths are constructed. They may be outside,
inside, crossing each other or circular paths.

(i) Paths outside the field

Let, ABCD be a rectangular field having length l and breadth b. A' d B'
A B
l d
Then, area of field = length × breadth = l × b b+2d d d
C
Again, let the field be surrounded outside by a path, of l+2d C'
uniform width 'd'.

D

Then, new length of field with path (D’C’) = l + 2d, and new D'
breadth of field with path (B'C') = b + 2d then,

Area of field with path = (l + 2d) (b + 2d)
So, Area of the path = Area of field with path – Area of field without path.

= Area of A'B'C'D' – Area of ABCD

= (l + 2d) (b + 2d) – l × b

= lb + 2ld + 2bd + 4d2 – lb

= 2ld + 2bd + 4d2

= 2d (l + b + 2d) A l
(ii) Paths inside the field A' d
d B
Let, A BCD be a rectangular field having length l - 2d B'
D'
(AB) = l and breadth (BC) = b D b-2d d b

Then, area of field ABCD = l × b l C'
C
d

A path of uniform width 'd' runs inside the field.

The new length of field without path (A'B') = l – 2d
The new breadth of field without path (B'C') = b – 2d
Area of the path = Area of ABCD – Area of A'B'C'D'

Oasis School Mathematics-9 97

= lb – (l – 2d) (b – 2d)

= 2d ( l + b – 2d) [∴ multiply and simplify]

(iii) Paths crossing each other

Let, ABCD be a rectangular field having length of field (AB) = l and breadth of field
(BC) = b.
A XY B

Also, let two paths, PQRS and XYZW with uniform width d

'd' cross each other in the middle of the field. P EF Q

Then, area of path PQRS = PQ × PS = l × d d db
SH GR

Area of XYZW = YZ × XY = b × d d

Area of EFGH = EF × FG = d × d = d2 D WZ C
l

∴ Total area of paths crossing each other.

= Area of PQRS + Area of XYZW – Area of EFGH

= ld + bd – d2

= d(l + b – d)
(iv) Path of a circular field:

Let, R and r be the radius of the outer and inner circles respectively. Then,

Area of outer circle = πR2

Area of inner circle = πr2 rR
then,

Area of circular path = π(R2 – r2)

Note: If d be width of circular path then R = r + d.

Worked Out Examples

Example: 1 20 cm
15 cm 2m
A rectangular garden 20m long and 15m broad has a
path 2m wide running outside it on all sides. Find the
area of the path.

Solution:

Here,

Length of garden (l) = 20m

Breath of garden (b) = 15m

Width of path (d) = 2m

Area of path (A) = ?

98 Oasis School Mathematics-9

S ince the path is running outside it on all sides, so,

Area of path (A) = 2d (l + b + 2d)
= 2 × 2 (20 + 15 + 2 × 2) m²
= 156m²

Example: 2

A park 40m long and 30m broad has a path 3m wide running inside it on all sides. Find

the area of the path.

Solution: Here, 40 m

Length of the park (l) = 40m 3m 30m
Breadth of the park (b) = 30m

Width of the path (d) = 3 m

Area of path (A) = ?

Since, the path is running inside it on all sides, so,

Area of path (A) = 2d ( l + b – 2d)
= 2 × 3 (40 + 30 – 2 × 3)

= 6 × 64 m2

= 384m2

Example: 3

A field in the form of a square of side 220m has two paths with uniform width 5m, from
the middle of the opposite sides, find the area of path.

Solution: Here,

Length = breadth 220m 220m
(l) = 220m 5m

Width of path (d) = 5m
Area of paths (A) = ?

Since the paths are crossing each other,

Area of path = d(l+ b – d)
= 5(220 + 220 – 5) m2 = 5 × 435 m2 = 2175 m2

Example: 4

The perimeter of a circular garden is 440 m. A path 1.4 m wide runs inside the park.
Find the cost of gravelling the path at the rate of Rs. 60 per square meter.
Solution: Here,

Oasis School Mathematics-9 99

Perimeter of a circular garden = 440m

Width of path (d) = 1.4 m

Area of inside circular Path (A) = ?

We have, circumference = 2πR

420 = 2 × 22 × R
7
440 × 7
or, R = 44 = 70 m.

Let 'r' be radius of circle without path,

then, r = R – d

= (70 – 1.4) 1.4 m

= 68.6 m

So, Area of circular path = π (R2 – r2)

= 22 [(70 )2 – (68.6)2]
7
= 609.84 m2

Since cost/m² for gravelling the path (C) = Rs. 60.

∴ Total cost (T) = A × C

= Rs. 609.84 × 60
= Rs. 36590.4

Exercise 5.3

1. (a) If 'l' and 'b' be the length and breadth of a rectangular field, 'd' be the width of
the path around the field, what is the area of the path?

(b) If 'l' and 'b' the length and breadth of a rectangular field, 'd' be the width of the
path inside the field, what is the area of the path?

(c) If 'l' and 'b' the length and breadth of a rectangular field, two paths of uniform
width 'd' cross each other in the middle of field, what is the area of the path?

(d) If 'R' and 'r' be the radii of two concentric outer and inner circles respectively,
what is the area between two circles?

2. (a) Find the area of shaded part of given figures:

(a) N (b) a

x Cb
y

100 Oasis School Mathematics-9

(c) a (d)

y T
R

x

3. From the given diagrams, find the area of the shaded regions.

(a) d (b) d (c) 20m
d

42 m d dd d 30m dd
20m
d = 2m 32m58m d
l d d = 3m
6m
d 7md = 4m
45 m 16m

4. Find the area of shaded portions in each of the following figures.

(a) (b) 24m (c)
dd
rR

d

R=14m, r = 10m dd 7m
d = 2m

(d) 28m 4m (f)
1m 1m
(e) 4m

28m

8m

40m
5. (a) A rectangular garden is 50m long and 30m broad. A path 2m wide is running inside

the garden. Calculate the area of the path.

(b) A rectangular ground 100m long is surrounded by a path 5m wide. Find the area of
the path if the breadth of the ground is 80 m.

(c) A square handkerchief having length 20 cm has a border of 3 cm. Find the area of
the border.

(d) Two paths of 2 m wide crossed the middle of a square garden of length 20m.
Calculate the area of the paths.

(e) The radius of a circular pond is 30m. Calculate the area if 3 metres wide path is
running outside the pond.

6. (a) A path of width 4m surrounds a field 55 m long. Find the perimeter of the field, if
the area of the path is 564 m2.

Oasis School Mathematics-9 101

(b) A 52 m long and 40 m broad rectangular field has a footpath of uniform width
running inside it. If the area of the path is 960 m2, find the width of the path.

(c) A square ground is 45m long. Two paths of uniform width are running across the
middle of the ground, if the area of the path is 176 m2, find the width of the path.

(d) A circular field of radius 42 m is surrounded by a path of uniform width. Find the

width of the path if the area of the path is 2002 m2.

7. (a) A rectangular ground having 100 m length and 80 m breadth has a 2 m wide path
within it. Find the total cost of plastering the path at the rate of Rs. 150 per square meter.

(b) A rectangular field having length 60m and breadth 40m is surrounded by a path of
uniform width 5m, find

(i) area of the path.

(ii) cost of plastering the path at the rate of Rs. 150 per square meter.

(c) A 2 m wide road surrounds a square garden of 4225 m2 in area. Find the cost of
paving stones of size 25 cm by 8 cm each on the road at the rate of Rs. 40 per stone.

(d) A rectangular garden is 50 m long and 30 m broad. A path of 2 m wide is running
inside the garden. Calculate:

(i) area of the path. (ii) area of the field except the path.

(iii) the cost of paving stone of size 20 cm × 16 cm on the path at the rate of Rs.
8 per stone.

(iv) the cost of growing grass on the remaining space at the rate of Rs. 50 per
square metre.

(e) A circular pond of area 7546 m2 is surrounded by a path of 3.5 m wide. Find,

(i) the area of the path.

(ii) the cost of paving stone of size 11cm by 5 cm on the path at the rate of Rs. 10
per stone.

(f) The circumference of a circular park is 308 m. A path 2.1 m wide path runs inside
the park. Find the cost of gravelling the path at Rs. 50 per sq. metre.

8. (a) A rectangular field is thrice as long as its breadth. A path of uniform width 2 m is
running outside the field. If the cost of plastering the path at the rate of Rs. 16 per
square meter is Rs. 1,536, find the area of the field.

(b) A rectangular field is twice as long as its breadth. A path of uniform width 2m is
running inside the field. If the cost of gravelling the path at the rate of Rs. 43 per
square meter is Rs. 1,4792, find:

(i) length and breadth of the field. (ii) area of the field.

(c) Length of a rectangular garden is three times its breadth. Two cross paths of uniform
width 5m are running across the middle of the garden. If the cost of gravelling the
path at Rs. 40 per square meter is Rs. 11,000, find the cost of growing grass on the
remaining space at the rate of Rs. 100 per square metre.

102 Oasis School Mathematics-9

Answer

1. Consult your teacher 2. Consult your teacher

3. (a) 332m² (b) 784m² (c) 99 m² 4. (a) 301.71 m² (b) 128 m²

(c) 28.87 m² (d) 168 m² (e) 264 m2 (f) 20m2 5. (a) 304 m²
(b) 1900 m² (c) 204 cm² (d) 76m²
(e) 594 m² 6. (a) 125m

(b) 6 m (c) 2m (d) 7 m 7. (a) Rs. 105,600 (b) (i) 1100m²

(ii) Rs. 1,65,000 (c) Rs. 10,72,000 (d) (i) 304 m² (ii) 1196 m² (iii) Rs.76,000

(iv) Rs. 59800 (e) (i) 1116.5 m² (ii) Rs. 20,30,000 (f) Rs. 31647

8. (a) 75 m² (b) (i) 60m, 30m (ii) 1800m² (c) Rs 40,000

5.5 Prism

Lets observe the following figures.



cb

a

Fig (i) Fig (ii) Fig (iii)

In figure (i) opposite faces are equal and parallel.

In figure (ii) opposite faces are not equal and parallel.

In figure (iii) area of front face and the back face are equal.

Figure (i) and (iii) are prism where as figure (ii) is not a prism.

Hence, the prism is a solid object whose at least one pair of opposite faces are equal

and parallel. A' B'
Area of cross section

Given figure is a prism as its shaded face and its opposite A B c' D'
face are equal and parallel. If we cut the prism parallel

to the shaded region, similar structure be obtained. The c E'
structure so obtained in the cross section of prism.

FE

Cross section of above prism.

Different types of prism and their cross section.

cross section cross section

cross section

Circular prism Rectangular prism

Note: Area of cross section of the prism = Area of base of the prism.

Oasis School Mathematics-9 103

Cross section area of rectangular and square base prism H F

Area of cross section i.e. area of base of prism is the area of A G

the geometrical shape present on its base. In rectangular and

square prism, the base is either rectangle or square, so their

area are given by: ED

Area of square base = a2 (where 'a' is the length of B C
one side of square base.)

Area of rectangular base = l × b

Where, l = length of rectangular base and

b = breadth of the rectangular base

Surface area (Total surface area) of rectangular and square base prism

In prism, all lateral faces have one dimension (height) common so the lateral surface

area of prism is given by (perimeter of base) × height.
In the adjoining figure, if the height of the prism,

AB = HE = GC = FD = h and BC = ED = AG = HF = l,
BE = CD = AH = GF = b

Its lateral surface area = Area of ABEH + Area of ABCG + Area of GCDF + Area of

HEDF = h × b + h × l + h × b + h l.


= 2lh × 2bh

= 2h (l + b)

= Perimeter of base × height.

Total surface area of prism = lateral surface area of prism + 2 × base area

= 2h (l + b) + 2 (l × b)

= 2 (l h + bh + l b)
Volume of prism

Volume of prism is given by V = base area × height = l × b × h

For cube V = l3

Prism Area of cross section LSA TSA Volume

Cube l2 4l2 6l2 l3
Cuboid l×b 2h (l+b) 2(lh+lb+bh) l×b×h

Remember !

• Area of cross section of prism = Area of base
• Lateral surface area of prism = Perimeter of the base × height
• Total surface area = LSA + 2 × Area of base
• Volume of prism = Area of base × height

104 Oasis School Mathematics-9

Worked Out Examples

Example: 1
Which of the following solids are prisms? Identify with reasons.

Fig.(i) Fig. (ii) Fig. (iii)

Solution:
Fig. (i) is a prism because it has one pair of opposite faces which are equal in area and
parallel. The slice taken out along these faces are also equal in area and parallel to the
bases.

Fig (ii) is not a prism because it has no equal and parallel faces.

Figure (iii) is a prism because the shaded face and its opposite faces are congruent.

Example: 2

Find the total surface area of a cube whose volume is 27cm³.

Solution:

Here, Volume of cube (V) = 27cm3

Length of each side (l) = ?

Total surface area (A) = ?

We have, (V) = l3
or, l³ = (3cm)3
or, l = 3 cm
Now, A = 6l2

= 6×3×3 4cm
5cm
= 54 cm2 10cm6cm

∴ The total surface area of the cube is 54 cm2.

10cm 12cm 5cm

Example: 3

Calculate the cross section area and volume of the given prism. 20cm
4cm
Solution:
6cm
Taking cross section only 5cm
10cm
Area of cross section (4 + 6) cm

= (4 × 5 + 10 × (4 + 6) + 5 × 20) cm2 10cm

= (20 + 100 + 100) cm2 = 220 cm² 5cm

20cm

Oasis School Mathematics-9 105

Now, Volume of prism

= cross section area × height

= 220 × 12 cm3

= 2640 cm3

Example: 4

Find the LSA and TSA of given prism. 2cm 3cm
Solution: 2cm
Area of cross section of given prism 2cm 3c1m0cm

= (2 × 3)cm2 + (6 × 3)cm2 3cm 2cm
= 6cm2 + 18cm2 = 24cm2 2cm 3cm

3cm 2cm 3cm
6cm

Lateral surface area of the prism (LSA)

= (2 + 3 + 2 + 3 + 6 + 3 + 2 + 3)cm × 10cm

= 24 cm × 10 cm

= 240 cm2

Total surface area (TSA) = LSA + 2 × area of cross section

= 240cm2 + 2 × 24cm2

= 240cm2 + 48cm2

= 288 cm2.

Example: 5

Three metallic cubes of edges 3 cm, 4 cm and 5 cm are melted and recast to form a new
cube. Find the length of edge and surface area of the new cube.
Solution:

Here, when three cubes are melted and recast into a new cube then the volume is constant,

so let a1, a2, a3, V1, V2 and V3 be the length of the edges of three cubes and volumes
respectively. Then the volume of new cube.

V = V1 + V2 + V3
or, a³ = a31 + a32 + a33 [Where a is the length of the edge of new cube]
or, a3 = 27 + 64 + 125

or, a = 3 216 = 6

∴ Length of the edge of new cube is 6cm .

Now, total surface area of new cube is (A) = 6 a2

= 6 × 62
= 216 cm2


106 Oasis School Mathematics-9

Exercise 5.4 A' B'

AB C' D'
E'
1. (a) In the given figure, write the names of the part of cross section CD

of this prism and its height. F E'

A' B'
B C'
(b) Which part of this prism represents cross section. Also, A
mention which part represents its height? D E'

G

PE

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

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

(e) Write the formula to calculate the volume of the prism.

(f) Write the formula to calculate the volume and TSA of cube.

(g) Write the formula to calculate the TSA of lidless cube.

(h) Write the formula to calculate the volume and TSA of cuboid.

2. Which of the following are prisms? Give reasons. (d)
(a) (b) (c)

3. (a) Find the volume and total surface area of a cube having each side 6cm.

(b) Length, breadth and height of a cuboid are 10cm, 8cm and 6cm respectively,
Find its volume and total surface area.

(c) Find the volume and TSA of given figures.

(i) (ii) (iii)

10cm 20cm
12cm
12cm

8cm 8cm 10cm

4. (a) Total surface area of a cube is 150cm2. Find its volume.

(b) Volume of a cube is 512 cm3. Find its total surface area.

(c) Length and breadth of a cuboid are 20cm and 8 cm respectively. If its volume is
1,600 cm3, find its total surface area.

(d) Length of a cuboid is thrice its breadth and the height is 5 cm. If the volume of
the cuboid is 1,500 cm3, find (i) its length and breadth. (ii) its TSA.

Oasis School Mathematics-9 107

5. (a) A lidless cubical box is 27 cm long. Find its outer total surface area.

(b) If the total surface area of a lidless cubical box is 80 cm2, find the length of its
edge.

6. (a) Find the lateral surface area of the prism if the perimeter of the base is 75cm
and the height is 20cm.

(b) If the lateral surface area of the prism is 630cm2. If its height is 15cm, find the
perimeter of the base.

(c) Perimeter of the base of the prism is 60cm and its lateral surface area is
1,200 cm2, find its height.

7. (a) Find the volume of the prism if the area of the base is 250 cm2 and the height is
15cm.

(b) The volume of the prism is 540 cm3, if its area of base is 30cm2, find its height.

8. (a) Find the total surface area of the prism having lateral surface area 250cm2 and
area of cross section 24cm2.

(b) If the total surface area of a prism is 550cm2 and lateral surface area is 350cm2,
find the area of cross section.

(c) If the total surface area of a prism is 750cm2 and area of base is 70cm2, find the
lateral-surface area.

9. Find the area of cross section, lateral surface area, total surface area and volume of
the following prisms.

(a) (b) (c)

12cm

6cm 8cm 3cm5cm 3cm
3cm 4cm 3cm 8cm
15cm 6cm 5cm
6cm 6cm 6cm
12cm
10cm

(d) (e) (f)

12cm25cm 2cm 3cm 2cm3cm
25cm 2cm 2cm 3cm
25cm 6cm 5cm 2cm 6cm 2cm
3cm 3cm
12cm 3cm



50cm 75cm

10. (a) Find the length and breadth of a cuboid whose height is 25 cm, volume
45,000 cm3 and length is double of its breadth.

108 Oasis School Mathematics-9

(b) Length of a cuboid is three times the breadth and two times the height. If the
volume of the cuboid is 7,776cm3, find its length, breadth and height.

11. (a) Water is poured into a cubical tank at dimension 2.5 m × 3m × 1.5m at the rate
of 10 litre per second. How much time does it take to fill the tank.

(b) To fill water in a cubical cistern at the rate of 8 lit./sec, it takes 4096 second.

(i) Find the inner volume of the cistern.

(ii) Find the total surface area of the inner part of the cistern.

12. (a) Cubes having each side 6cm, 8cm and 10cm are melted and recast into a large
cube. Find the length of a side of new cube.

(b) A metal cube of edge 6 cm is melted and cast out into three smaller cubes. If the
edges of two smaller cubes are 3 cm and 4 cm respectively, find the edge of the
third cube.

Answer
1. Consult your teacher 2. (a) prism (b) not prism (c) not prism (d) prism
3. (a) 216cm³, 216 cm² (b) 480cm³, 376cm² (c) i) 640 cm3, 448cm² (ii) 1200cm³, 680cm²

(iii) 2880cm3, 1248cm2 4. (a) 125cm³ (b) 384cm² (c) 880cm² (d) (i) 30cm, 10cm (ii) 1000cm²
5. (a) 3645cm² (b) 4cm 6. (a) 1500cm2 (b) 42cm (c) 20cm 7. 3750cm2 (b) 18cm (c) 32cm2
8. (a) 298cm2 (b) 100cm2 (c) 610cm2 9. (a) 36cm², 144cm2, 216cm2, 216cm3, (b) 92cm², 720cm2,
904cm2 1380cm³ (c)144cm², 512cm2, 800cm2, 1152cm³. (d) 2850cm2, 18750 cm², 24450cm2,
214750cm3, (e) 30cm2, 150cm2, 210cm2, 150cm3­ (f) 32cm2, 180cm², 244 cm², 192cm3
10. (a) 60cm, 30cm, (b) 36cm, 12cm, 18cm, 11. (a) 18min, 75 sec. (b) 32.768m³, 61.44m2
12. (a) 12cm (b) 5cm

5.6 Area of 4 Walls

i. Area of 4 Walls, Ceiling and Floor of a Room

A rectangular room contains 4 walls, a ceiling and a floor. If the room is rectangular,
2 walls opposite to each other are equal in area.

Here, ABCD represents floor. EFGH represents ceiling. ABGF and CDEH are equal
in area. BCHG and ADEF are equal in area.

∴ ABGF, CDEH, BCHG and ADEF are its four walls. F E
Now, Area of floor ABCD = l × b G H

h

Area of ceiling EFGH = l×b AD
Area of wall BCHG = l×h
Area of wall ADEF = l×h B l Cb

Area of wall ABGF = b × h

Oasis School Mathematics-9 109

Area of wall CDEH = b × h

∴ Area of four walls = l × h + l × h + b × h + b × h

= 2hl + 2bh

= 2h(l + b)

∴ Area of four walls = 2h (l + b)

Again, area of four walls + floor = 2h(l + b) + lb

area of four walls + floor + ceiling = 2h (l + b) + 2lb = 2(lb + bh + lh)

ii. Area of 4 Walls Excluding Doors and Windows

As we know that,

area of 4 walls = 2h(l + b)

Let, there are n doors having dimension l1 × b1 and 'm' windows having
dimensions l2 × b2.

Area of n doors = n(l1 × b1)

Area of m windows = m(l2 × b2)

∴ Area of walls excluding doors and windows

= Area of four walls – area of doors – area of windows

= 2h (l + b) – n(l1 × b1) – m(l2 × b2)

Remember !

• Area of floor = l × b • Area of four walls + ceiling = 2h(l+b)+lb

• Area of ceiling = l × b • Area of four walls + ceiling + floor

• Area of four walls = 2h (l + b) = 2h(l+b)+2lb = 2(lb+lh+bh)
= Perimeter of floor × height

Worked Out Examples

Example: 1

Find the area of four walls of a room having dimensions 5m × 4m × 3m.

Solution:

Here, Length of the room (l) = 5m
Breadth of the room (b) = 4m

We have, Height of the room (h) = 3m

Area of four walls = 2h(l + b)
= 2 × 3 (5 + 4)m2

= 6 × 9m2
= 54m2

110 Oasis School Mathematics-9

Example: 2

A square room has length 6m and its height is 3.5m. It has 2 doors of size 2.5m × 1m and
3 windows of size 2m × 1.5m. Find the area of 4 walls excluding doors and windows.

Solution:

Here, Length of the room (l) = 6m
Breadth of the room (b) = 6m

Height of the room (h) = 3.5m

We have, Area of 4 walls = 2h (l + b)
= 2 × 3.5 (6 + 6)

= 7 × 12 = 84m2

Again, Area of 2 doors = 2(2.5 × 1)m2 = 5m2

Area of 3 windows = 3(2 × 1.5)m2 = 9m2

∴ Area of 4 walls excluding doors and windows

= 84m2 – 5m2 – 9m2 = 70m2

Exercise 5.5

1. (a) Write the formula to calculate the area of four walls.
(b) If 'l' 'b' and 'h' be the length, breadth and height of a room, write the formula to

calculate the area of 4 walls and ceiling.

(c) If 'x', 'y' and 'z' be the length, breadth and height of a room, what is the area of
4 walls, ceiling and floor.

(d) If 'a' be the area of 4 walls of a room, 'a' be the area of a window and 'b' be the
area of a door. If a room contains 3 windows and 2 doors, what is the area of
walls without doors and windows?

(e) If 'x' be a side of a square room, 'y' be its height, what is the area of 4 walls?

2. Find the area of 4 walls of a room having

(a) length 7 m, breadth 5 m and height 4 m.

(b) length 5.75 m, breadth 4.25 m and height 3.65 m.

(c) length 608 cm, breadth 510 cm and height 3 m.

3. Find the area of 4 walls and ceiling of the room having following dimensions.

(a) length 5 m, breadth 4.5 m and height 3.5 m.

Oasis School Mathematics-9 111

(b) length 8 m 25 cm, breadth 6 m and height 350 cm.

4. Find the total surface area of 4 walls, ceiling and floor of the room having the
following dimensions.

(a) length 12 m, breadth 10 m and height 4 m.

(b) length 6.25 m, breadth 6 m and height 3.5 m.

(c) square room of length 4 m and height 3 m.

(d) cubical room of length 4 m.

5. (a) Find the length of a room, which is 6 m broad, 4 m high and the area of 4 walls
is 208m2.

(b) Find the height of the room which is 8 m long, 5 m broad and area of 4 walls is
308 m2.

(c) Find the height of a square room whose area of 4 walls, ceiling and floor is 112.5
m2 and the ceiling is 4.5 m. wide.

(d) The perimeter of room is 28 m. and the height is 3.5 m. Find the area of 4 walls.

6. Find the area of 4 walls of a room having given dimensions and given number of
doors and windows of given dimensions.

(a) l = 6m, b = 4m h = 3m,

Number of windows = 2, having dimensions 2.5m × 1m

Number of doors = 1, having dimension 3m × 1m

(b) l = 15m, b = 5m h = 4m

Number of windows = 5, having dimensions 2m × 1.25m

Numbers of doors = 3, having dimension 3m × 1m

(c) l = 8m, b = 5m, h = 4m

Number of windows = 2, having dimensions 1.9m × 1.4m

Number of doors = 1, having dimension = 2.2m × 1.6m



Answer 3. (a) 89 m² (b) 149.25 m²
1. Consult your teacher 2. (a) 96 m², (b) 73 m² (c) 67.08 m² 5. (a) 20 m. (c) 95.16m²
(b) 138.5m²
4. (a) 416 m² (b) 160.75 m² (c) 80 m² (d) 96 m²

(b) 11.85 m. (c) 4 m. (d) 98 m² 6. (a) 52m2

112 Oasis School Mathematics-9

5.7 Cost Estimation

The dimension of a room of a house is 15ft × 14ft × 9 ft.

To plaster its 4 walls and ceiling how much money is needed?

As, we know that, area of 4 walls = 2h (l+b)

=2 × 9 (15+14)
=18 × 29 ft2 = 522 ft2
If the cost of plastering its wall at the rate of Rs. 35 per square feet, what will be the cost?

Now, cost per square feet of plastering is Rs. 35

Cost of plastering 522 ft2 is Rs. 522 × 35

= Rs. 18,270
Here, 522 ft2 = A.

Rs. 35 per sq. feet = C

Total cost = T

Hence,

Total cost (T) = A × C.


Worked Out Examples

Example: 1

A room is 8 m long, 5m broad and 4 m high. It also contains a door of dimensions 2.5 m
× 1 m and 2 windows of dimensions 3 m × 2 m. Find the cost of colouring its wall at the
rate of Rs 240 per square meter.

Solution: Here,

Length of the room (l) = 8 m
Breadth of the room (b) = 5 m

Height of the room (h) = 4 m

We have, area of the wall = 2h(l + b)

= 2 × 4(8 + 5) m2

= 8 (8 + 5) m2

= 8 × 13 m2 = 104 m2

Area of a door = 2.5 m × 1 m = 2.5m2

Area of 2 windows = 2 (3 × 2) m2 = 12 m2

Area of the wall excluding doors and windows (A)

= (104 – 2.5 – 12)m2 = 89.5 m2

Cost per square meter (C) = Rs 240

We have, Total cost (T) = A × C = Rs. 89.5 × 240 = Rs 21,480

Oasis School Mathematics-9 113

Example: 2

The length of a room is 7m and its height is 3.5 m. If the cost of papering its wall at
Rs. 150 per square meter is Rs. 11,550, find the breadth of the room.

Solution:

Here,

Total cost of papering (T) = Rs. 11,550

Unit cost of papering (C) = Rs. 150 per meter2

Area of wall paper (A) = T = 111,55050 = 77 m2
C

Breadth of room (b) = ?

Length of room (l) = 7 m

Height of room (h) = 3.5 m

Area of 4 walls of a room = Area of wall paper

2h (l + b) = 77

or, 2 × 3.5 (7 + b) = 77

or, 7 ( 7 + b) = 77
or,
7 + b = 77 m
or, 7

b = (11 – 7) m = 4 m.

∴ Breadth of the room is 4 m.

Example: 3

The cost of covering the floor of a square room at Rs. 370 per square meter is Rs. 1,19,880.
The cost of plastering the walls at Rs. 350 per square meter is Rs.1,00,800. Find the
height of the room.

Solution:

Here, while covering the square room,

Total cost (T) = Rs 1,19,880

Cost per square meter (C) = Rs 370

We have, T = A×C

A = T = 119880 = 324 m2
or, C 370

= 324 m2 of floor

∴ Area of the square room = 324 m2

114 Oasis School Mathematics-9

l² = 324 m2

l² = (18)2

or, l = 18 m

∴ l = b = 18 m

Again, while plastering the walls,

Total cost (T) = Rs. 1,00,800

Cost per sq. m (C) = Rs 350

We have, T = A × C

A = T = 1,00,880
C 350

= 288 m2

∴ Area of 4 walls = 288 m2

2h (l + b) = 288
2 × h (18 + 18) = 288

72h = 288

h = 4 m.

∴ The height of the room is 4 m.

Exercise 5.6

1. (a) If 'T' be the total cost of painting the 4 walls of a room and 'C' be the cost per
square meter of painting the wall , what is the area of 4 walls in terms of 'T'
and 'C'?

(b) Area of 4 walls excluding doors and windows is 'A' and the number of papers
required to paste on the wall is 'n', what is the area of the piece of paper?

(c) 20 piece of papers are required to paste it on the wall. If cost per piece of paper
is Rs. x, find the total cost to buy the paper.

(d) Let 'x' be the area of four walls, y be the area covered by doors and windows,
the rate of Rs. z per square meter, what is the cost of painting the walls?

2. (a) Length, breadth and height of a room are 10m, 8m and 4m respectively. Find the
cost of plastering its four walls at the rate of Rs. 75 per square meter.

(b) Length of a square room is 8m and its height is 3.5m. Find the cost of plastering
its four walls and ceiling at the rate of Rs. 45 per square metre.

(c) Find the cost of plastering the 4 walls of a room having dimensions 9m × 8m
× 4m at the rate of Rs. 75 per square metre.

Oasis School Mathematics-9 115

3. (a) A room is 13 m long, 7 m broad and 6 m high. It has two doors each measuring
1.6 m by 2.5 m. Find the cost of painting its walls at Rs. 30 per sq. meter.

(b) A room is 8m long 6m broad and 3m high. It has a door of size 3m × 1m and 2
windows of size 2.5m × 1.5m each. Find

(i) the area of four walls excluding doors and windows.

(ii) cost of plastering the wall at the rate of Rs. 40 per square metre.

4. (a) The length and height of a room are 5m and 3m respectively. If the cost of
plastering its wall at the rate of Rs. 150 per square meter is Rs. 8,100, find the
breadth of the room.

(b) The height of a square room is 4m, the cost of plastering its wall at the rate of
Rs. 120 per square meter is Rs. 9,600, find the length of each side of the room.

5. (a) A room is three times as long as it is broad and its height is 4.6 meters. If the
cost of carpeting its floor at Rs. 240 per sq. meter is Rs. 18,000, find the cost of
painting on its walls at Rs. 24 per sq. meter.

(b) A room is 3 times as long as it is broad and the height is 3 m. If the cost of
carpeting its floor at Rs. 60 per square meter is Rs. 4,500, find the cost of painting
its wall and ceiling at Rs. 18 per square meter.

(c) A square room is 6 m high. The cost of carpeting its floor at the rate of Rs. 40 per.
sq. meter is Rs. 4,000. Find the cost of painting its wall at the rate of Rs. 30 per
square meter.

6. (a) The length of room is two times its breadth and three times its height. If the cost
of painting its walls at Rs. 90 per sq. meter is Rs. 12,960, find the cost of carpeting
its floor at Rs. 360 per sq. metre.

(b) The length of a room is twice the breadth and three times its height, the cost of
plastering its 4 walls at the rate of Rs. 10 per square meter is Rs. 1,440. Find the
cost of carpeting its floor at the rate of Rs. 120 per square meter.

7. (a) The cost of carpeting a square hall at Rs. 45 per square meter is Rs. 288,000. If the
cost is Rs. 43,200 for painting the four walls and ceiling at Rs. 5 per sq. m, find
the height of the room.

(b) The cost of carpeting a room having length 6m at rate of Rs. 250 per square meter
is Rs. 6,000. The cost of colouring its four walls at the rate of Rs. 150 per square
meter is Rs. 9,000, find the height of the room.

116 Oasis School Mathematics-9

Answer (b) Rs. 7920 (c) Rs. 10,200
1. Consult your teacher 2. (a) Rs. 10,800 (ii) Rs. 2940 4. (a) 4m (b) 5m.
(c) Rs. 7200
3. (a) Rs. 6960 (b) (i) 73.5m2 7. (a) 7m (b) 3m.

5. (a) Rs. 4416 (b) Rs. 3510

6. (a) Rs. 25920 (b) Rs. 8640

5.8 Area and Volume of Walls and Estimations of Number of Bricks, Cost

of Making Walls

Let's take a cuboid of any dimension. Let it be 20cm × 15cm × 12cm. Let's keep the
match box of size 2cm × 1cm . How many match boxes can be kept in it?

Let us suppose that n match boxes can be kept in it. Then n match boxes occupy the
same space as 1 box occupies.

If 'V' be the volume of a box

v be the volume of a match box

Then Volume of a box = Volume of N match boxes

N = V = Volume of a bigger box
v volume of a match box

Similarly, while calculating the number of bricks required to construct the wall,

Number of bricks (N) = Volume of the wall
volume of a brick

If a wall contains doors and windows, to calculate the number of bricks, subtract the
space occupied by doors and windows from the total area of the wall. In this case,
width of doors and windows = width of the wall.

Cost Estimation
If V be the volume of the wall

v be the volume of a brick

We know that
Number of bricks (N) = Vv


Again, if the cost of 1 brick = Rs. c

The cost of N bricks = Rs. N × c

Hence, Total cost (T) = N × c

Oasis School Mathematics-9 117

Remember !

• Volume of wall (V) = l × b × h

• Volume of bricks (v) = l1 × b1 × h1
• Number of bricks (N) = V
v
• Volume of wall excepts doors and windows

= Volume of wall – Volume of doors and windows

• Width of wall = width of doors and windows

Worked Out Examples

Example: 1

Find the number of bricks required to construct a wall of length 20m, breadth 50cm and
height 2m with the bricks having volume 0.2 m3 each.

Solution:

Here, Length of wall = 20m

Breadth of the wall = 50cm = 0.5m

Height of the wall = 2m

We have, Volume (V) = l × b × h
= 20 × 0.5 × 2

= 20m2

Volume of a brick (v) = 0.2m3

We have, N = V
v

= 20
0.2

= 200 = 100.
∴ No. of required bricks is 100. 2

Example: 2

Find the number of bricks required to construct a wall 30m long, 80cm broad and 5m
high with the bricks of size 12cm × 5cm × 4cm each. Also find the total cost to construct
it if the cost per brick is Rs. 15.

Solution: Length of the wall (l) = 20 m
Breadth of the wall (b) = 80cm = 0.8m

Height of the wall = 5m

118 Oasis School Mathematics-9

We have, Volume of the wall = l×b×h
= (30 × 0.8 × 5) m3

Again, = 210 m3
Volume of a brick (v)
= 12cm × 5cm × 4cm

= 0.12m × 0.05m × 0.04m = 0.000240m3

We have, N = V = 120 = 1,20,00,000
v 0.000240 24

= 5,00,000

Again, Cost per brick (c) = Rs. 15

Now, Total cost (T) = N × C
= 5,00,000 × 15 = Rs. 75,00,000

Example: 3

Total population of Kusunde village is 1200. Each person needs average 20 litres of
water per day. A water tank is 20m × 15m × 10m. Find

(i) volume of water used per day.
(ii) for how many days full tank of water is enough for the villagers.
Solution:

Total population of the village = 1,200
(i) Total consumption of water per day = 1,200 × 20 litre

or, (v) = 24,000 litre

Again, Volume of the tank (V) = 20m × 15m × 10m = 3,000m3

= 3,000 × 1000 litre

= 30,00,000 litre
∴ (ii) Number of days (N)
= V
v
30,00,000
= 24,000 = 125

∴ A tank of water is enough for 125 days.

Example: 4

A wall is 14 m long, 3 m high and 50 cm thick. It contains 3 windows of dimension
2 m × 1 m and a door of dimension 3 m × 1.5 m. Find,

(i) the number of stones of dimension 50 cm × 20 cm × 30 cm required to construct the
wall.

(ii) total cost of stones if cost per stone is Rs 12.

Solution: Length of the wall (l) = 14 m

Breadth of the wall (b) = 3 m

Oasis School Mathematics-9 119

Height of the wall (h) = 50 cm = 50 m = 0.5 m.
We have, 100
Volume of the wall (V) = l × b × h

= (14 × 3 × 0.5) m3

= 21 m3

Again, thickness of the wall = thickness of doors and windows.

∴ Volume occupied by a door = 3 m × 1.5 m × 0.5 m = 2.25 m3

Volume occupied by 3 windows = 3(2 m × 1 m × 0.5 m) = 3 m3

Volume of the wall except doors and windows (V).

= (21 – 2.25 – 3) m3 = 15.75 m3

Again, Volume of a stone (v) = 50 cm × 20 cm × 30 cm

= 0.5 m × 0.2 m × 0.3 m

= 0.03 m3
We have,
Again, Number of stones (N) = V = 15.75 = 525
v 0.03

Cost per stone (C) = Rs. 12.

We have, Total cost (T) = N × C = Rs. 525 × Rs. 12 = Rs 6,300.

Example: 5

A square room contains 180 m³ of air. It costs Rs. 2400 to plaster 4 walls of the room at
the rate of Rs. 20 per sq. metre. Find the height of the room.

Solution: Let, Length (l) = the breadth (b) = x metre

Here, Volume of the room = volume of the air

i.e. l × b × h = 180 m3

or, x × x × h = 180 m3

or, h = 180 __________ (i)
x2
Again, on plastering 4 walls,

Total cost (T) = Rs. 2400

Cost per m2 (C) = Rs. 20
We have,
or, A = T
or, C

2h (l + b) = 2400 m2
20

2h (x + x) = 120 m2

or, 4xh = 120 m2
or,
or, h = 120
4x
30
h = x m __________ (ii)

120 Oasis School Mathematics-9

From equations (i) and (ii), we get,

180 = 30
or, x2 x

30x = 180

or, x = 13800 = 6 m

Now, putting the value of x in equation (ii) we get,
30
h = 6 m = 5 m

So, the required height of the room is 5 m.

Exercise 5.7

1. (a) If 'V' be the volume of a wall and 'v' be the volume of a brick, find the number
of bricks required to construct the wall.

(b) If 'x' bricks are required to construct the wall and cost per brick is Rs. 'y', find
the total cost of bricks.

(c) If the volume of the wall is 'x' m3 and volume of a brick is 'y' cm3, find the
number of bricks required to construct the wall.

(d) If 1,000 bricks are required to construct a wall and cost per brick is Rs. 5, what
is the total cost of construction of the wall?

2. (a) Find the volume of a brick whose dimensions is 20 cm × 14 cm × 10 cm.

(b) Find the volume of block of length 30 cm, breadth 20 cm and height 10 cm.

(c) Find the volume of a wall 20 m long, 3 m high and 50 cm thick.

3. (a) Find the volume of a wall without doors, if the wall is 6 m long, 3 m high and
30 cm thick and it contains 2 doors of 1.5 m by 1m each.

(b) A wall is 24 m long, 3 m high and 75 cm thick. It contains 4 rectangular windows
of dimensions 2 m × 1.5 m and a door of dimensions 2.5 m × 1 m. Find the
volume of wall except doors and windows.

4. (a) Find the number of bricks of size 20 cm × 12 cm × 5 cm required to build a wall
whose volume is 60 m3.

(b) How many bricks each of volume 900 cm3 will be required to construct a wall
of 20 m × 15 m × 60 cm?

(c) 15000 stones are required to build a wall 12 m × 2.5 m × 50 cm. Find the volume
of each stone.

5. (a) Find the number of bricks required to construct a wall of 20 m × 10 m × 3 m
using the brick of size 20 cm × 10 cm × 12 cm.

(b) How many bricks, each 24 cm × 12 cm × 6 cm will be required for a wall 33 m
long, 3 m high and 36 cm thick, leaving in it 3 gate ways each 3 m × 2 m?

(c) A wall 33 m long, 2.2 m high and 42 cm thick is to be built with bricks of size
1
21 cm × 11cm × 5.5 cm. If the clay joints occupy 5 th of the space of the wall, find
the required number of bricks.

Oasis School Mathematics-9 121

6. (a) Total population of Barpak village is 3,000. Each person needs average 20 litre
of water per day. If a water tank is 20m × 10m × 5m. Find:

(i) total consumption of water in a day.

(ii) for how many days full tank of water is enough for the village?

(b) The total number of household in Giranchaur Namuna Basti is 150. Each
household needs average 500 litres water per day. If a water tank has dimension
12m × 10m × 8m, find

(i) total consumption of water in a day.

(ii) volume of the water tank.

(iii) the number of days, a full tank of water is enough for the village.

7. (a) A wall is 24 m long, 2.5 m high and 40 cm thick. It contains 4 square windows of
length 1.5 m and two gates of 1 m × 2 . Then,

(i) find the number of bricks of size 10 cm × 4.7 cm × 5 cm required to build the wall.
(ii) find the cost of bricks if cost per brick is Rs. 5.

(b) A wall is 30 m long, 4 m high and 80 cm thick. It contains 3 square windows
having each side 2 m and a door of dimensions 2.5 m × 1 m.

(i) Find the number of bricks with size 20 cm × 5 cm × 10 cm to construct the
1
wall if 4 part of the wall is occupied by clay joint.

(ii) Find the total cost of the bricks if cost per brick is Rs, 10.

8. (a) A room contains 960 m3 of air. If its length and breadth are 16 m and 12 m
respectively, find the cost of plastering its wall at Rs. 60 per square meter.

(b) Volume of a square room is 392 m3. If its height is 8 m. Find the cost of carpeting
its floor at the rate of Rs. 30 per square meter.

(c) The length of a room is twice the breadth and it contains 490 m3 of air. If the

cost of plastering its floor at Rs. 150 per square meter is Rs. 14700. Find the cost
of painting its 4 walls at the rate of Rs. 20 per square meter.

Answer 2. (a) 2,800 cm³ (b) 6,000 cm³ (c) 30 m³


1. Consult your teacher

3. (a) 4.5 m³ (b) 43.125 m³

4. (a) 50,000 (b) 2,00,000 (c) 15,00,000 (d) 1,000 cm³
2
5. (a) Rs. 2,50,000 (b) 16,875 6. (a) (i) 60,000l (ii) 16 3 days.
(b) (i) 75,000 l (ii) 960m3/9,60,000 days
l (iii) 12 4
5

7. (a) (i) 80,000 (ii) Rs. 4,00,000 (b) (i) 60,400 (ii) Rs. 2,11,000

8. (a) Rs. 16,800 (b) Rs. 1,470 (c) Rs. 4,200

122 Oasis School Mathematics-9

Assessment Test Paper

Full Marks : 40

Attempt all the questions

Group 'A' [3 × 1 = 3]

1. (a) Find the perimeter of given circle. 14cm

(b) Find the cost of carpeting a room at the rate of Rs. 120 per meter. If the length
of carpet required is 6m.

(c) Find the total surface area of prism if the area of base is 24cm2 and the lateral
surface area is 80cm2.

Group 'B' [8 × 2 = 16]

1. (a) Find the area of given figure. 20 cm 16 cm
2 cm
14 cm
(b) Find the area of shaded part in the given figure.

2. (a) Length, breadth and height of a room are 8m, 4 cm
6m and 4m respectively. Find the area of 4 walls. 4 cm
4 cm
4 cm

(b) Find the area of cross section of given prism. 20 cm

3. (a) Find the number of stones each having area 12cm2 required to pave a ground
having area 96m2.

(b) Perimeter of a ground is 60m. Find the cost of fencing it at the rate of Rs. 250 per
meter.

4. (a) Volume of a wall is 900m3. If 18,000 stones are required to construct it., find the
volume of each stone.

(b) Volume of a cube is 512cm3. Find its total surface area.

Group 'C' [4 × 4 = 16]
5. A room is twice as long as it is broad. Its perimeter is 36m. Find the cost of carpeting

its floor at the rate of Rs. 250 per square metre.

Oasis School Mathematics-9 123

6. A room is three times as long as it is broad. Its height is 4.6m. If the cost of carpeting
its floor at the rate of Rs. 240 per square metre is Rs. 18,000, find the cost of painting
its wall at the rate of Rs. 30 per square meter.

7. How many bricks each 22cm × 12cm × 5cm are required to build a wall of 50m long,
2m hight and 45 cm thick, leaving 2 doors of dimension 3m×2m?

8. Volume of a cubical room is 1,000m3. It has a door of 3m×2m and 3 windows of
3m× 1m. Find the cost of plastering its surface at the rate of Rs.60 per square metre.

Group 'D' [1 × 5 = 5]

9. A rectangular garden is 20m long and 16m broad. A 2m wide path is running around
it. Find

(i) area of the path. (ii) cost of gravelling the path at the rate of Rs.150 per square
metre.

(iii) cost of growing grass in the remaining space at the rate of Rs. 60 per square metre.

Project work
Measure the length, breadth and height of your room. Measure the dimensions of
doors and windows
(i) Ask the rate of plastering the floor, rate of plastering the walls, rate of plastering

the ceiling and estimate cost.
(ii) Calculate the volume of wall, volume of doors and windows and find the number

of bricks required to construct the wall. Ask the cost per brick and estimate the
total cost.
Objective: To estimate the cost of plastering and cost of constructing the wall.
Material required: Measuring tape.

Estimation of cost of plastering the floor

Length of Breadth of Height of Area of Rate of Total cost
room room room floor plastering

Estimation of cost of plastering 4 walls

Length of Breadth Height of Area of Area cov- Exact Rate Total cost
room of room room 4 walls ered by area

doors and of 4
windows walls

Similarly estimate the total cost of construing the wall.
124 Oasis School Mathematics-9

Algebra

32Estimated Teaching Hours

Contents

• Factorisation
• Indices
• Ratio and proportion
• Simultaneous equations of two variables
• Quadratic equation

Expected Learning Outcomes

At the end of this unit, students will be able to develop the
following competencies:
• To factorise the algebraic expressions having common factors,

algebraic expressions in the form of a2 – b2, a3 – b3, a3 + b3
• To factorise the algebraic expressions like ax2 + bx + c
and a4 + a2b2 + b4.
• To simplify the simple problems on indices involving
• To solve the negative and simple fractional exponents the

simple exponential equation
• To solve the simultaneous equations of two variables using

elimination method, substitution method and graphical
method
• To solve the quadratic equations by factorisation method,
completing square method and using formula

Teaching Materials

• Chart paper, A4 size paper, graph, etc.

Oasis School Mathematics-9 125

Unit

6 Factorisation

6.1 Warm-up Activities

Discuss the following in your class and draw out the conclusion.

• What is the formula of a2–b2? Aa GbB
• What is the formula of a3–b3?
IF
• What is the formula of a3+b3? a HC

• in the given figure, what is the area of E
(i) square AEIG (ii) square IFCH
(iii) rectangle EIHD (iv) rectangle GBFI? b

D

• What is the sum of their area?

• What is the area of square ABCD?

• What is your conclusion?

6.2 Factorisation of various types of expressions

Look at the following examples.

2x(a + b) = 2ax + 2bx

and, 2ax + 2bx = 2x (a + b)

the product of 2x and (a + b) = 2 × (a + b)

= 2ax + 2bx

So, 2 x and (a + b) are the factors of (2ax + 2bx).

Again, (x + y) (a + b) = ax + bx + ay + by

∴ ax + bx + ay + by = (x + y) (a + b)

The product of (x + y) and (a + b) = (x + y) (a + b)

= ax + bx + ay + by

So, (x + y) and (a + b) are the factors of (ax + bx + ay + by)
Thus, the process of expressing the given expression in terms of the product of two
or more factors is called factorisation.

And, when an expression is resolved into factors, the given expression is the product
of their factors.

126 Oasis School Mathematics-9

Worked Out Examples

Examples: 1 Factorize the given expressions:

(i) ax – 4ax2 + 3a2x2 (iii) a(x – y) – b(y – x)
co mamx oisnthfaector Solution: a(x–5) – b (y–x)
Solution: Ax – 4ax2 + 3a2x2 = a (x – y) + b ( x – y)

= ax(1 – 4x + 3ax)

= (x– y) (a + b) (x – y) is
(ii) 2a(x+y) – 3b(x+y) – c2 (x + y) (x + y) is the the common
common factor
Solution: 2a (x+y) –3b (x+y) –c2 (x+y) factor

= (x+y) (2a – 3b – c2)

Factorising with a common factor by regrouping the terms:
Let the given expression be ax + ay + bx + by
Let's arrange them to take common factor.

ax + ay + bx + by
= a (x + y) + b (x + y)

= (x + y) (a + b)

Examples: 2 Factorize the given expressions.

(i) ab (x2+1) – x (a2+b2) Open the bracket
= abx2 + ab – a2x – b2x Rearrange the terms
= abx2 – a2x + ab – b2x. Take common
= ax(bx – a) + b (a – bx)
= ax (bx – a) – b (bx–a)
= (bx – a) (ax – b)

(i) x2 – x (2y–z) – 2xz.
Solution: x2 – x(2y–z) – 2yz
= x2 –2xy + xz – 2yz
= (x(x–2y) + z ( x– 2y)
= (x–2y) (x + z)

Exercise 6.1

1. (a) What are the factors of expression x2 + 2x?
(b) Find out the factors of algebraic expression x2 + xy.
(c) Factorise the expression (a+b) (a–b) + a(a+b).

Oasis School Mathematics-9 127

2. Factorise:

(a) 4a + 8ax + 10 ax2 (b) ax – 2ay + 3az

(c) 27 xy – 9x2z (d) 7a3b2 + 49a2b3

(e) 5a (x–3y) – 2b (3y–x) (f) 9xy (2x – 3y) – 12yz (3y – 2x)

(g) 6x2y (a – 5b) – 14 by2(5b – a) (h) 9a2b (x + y) – 15ab2 (x + y)

(i) 2x (3x – 5y) + y (5y – 3x) + 2(3x – 5y) (j) a(a – 2b + c) – 3b (2b – a – c)

3. Factorise: (b) a2 – ab + ac – bc.
(a) x3 + x2 + x + 1

(c) x2 – x (2a – b) – 2ab (d) x2 – 6x + 6y – y2

(e) p2 – p(q + r) + qr (f) a2 – a(2x – y) – 2xy

(g) a2 – (b – 3)a – 3b (h) ab (c2 + 1) – c(a2 + b2)

(i) c(a2–b2) – a(b2–c2) (j) xy (a2 + b2) – ab(x2 + y2)

(k) a(a + c) + b(c + a) + ab + b2 (l) ab (p2 – q2) + pq (a2 – b2)



Answer

1. Consult your teacher. 2. (a) 2a(2 + 4x + 5x²) (b) a(x – 2y + 3z) (c) 9x (3y – xz)

(d) 7a²b²(a + 7b) (e) (x – 3y) (5a + 2b) (f) 3y(3x + 4z)(2x – 3y) (g) 2y(3x² + 7by) (a – 5b)

(h) 3ab ( 3a – 5b) ( x + y) (i) (3x–5y) (2x – y+2) (j) (a–2b+c) (a + 3b)

3. (a) (x² + 1) (x + 1) (b) (a – b)(a + c) (c) (x – 2a) ( x + b)

(d) (x–y) (x + y – 6) (e) (p – q) (p – r) (f) (a – 2x) (a + y)

(g) (a – b) (a + 3) (h) (ca – b) (bc – a) (i) (a + c) ( ca – b²)

(j) (ay – bx) (ax – by) (k) (a + b) ( a + b + c) (l) (aq + bp)(ap –bq)

6.3 Factorisation of the expression of the type (a² – b²):

Geometrical Interpretation of a2 – b2

• Take a square sheet of paper having a side 'a' units.

Its area is a2 a
128 Oasis School Mathematics-9 a

• From the square sheet take 'b' units from each side as shown in the figure.



ba a

b
a

Area of that part = b2

Area of the shaded part = a2 – b2 .............. (i)

• Cut the shaded part. b a

• Again, cut it along the dotted line. b
a–b
a

a–b

a
bb

a–b

• Join these two parts as shown in the figure.

ab

a–b a–b

ba

Then the new sheet is a rectangle having two sides (a + b) and (a – b)
ab

a–b a–b

ba

Area of shaded part = (a + b) (a – b)

∴ ∴ a2 – b2 = (a + b) (a – b)

Alternative method

Worked Out Examples 9x3 – x
9

( )= x 1
Example: 1 Example: 2 9x2 – 9
Factorise: 4x2 – 25y2
Solution : x [ ( ) ]= x 12
= (2x)2 – (5y)2 Factorise: 9x3 – 9 (3x)2 – 3
= (2x + 5y) (2x – 5y)
Solution : = x (3x + 1 ) (3x – 1 )
3 3
= 81x3 – x
9 =x (9x+1) . (9x–1)
3 3
= x (81x2 – 1) x
9 = 9 (9x+1) (9x-1)
= x – 1)
9 (9x + 1) (9x

Oasis School Mathematics-9 129

Example: 3 Example: 4

Factorise: Factorise: a2 – b2 – 3a + 3b
16 (x – 2y)2 – 25(y – 2x)2 = {4(x – 2y)}2 – {5 (y – 2x)}2 Solution :
Solution : = (a + b) (a – b) – 3 (a – b)
= (a – b) ( a + b – 3)
= {4(x – 2y)}2 – {5 (y – 2x)}2
= [4 (x – 2y) + 5(y – 2x)] [4(x – 2y) – 5 (y – 2x)]
= (4x – 8y + 5y – 10x) (4x – 8y – 5y + 10x).
= (– 6x – 3y) (14x – 13y).
= – 3 (2x + y) (14x – 13y).

Example: 5 Example: 6

Factorise: a4 – a2 + 2a – 1 Factorise: 9x2 – 4y2 – 4yz – z2
Solution : Solution :
= (a2)2 – (a2 – 2a + 1) = 9x2 – (4y2 + 4yz + z2)
= (a2)2 – (a – 1)2 = 9x2 – [(2y)2 + 2.2y.z + z2]
= (a2)2 – (a – 1)2 = 9x2 – (2y + z)2
= (a2 + a – 1) (a2 – a + 1) = (3x)­2 – (2y + z)2
= [3x + (2y + z) [3x – (2y + z)]
Example: 7 = (3x + 2y + z) (3x – 2y – z)

Factorise: 16x4 + 40x2 – 24 – 42y2 – 9y4
Solution :
= 16x4 + 40x2 – 24 – 42y2 – 9y4
= (4x2)2 + 2. 4x2 . (5) + (5)2 – 25 – 24 – 42y2 – 9y4
= (4x2 + 5)2 – 49 – 42y2 – 9y4
= (4x2 + 5)2 – ((49 + 42y2 + 9y4)
= (4x2 + 5)2 – [7 + 2. 7. 3y2 + (3y2)2]
= (4x2 + 5)2 – (7 + 3y2)2

= [(4x2 + 5) + (7 + 3y2)] [(4x2 + 5) – (7 + 3y2)]
= (4x2 + 5 + 7 + 3y2) (4x2 + 5 – 7 – 3y2)

= (4x2 + 3y2 + 12) (4x2 – 3y2 – 2)

130 Oasis School Mathematics-9

Exercise 6.2

1. (a) Write the formula of a2 – b2. b a
(b) What are the factors of an algebraic expression x2 – y2?
(c) What is the area of shaded part in the given figure? b
a
2. Resolve into factors:

(a) 9a2 – 25 (b) x2y2 – 25 (c) 121 – 25a2 (d) x4 – y4
(e) 3x3 – 12x
3. Factorise: (f) 3a2 – 75 (g) a+ 1 2 – a – 1 2
b2 b b

(a) 2 (x + y) – x2 + y2 (b) (a + b)2 – a2 + b2

(c) 2a + 3b + 4a2 – 9b2 (d) 9(a2 + 1)2 – 4 (b2 – 1)2

4. Factorise: (b) x2 – 2xy + y2 – z2
(a) x2 – y2 + 2x + 1 (d) 2bc + a2 – b2 – c2
(c) 3a + 3b – 2a2 – 2b2 – 4ab (f) a2 – 4b2 – 25c2 + 20bc
(e) 9a2 – 4b2 + 6a + 1

5. Factorise: (b) 4x2 – 4xy – 3y2 + 12yz – 9z2
(a) a2 – 10a + 24 + 6y – 9y2 (d) x4 – 6x2 – 7 – 8x –x2
(c) x2 – 10xy + 16y2 – z2 + 6yz

(e) (l – x2) (1 – y2) + 4xy (f) (a2 – c2) (b2 – d2)+ 4abcd

6. Using factorisation, find the value of:

(a) (44)2 – (34)2 (b) (100)2 – (90)2 (c) 99 × 101 (d) 110 × 90

Answer
1. Co nsult your teacher. 2. (a) (3a + 5) (3a – 5) (b) (xy + 5) (xy – 5) (c) (11 + 5a) ( 11 – 5a)

( ) ( ) (x²+ y²) (e) 3x a– 5 a+ 5
3. x + y) b b
4a
(d) b (x+ y) ( x – y) ( x + 2) ( x – 2) (f) 3
(g) (a) (x + y) ( 2 – (b) 2b (a + b)
(c) (2a+3b) (2a–3b+1)

(d) (3a² + 2b² + 1) (3a² – 2b² + 5)

4. (a) (x + y + 1) ( x – y + 1) (b) (x – y + z) ( x – y – z)

(c) (a + b) ( 3 – 2a – 2b) (d)( a + b – c) ( a – b + c) (e) (3a + 2b+1) (3a – 2b + 1)

(f) (a +2b– 5c) (a–2b+5c) 5. (a) (a– 3y – 4) (a + 3y – 6) (b) (2x+y–3z) (2x–3y+3z)

(c) (x – 2y – z) (x – 8y + z) (d) x² + x + 1) (x² – x – 7) (e) (1+xy+x–y) (1+x y–x + y)

(f) (ab+cd+bc–ad)(ab+cd–bc+ad) 6. (a) 780 (b) 1900 (c) 9999 (d) 9900

Oasis School Mathematics-9 131

6.4 Factorising the expression of the type a4 + a²b² + b4

To factorise the expression of the type a4 + a2b2 + b4

We make it in the form of perfect square and use the identity a2 – b2 = (a + b) (a – b).

a4 + a2 b2 + b4 = (a2)2 + 2. a2. b2 + (b2)2 – a2 . b2 Alternative method:
a4+b 4+a2b2
= (a2+ b2)2 – (ab)2

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

=(a2+b2)2–(ab)2

Worked Out Examples =(a2+b2+ab) (a2+b2–ab)
= (a2+ab+b2) (a2–ab+b2)

Example: 1 Example : 2

Resolve into factors: 4a4 + 3a2 + 9 Resolve into factors: 64x4 + y4

Solution : Solution :

4a4 + 3a2 + 9 16x4 + y4

= (2a2)2 + 2.2a2.3 + (3)2 – 9a2 = (8x2)2 + 2.8x2.y2 + (y2)2 – 16x2y2

= (2a2 + 3)2 – (3a)2 = (8x2 + y2)2 – (4xy)2

= (2a2 + 3 + 3a) (2a2 + 3 – 3a) = (8x2 + y2 + 4xy) (8x2 + y2 – 4xy)

= (2a2 + 3a + 3) (2a2 – 3a + 3) = (8x2 + 4xy + y2) (8x2 – 4xy + y2)

Exercise 6.3

1. (a) Write the formula for a2 + b2 in terms of (a+b)2.
(b) Write the formula for a2 + b2 in terms of (a–b)2.

(c) Write the relation among (a+b)2, (a–b)2 and ab.

2. Resolve into factors. (b) a2 + 1 + 1 (c) a4 + 64 (d) a4 – 7a2 + 9
(a) a4 + a2 + 1 a2
(e) 4x4 – 25x2 + 36 (f) 9a4 – 19a2b2 + 25b4 (g) x4 – 7x2 + 1
(h) x4 – 6x2+ 1
3. Factorise: (i) 16x4 – 17x2 + 1 (j) 25x4y – 5x2y3 + 4y5
y4 y2

(a) m5 – 15m3n2 + 9mn4 (b) 81x4 + 4y4 (c) 49a4 – 154a2b2 + 9b4

(d) 25x4 – 34x2y2 + 9y4 (e) 49a4 – 11a2b2 + 25b4 (f) x4 + 1 + y4
y4 x4

132 Oasis School Mathematics-9



Answer


( ) ( ) 1.
a + 1 + 1 a – 1 + 1
Consult your teacher 2. (a) (a² + a + 1) ( a² – a + 1) (b) a a

(c) (a² + 4a + 8) ( a² – 4a + 8) (d) (a² – a – 3) (a² + a – 3) (e) (2x² – x – 6) (2x² + x – 6)

(f) (3a² + 7ab + 5b²)(3a² – 7ab + 5b²) (g) (x² + 3x + 1) (x² – 3x + 1) (h) (x² + 2x – 1) (x² – 2x – 1)

( )( ) (i) 4x² + 5x + 1 4x² – 5x + 1 (j) y(5x² + 5xy + 2y²) (5x² – 5xy + 2y²)
y² y y² y

3. (a) m (m2+3mn – 3n2) (m2 – 3mn – 3n2)

(b) (9x2 + 6xy + 2y2) (9x2 – 6xy + 2y2) (c) (7a2 + 14ab + 3b2) (7a2 – 14ab + 3b2)

(d) (5x2 + 8xy+3y2) (5x2 – 8xy+3y2) (e) (7a2 + 5b2 + 9ab) (7a2 + 5b2 – 9ab)

( ) ( ) ( ) x y x y x2 + y2
(f) y –1+ x y +1+ x y2 –1 x2

6.5 Factorisation of the expression of the type a³ + b³ and a³ – b³ :

T o factorise the expression of the type a3 + b3 and a3 – b3, we use the identities,

i) (a + b)3 = a3 + b3 + 3ab (a+b)

or, (a + b)3 – 3ab ( a + b) = a3 + b3

or, (a + b) {(a + b)2 – 3ab } = a3 + b3

or, (a + b) (a2 + 2ab + b2 – 3ab) = a3 + b3

or, (a + b) (a2 – ab + b2) = a3 + b3

∴ a3 + b3 = (a + b) (a2 – ab + b2) …….. (i)

(ii) Similarly, (ii)

(a – b)3 = a3 – b3– 3ab(a–b)

or, (a – b)3 + 3ab (a – b) = a3 – b3

or, (a – b) { (a – b)2 + 3ab} = a3 – b3

or, (a – b) (a2 – 2ab + b2 + 3ab) = a3 – b3

or, (a – b) (a2 + ab + b2) = a3 – b3

∴ a3 – b3 = (a – b) (a2 + ab + b2) ….……. (ii)

(i) and (ii) are the formula. By using these formula, we can solve the following examples.

Worked Out Examples

Example: 1 Example: 2
Factorise : 8a3 + b3
Factorise : 81x3y + 24y4

Solution: Solution:
= (2a)3 + (b)3
= (2a + b) { (2a)2 – 2a.b + b2 } = 3y (27x3 + 8y3)

= 3y { (3x)3 + (2y)3)}

= (2a + b) (4a2 – 2ab + b2 ) = 3y (3x + 2y) { (3x)2 – 3x . 2y + (2y)2}

= 3y ( 3x + 2y) (9x2 – 6xy + 4y2)

Oasis School Mathematics-9 133

Example: 3 Example: 4
( )Resolve into factors : 2 1 Resolve into factors : 125a3 – 64b3 – 5a + 4b
a3 – 8a3
Solution:
Solution:

( ) 1 = {(5a)3 – (4b)3)} – 5a + 4b
= 2 a3 – 8a3

([ ()[ ) ] ( ) ] =2(a)3 –13 = (5a – 4b) {(5a)2 + 5a. 4b + (4b)2 } – 1(5a – 4b)
2a = (5a – 4b) (25a2 + 20ab + 16b2 – 1)

= 2 a – 1 (a)2 – a . 1 + 12
2a 2a 2a

( ) ( )= 2 1 1 + 1
a– 2a a2 + 2 4a2

Exercise 6.4

1. (a) Write the formula for a3 + b3.
(a) Write the formula for a3 – b3.
(c) What are the factors of algebraic expression x3 + (2)3?

(d) Which algebraic expression is equivalent to the product of (x–3) (x2+3x+32)?

2. Factorise:

(a) 64 + x3 (b) 27a3 + b3 (c) 2x4 + 16x
(e) 27x3 – 8y3 (f) a3b3 – c3
(d) a3 + 1 (h) (x + 2)3 – 125 (i) x6 + 64
(g) 1 512 (k) (x + y)3 – (x – y)3
8a3 – b3

(j) 64 (x + 1)3 – 343

3. Factorise: (b) p2 – q2 + p3 – q3 (c) a + b + a3 + b3
(a) p3 + 8q3 + 2p + 4q (e) a – 2b – a3 + 8b3 (f) 8a3 +27b3 – 2a – 3b.
(d) a3 – b3 – a + b



Answer
1. Consult your teacher 2. (a) (4 + x) (16 – 4x + x²) (b) ( 3a + b) ( 9a² – 3ab + b²)

( ) ( ) (d) 1 a 1
(c) 2x ( x + 2) ( x² – 2x + 4) a + 8 a² – 8 + 64 (e) (3x – 2y) ( 9x² + 6xy + 4y²)

( ) ( ) (f) (ab – c) ( a²b² + abc + c²) (g) 1 – b 1 + b + b² (h) (x – 3) (x² + 9x + 39)
2a 4a² 2a

(i) ( x² + 4) (x4–4x²+16) (j) (4x – 3) ( 16x² + 60x + 93) (k) 2y ( 3x² + y²)

2. (a) (p + 2q ) (p² – 2pq + 4q² + 2) (b) (p – q) (p² + pq + q² + p + q) (c) (a + b) ( a² – ab + b² + 1)

(d) (a – b) ( a² + ab + b² – 1) (e) ( a – 2b ) ( 1 – a² – 2ab – 4b²) (f) (2a+3b) (4a2– 6ab +9b2–1)

134 Oasis School Mathematics-9

6.6 Factorisation of the expression of the type ax² + bx + c, a ≠ 0 :

x2 + 5x + 6, x2 + 3x – 4, x2 – x – 72 are the expressions (with three terms i.e. trionomial
of the form ax2 + bx + c).

To factorise such expressions, we must have two numbers whose product is ac and
whose sum or difference is b. We have to follow the following steps to factorise the
expressions of the type ax2 + bx + c;

Steps:
• Find the product ac.
• Write all the possible factors of ac.
• Find the part of factors such that their sum or difference is b.
• Break bx into two factors.
• Regroup the terms and factorise.

Worked Out Examples Here, the product of the numbers = 6 and
the sum of numbers = 5. We have to find
Example 1: Factorise: x² + 5x + 6 two numbers whose product is 6 and sum
Solution: x2 + 5x + 6 = x2 + (3 + 2)x + 6 is 5.
= x2 + 3x + 2x + 6
= x(x + 3) + 2(x + 3) Since, the factors of 6 are 6 × 1 and
= (x + 3) (x + 2) 3 × 2. And the sum of 3 and 2 is 5 and the
Example 2: Factorise: x² + 5x – 6 product of 3 and 2 is 6, so 3 and 2 are the
Solution: required numbers. Since the sign of 6 is
positive, so both the numbers are positive.

We find the two numbers whose product is 6 and difference is 5.

The two numbers are 6 and 1 as 6 × 1 = 6 and 6 – 1 = 5 .

x2 + 5x – 6 = x2 + (6 – 1)x – 6

= x2 + 6x – x – 6

= x (x + 6) – 1(x + 6)

= (x + 6) (x – 1)

Example 3: Factorise: x² – 15x + 56
Solution:

We find the two numbers whose product is 56 and sum is 15, the two numbers are
8 and 7 as 8 × 7 = 56 and (8)+(7) = 15

x2 – 15x + 56 = x2 – (8 + 7) x + 56

= x2 – 8x – 7x + 56

= x ( x – 8) – 7 (x – 8)

= (x – 8) (x – 7)

Oasis School Mathematics-9 135

Example 4: Factorise : 2x² – x – 3

Solution:

We find the two numbers whose product is 6 and difference is 1, the two numbers
are 3 and 2 as 3 × 2 = 6 and 3 – 2 = 1

2x2 – x – 3 = 2x2 – (3 – 2)x – 3

= 2x2 – 3x + 2x – 3

= x (2x – 3) + 1 (2x – 3)

= (2x – 3) (x + 1)

Exercise 6.5

1. (a) What are the two positive numbers whose sum is 7 and the product is 12?
(b) What are the two positive numbers whose product is 56 and the difference is 1?

(c) Which two positive numbers has the sum 9 and the product 18?

2. Resolve into factors.

(a) x2 + 8x + 12 (b) x2 + 9x + 18 (c) 2y2 + 7y + 6

(d) 4x2 – 12xy + 8y2 (e) x2 – 27x + 180 (f) 3x2 + x – 4

(g) 4x2 – 16x – 9 (h) 9a3bx + 12a2b2x – 5ab3x

(i) 12 yx22 + x – 20 (j) 2 yx22 – 3 + yx22
y

3. Factorise:

(a) 2(x + y)2 + 5(x + y) + 3 (b) 9(x +y)2 + (x + y) – 8

(c) (x – 1 )2 + 15 (x – 1 ) + 26 (d) (x2 + 3x)2 + 3(x2 + 3x) +2
x x



Answer

1. Consult your teacher 2. (a) ( x + 6) ( x + 2) (b) ( x + 3) ( x + 6) (c) (2y + 3) (y + 2)

(d) 4 ( x – y) ( x – 2y) (e) (x – 15) ( x – 12) (f) (3x + 4) ( x – 1) (g) (2x–9) (2x+1)
(h) abx (3a – b) (3a + 5b)
( ) ( ) ( ) ( )(i) 4x – 5 3x + 4 (j) 2y – x y – x
yy x y xy

3. (a) (x+y+1) (2x+2y+3) (b) (x + y + 1) (9x + 9y – 8)

(c) (x– 1 + 13) (x – 1 + 2) (d) (x +1)(x+2)(x2+ 3x +1)
x x

136 Oasis School Mathematics-9

Project work

Factorise x2+3x+2 and interpret it geometrically.

x 1 11 1

x2 x x x 1
1

1

• Take a square having each side 'x' units.

• Take 3 stripes having length x units and breadth 1 units

• Take 2 small square having each side 1 unit.

Now, the area of first square = x2

the area of 3 stripes = 3x

the area of 2 small square = 2 x 1
Total area = x2 + 3x + 2
Rearrange these to give a structure of rectangle. x

See the length & breadth of this rectangle. 1

length = (x+2)

breadth = (x+1)

Conclusion : Hence, the factors of x2+3x+2 are (x+2) & (x+1)

Oasis School Mathematics-9 137

Unit

7 Indices

1
x-4
x-2
x-3

a3
x2

Laws of indices

(i) am × an = am + n [Product law] (ii) am ÷ an = am – n [Quotient law]
(iii) (am)n = amn
[Product law] (iv) n am=amn [Root law of index]

am = (am)½ = am/2

3 an = (an) 1 = an
3 3

(v) aº = 1, i.e. 2º = 1, 5º = 1, 10º = 1, etc. [Law of zero index]

( )Note: • a m = am • (ab)m = ambm
b bm
( ) ( ) • am –p = bn p • a–m = 1
bn am am

138 Oasis School Mathematics-9

Worked Out Examples

8

11 38 1 1
46 4096 (243)35x× 32x+1 3
(23)3
3x+2 × 3x–2

Oasis School Mathematics-9 139

3x
(243) 5 × 32x+1
(f) 9x+2 × 33x–2

(3)5 × 3x × 32x+1
5

32(x+2) × 33x–2

(3)3x × 32x+1 3 5x+1
32x+4 × 33x–2
(3) 3x+2x+1

32x+4+3x–2 55x+2
11

55x+2-5x-1 3

1 m 1 n Example: 5
b b
a+ – a Simplify: xy ax × yz ay × zx az
ay az ax
b + 1 m 1 – b n Solution:
a a
ax . 1 ay . 1 az . 1
ay . xy yz zx
= 1 × ×
xy az . 1 ax . 1
m n yz zx
1 1
a + b b – a ay1 a 1 × ax1
z
= ×
b+ 1 m 1 – b n 1 1 1
a a x y z
a a a

1 m 1 n = a1 + 1 + 1 – 1 – 1 – 1
b b y z x x y z
a + . – a
= a0
b + 1 1 – b
a a = 1

140 Oasis School Mathematics-9

ab + 1 m 1 – ab n
= abb+ 1 b
. 1 – ab

aa

= ab + 1× a m. 1 – ab × a n
b ab + 1 b 1 – ab

= a m. a n
b b

= a m +n
b

Example: 6

Simplify: 1 + 1 + 1 =1
1 + ap–q + aP –r 1 + aq–p + aq–r 1 + ar–p + ar–q

Solution:

= 1 + 1 + 1 aq–r + 1
ap–q + aP –r 1 + aq–p + 1 + ar–p + ar–q

= 1 + 1 + a–­r + 1 + 1 + a–­r + 1 + 1 + a–­q
a­–q a–p a­–p a–q a­–p a–r
a–p a–q a–r

= 1 + 1 + 1
a­–p + a–q + a–r a–­q + a–p + a–r a–­r + a–p + a–q
a–p a–q a–p
a–p a–q a–r
= a–p + a–q + a–r + a–q + a–p + a–r + a–r + a–p + a–q

= a–p + a–q + a–r
a–p + a–q + a–r

=1

Exercise 7.1

1. (a) If am × an = ax, find the value of x.

(b) If ap × aq = ay, find the value of y.
ar

(c) If (ap)m × (aq)n = az, find the value of 7z.

(d) What is the value of 100?

(e) Find the value of : 51 – 50.
(f) What is the value of 3 23/2 ?

Oasis School Mathematics-9 141

(g) What is the root law of indices?

(h) Convert ( a )m in terms of the power of b .
b a

(i) If ( a )x = ( b )y , write the relation between x and y.
b a

2. Express with positive indices: 1 1
1 27 3–1
(a) 54 × 5–3 (b) 23 × 42 × 8 (c) 32 × × (d) 73 × 7–4 × 72

3. Find the numerical value of :
( ) 11
(a) (27) 1 (b) (64) 1 (c) 3
3 3 4 × 45


(d) (25)– 1 (e) 8 1 (f) 81 –1
2 27 16
3 3

(g) 126––54÷ 8–4 2 (h) 3–2 (i) 1296 –3
4³ 3–3 + 3–4 625
4

(j) 52 × 254 (k) 2 9 . 27 2 . 4 –3 (l) 32 9 3 64 –2
3 8 9 4 × 16 × 27
1 ÷ 1
625 125


(m) 2 –2 4 1 4 1 (n) (20)3 × (30)2 (o) 94 × 25 × 15²
7 49 49 2 (40)2 × (10)2 27³× 45²× 75
5 × 5÷

(p) 27 1 243 2 ÷ 3 –2 (q) 8 1 81 1 4 –1
3 32 5 4 27 16 25 2
3. 4÷
8

4. Simplify: (b) a3b5c–7 × a–5 – b–6 c³ (c) 3 2x5y4 × 3 4x7y5
(a) 4 a6b–2 c4 × 4 a6b–6 c8
(e) (8a3 ÷ 27b–3)–2/3 (f) (125a3 ÷ 27b–3)–2/3
(d) 3 a–3b–5 ÷ 3 a6b13

(g) 2n+2 × (2n–1)n +1 ÷ 22n (h) (243) 25n 3×2(3n–22n)+1 (i) (27)3n+1. (243)– 45n
2n(n–1) 9n+1 × 9n+5 . 33n–7

(j) 7x+2 – 28 × 7x–1 (k) 8n+2 + 9 × 8n (l) 5a+3 – 15 × 5a– 1
7x × 45 8n+1× 10 – 7 × 8n 61 × 5a

(m) 1 –1qx–y + 1 (n) 1 + 1
1 – qy–x 1 + ax–y 1 + ay–x

5. Simplify : (b) xa 1 xb 1 xc 1
(a) (xa–b)a + b . (xb+c)b–c . (xc+a)c–a ­xb ­xc x­ a ca
ab . bc .

m m2 + mn + n² n n2 + nl + l²
­n ­l
(c) l l²+ lm­ + m²

xx . xx . xx­ m

b2 + c2 b + c c2 + a2 c + a .q)2 (xq + r­­)2 . (xr p)2
­ bc ­ ac (xp. xq . xr)4
(d) a²+ b² a + b (e) (xp + +

xx . xx . xx ­ ab

142 Oasis School Mathematics-9

(f) ­aayx x+y–z ay y + z –x az z+x–y (g) x .a + b a2 – b² b + c x .b2 – c² c + a xc² – a²
­az ­ax
. .

(h) xy ax . yz ay . zx az (i) ab xa/b bc xb/c ca xc/a
ay az ax xb/a xc/b xa/c
× ×

1 ayx 1 ayz 1 axz
y z x
(j) xy . yz . zx

ax ay az

6. Simplify:

x + 1 a. 1 – x b b + 1 3. 1 – b 3
y a. y b a 3. a 3
(a) (b)
y + 1 1 – y a + 1 1 – a
x x b b

6. Prove that:

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

(b) 1 + 1 + 1 + xc 1 + xc – b = 1
1 + xa – b + xa – c 1 + xb – c + xb – a
–a

7. Simplify:

(a) m + (mn2)1/3 + (m2n)1/3 1 – n1/3 (b) x – (x2y²)1/3 + (xy2)1/3 1+ y1/3
m–n m1/3 x+y x1/3

8. If x = aw.bu+v, y = av.bu+w, z = a4.bv+w, prove that xu-v.yw-u.zv-w = 1



Answer

1. Consult your teacher. 2. (a) 51 (b) 4 (c) 1 (d) 7 3. (a) 3 (b) 4

(c) 1 (d) 1 ( )(e) 2 2 4/3 (g) 14 (h) 9 125
16 5 3 (f) 3 4 (i) 216

(j) 511 (k) 27 ( )(l) 3 14 (m) 7 (n) 45 (o) 1
8 4 2 81

(p) 243 (q) 2 4. (a) a3. c3 (b) 1 (c) 2x4y3 (d) 1 (e) 9
128 5 b2 a3b6 4a2b2
1
9
25a2b2 ab2c3

(f) (g) 2 (h) 27 (i) 1 (j) 1 (k) 1 (l) 2 (m) 1 (n) 1

5. (a) 1 (b) 1 (c) 1 (d) x2(a3 + b3 + c3) (e) 1 (f) 1 (g) 1
(h) 1 (i) 1 (j) 1 9b)
( ) ( )6. (a)x a+b (b) b6 7. (a) 1
y a

Oasis School Mathematics-9 143

7. 3 Exponential Equation

Let's take terms 4x, ax, 5y, etc.

In each of the term, constant like 4, a and 5 are base and the power is unknown
quantity (variable). Such terms are exponential terms.

An equation with any one term being exponential is called exponential equation.
For example,

2n = 4, 3y . 4z = 48 etc.

In order to solve the exponential equations, we maintain either the bases on both
sides of it to be equal or their exponents. For example,

If 5x = 52 ⇒ x = 2 and,

x5 = xy ⇒ y = 5

Similarly,

if xa = xb ⇒ a = b and,

ax = ay ⇒ x = y

Worked Out Examples

Example: 1 2x + 1 x+1
Solve:
(a) 3x = 81 =
Solution: (b) 22x + 1 = 42x – 1 3 16( ) ( )(c) 5 64 (d) 4x+1 + 4x = 80
(a)
or, 3x = 81 (b) 22x + 1 = 42x – 1
∴
3x = 34 or, 22x+1 = 22(2x– 1)
(c)
x = 4 ∴ 2x + 1 = 4x – 2
or,
( ) ( )3 16 2x + 1 = or, 4x – 2x = 1 + 2
or, 5 64 x + 1 or,
∴ 2x = 3
∴ 3
or, (16) 2x + 1 = (64) x+1 x = 2
or, 3 5
or,
(4) 2 2x + 1 = (4) 3 x+1
3 5

4x + 2 = 3x + 3
3 5

20x + 10 = 9x + 9

20x – 9x = 9 – 10

11x = –1

144 Oasis School Mathematics-9