Vedanta Excel in Mathematcs Book -10 Final (2078)

Mensuration (II): Cylinder and Sphere

b) Vegetable ghee is stored in a cylindrical vessel of internal radius. 1.4 m and height
1.5 m. If it is transferred into the rectangular tin cans 33 cm × 10 cm × 5 cm, how
14. a) many cans are required to empty the vessel?

b) An iron pipe of internal diameter 2.8 cm and uniform thickness 1 mm is melted
and a solid cylindrical rod of the same length is formed. Find the diameter of the
15. a) rod.
b)
A hollow cylindrical iron pipe with external and internal radii 8 cm and 6 cm
16. a) respectively and length 35 cm is melted and recast into a solid wire of thickness
b) 2.8 cm. Find the length of the wire.
c)
If the surface area of a sphere is 5,544 cm2, find its volume.
17. a)
If the volume of a spherical ball is 38,808 cm3, find its surface area.
b)
If the circumference of the great circle of a solid hemisphere is 132 cm, find
18. a) (i) its curved surface area (ii) total surface area (iii) volume
b)
If the total surface area of a hemisphere is 1,848 cm2, find:
19. a)
b) (i) its curved surface area (ii) volume
c)
d) If the volume of a hemisphere is 686S cm3, find its curved surface area and total
surface area. 3
20. a)
There are two hemispherical objects made of mud having the same shape and size.
b) If the total surface area of one hemispherical object is 462 sq.cm, find the surface
area of the sphere which is formed by attaching those two hemispherical objects.

The total surface area of a solid sphere made of mud is 616 sq.cm, two hemispheres
are formed by cutting it into two equal parts. Find the total surface area of each
hemisphere.

The surface area of a sphere is S sq.cm. If its radius is doubled, by how much does
its surface area increase?

If the radius of a sphere is reduced to one-third, by how many times does its
volume decrease?

Three metallic spheres of radii 2 cm, 12 cm and 16 cm respectively are melted and
re-formed to a single sphere. Find the radius of the new sphere.

A solid metallic sphere of radius 3 cm is melted and drawn into a cylindrical wire
of radius 5 mm. Find the length of the wire.

The diameter of a solid spherical metallic ball is 12 cm. If it is melted and drawn
into a cylindrical wire of length 288 cm, find the thickness of the wire.

How many solid spheres each 6 cm diameter can be made from a solid metallic
cylinder of diameter 4 cm and height 45 cm?

A solid iron sphere of diameter 42 cm is dropped into a cylindrical drum partly
filled with water. If the radius of the drum is 1.4 m, by how much is the surface of
the water raised?

A cylindrical jar of radius 6 cm contains water. How many iron solid spheres each
of radius 1.5 cm are required to immerse into the jar to raise the level of water by
2 cm?

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Mensuration (II): Cylinder and Sphere

21. Calculate the (i) curved surface area (C.S.A.) (ii) total surface area (T.S.A.) and
(iii) volume of the following solids:

a) b) 54 cm c) d)

10.5 cm 13 cm
20 cm
40 cm 14cm

Creative section - B 40 cm

22. a) Given figure is a solid composed of a cylinder with hemisphere 14 cm
at one end. If the total surface area and the height of the solid
b) are 770 sq.cm and 14 cm respectively, find the height of the
23. a) cylinder.

b) (Hints: Sr2 + 2Srh + 2Sr2 = 770 and put r = 14 – h)
c)
d) A combined solid made up of a cylinder of radius 3 cm and length h cm and a
hemisphere with the same radius as the cylinder has volume 792 cm3. Find the
value of h.

A roller of diameter 112 cm and length 150 cm takes 550 complete revolutions to
level a compound. Calculate the cost of levelling the compound at Rs 9 per square
metre.

The external and the internal radii of a hollow cylindrical metallic vessel 56 cm
long are 10.5 cm and 10.1 cm respectively. Find the cost of the metal contained by
the vessel at Rs 2 per cubic cm. Also, find the cost of polishing it's outer surface at
20 paisa per square cm.

A wooden solid is made up of a cylinder with hemispherical ends. If the whole
length of the solid is 84 cm and radius of each hemispherical end is 7 cm, find the
cost of polishing its surface at the rate of 50 paisa per sq. cm.

A water tank is made up of the combination of a cylinder and a hemisphere. The
height of the tank is 6.4 m and the area of the base is 6.16 m2. Find the cost of
filling the tank with water at 40 paisa per litre.

e) A gate has two cylindrical pillars with a hemispherical end on the top of each
pillar. The height of each pillar is 9.96 m and the height of each cylinder is 9.75 m.
Find the cost of colouring the surface of both pillars at Rs 500 per sq. m.

Project work

24. a) Measure the different dimensions (such as diameter, circumference, height) of
water tank, drum, cylindrical bucket, etc. in your house or in school and calculate
b) their curved surface area, total surface area and capacity.
c) Measure the circumference of the circular base of a mineral water bottle, then find
its radius. Calculate the approximate capacity of the bottle.
Measure the length and breadth of a rectangular sheet of chart paper and find its
area. Now, roll the paper to form a cylinder and find its curved surface area. Then
compare the two areas.

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Unit Mensuration (III): Prism and Pyramid

7

7.1 Prism and Pyramid

A prism is a solid (3-D) geometrical polyhedron that has two parallel and congruent
bases. The lateral faces of a prism are rectangles (or parallelogram).

Triangular prism Square prism Pentagonal prism

A pyramid is a polyhedron formed by connecting a polygonal base and a point, called
the apex. The lateral faces of a pyramid are triangles which are formed by each base
edge and apex.

Triangular pyramid Square pyramid Rectangular pyramid

In this way, a prism has two congruent and parallel polygonal bases, whereas a
pyramid has only one polygonal base. A prism has rectangular lateral faces and a
pyramid has triangular lateral faces.

7.2 Surface area and volume of triangular prisms Cross section

A triangular prism has two congruent and parallel triangular bases.
It has 3 rectangular lateral faces.

In the adjoining triangular prism, Its base is a triangle.
(i) Cross sectional area = Area of triangular base So, it is a triangular prism.

(ii) Lateral surface area = Area of 3 rectangular surface a c
= l.a. + l.b + l.c
b

= l (a + b + c) l

= perimeter of triangular base × length

= p.l

(iii) Total surface area = Lateral surface area + 2 × Area of triangular base

= p.l + 2'

(iv) Volume of prism = Area of triangular base × length (or height of prism)

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Mensuration (III): Prism and Pyramid

Worked-out examples

Example 1: Find the total surface area and volume of the following triangular prisms.
a) b)

8 cm

Solution: 6 cm 15 cm 6 cm 15 3 cm

a) The triangular base is a right angled triangle.

? Hypotenuse = 82 + 62 = 64 + 36 = 100 = 10 cm

Now, the total surface area of the prism = Perimeter of the triangular base u l + 2 '
1
= (8 + 6 + 10) cm u 15 cm + 2 u 2 u 8 u 6 cm2
= 360 cm2 + 48 cm2 = 408 cm2

Again, volume of the prism = Area of triangular base u length of prism
1
= 2 u 8 u 6 cm2 u 15 cm = 360 cm3

b) The triangular base is an equilateral triangle.

? The total surface area of the prism = p.l + 2 ∆

= (6 + 6 + 6) cm u 15 3 cm + 2 u 43 (side)2
cm2 + 23
= 270 3 cm2 + 18 u 6 u 6 cm2 3 cm2
= 270 3 3 cm2 = 288

Again, volume of the prism = Area of triangular base u length of prism

= 43 (side)2 u 15 3 cm = 34 u 6 cm u 6 cm u 15 3 cm
= 405 cm3

EXERCISE 7.1
General section

1. a) If x, y, z are the three sides of a triangle of a triangular base prism of length l, write
the formulae to find its (i) cross sectional area (ii) lateral surface area (iii) total
surface area

b) A prism has a right-angled triangular base with perpendicular 8 cm and base 6 cm.
Find its cross sectional area.

c) A right prism base is a triangle whose sides are 13 cm, 20 cm and 21 cm. Find the
area of its cross section.

d) If the perimeter of the triangular base of a prism is 18 cm and its length is 15 cm,
find its lateral surface area.

e) The perimeter of the triangular base of a prism is 14.5 cm and its lateral surface
area is 290 cm2. Find the length of the prism.

f) The area of triangular base of a prism is 32.5 cm2 and its lateral surface area is
300 cm2. Find the total surface area of the prism.

g) The perimeter of the triangular base of a prism is 24 cm and the area of its cross
section is 24 cm2. If the prism is 10 cm long, find its total surface area.

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Mensuration (III): Prism and Pyramid

h) If the area of the triangular base of a prism 25 cm long is 16.4 cm2, find the volume
of the prism.

i) A triangular base prism is 30 cm long. If the length of the sides of its triangular
base are 3 cm, 4 cm and 5 cm, find its volume.
A
j) What height of the prism shown in the figure makes its 8cm 3cm A'
volume 48 cm3? B'
B 4cm
C'
C

k) The given diagram is a solid prism of triangular base. If the B'
volume of the prism is 480 cm3, find its height.
10cm
10 cm
Creative section
2. Find the area of cross section, the lateral surface area, the total surface area and volume

of the following prisms.

a) b) c)

8cm 3cm 4cm 20 cm 8 cm 15 cm 6 cm 10 3 cm

5 cm 10 cm

d) e) f) 15 cm 5cm

20 5 cm 6 cm 35 cm 12cm 15cm

9 cm 13 cm

3. a) If a prism is 15 cm high with its base a triangle having sides 6 cm, 8 cm, and 10
cm, find its:

(i) area of cross section (ii) lateral surface area (iii) total surface area (iv) volume

b) The area of cross section of a triangular prism is 126 cm2 and its volume is
5,040 cm3. If the perimeter of its base is 54 cm,

(i) find its lateral surface area (ii) find its total surface area.

c) The volume of an isosceles right-angled triangular prism is 1,000 cm3. If the length
of the prism is 20 cm, find: (i) its lateral surface area (ii) its total surface area.

7.3 Pyramids

A pyramid is a solid having a polygonal base and triangular faces. The triangular
faces meet at a common point called vertex (or apex). The polygonal base of a pyramid
may be a triangle, rectangle, square, pentagon, hexagon, etc. On the basis of the types
of polygonal bases, there are different types of pyramids.

Its base is a triangle. Its base is a rectangle. Its base is a square.
A triangular pyramid or a A rectangular pyramid A square-based pyramid
tetrahedron

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Mensuration (III): Prism and Pyramid

7.4 Surface area and volume of pyramids O Vertex

The pyramid given alongside is a square-based Height Slant
pyramid. All the triangular faces of a square-based height
pyramid are congruent. The perpendicular drawn h l
from the vertex to the base is the height of the
pyramid. OP = h is the height of the pyramid. The P Q
perpendicular drawn from the vertex to the opposite a
side of each triangular face is called the slant height
of the pyramid. OQ = l is the slant height of the
pyramid.

1. The lateral surface area (L.S.A) of a pyramid = sum of the area of triangular faces.

(i) L.S.A. of a square-based pyramid = 4 × area of a triangle l

1 l
2
= 4 × × a × l

= 2al a

(ii) L.S.A. of a rectangular pyramid a

= 2× area of ' along length + 2 × area of ' along breadth l1 l2
b
= 2 × 1 a × l1 + 2 × 1 b × l2
2 2

= al1 + bl2

a

(iii) L.S.A. of an equilateral triangular-based pyramid = 3 × area of triangular faces

= 3 × 1 a × l l
2

= 3 al a
2

aa

2. The total surface area (T.S.A) of a pyramid = Area of base + Area of triangular faces

(i) T.S.A. of a square-based pyramid = (Side)2 + 4 × area of triangular faces

= a2 + 4 × 1 a × l
2
= a2 + 2al
(ii) T.S.A. of rectangular pyramid

= Area of rectangular base + (2 Area of ' along length + 2 × Area of ' along breadth)

= a × b + 2 × 1 a × l1 + 2 × 1 b × l2 l1 l2
2 2 a b

= ab + al1 + bl2

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Mensuration (III): Prism and Pyramid

(iii) TSA of an equilateral triangular - based pyramid
= Area of equilateral triangular base + 3 × Area of triangular faces

= 3 a2 + 3 × 1 a×l = 3 a2 + 3 al a l
4 2 4 2 a a

3. In the adjoining square-based pyramid, OP is the O
perpendicular drawn from the vertex to the surface of the
base. So, OP is the height (or altitude) of the pyramid.

The volume of a pyramid = 1 Area of the base × height h
3

(i) The volume of a square-based pyramid = 1 a2 × h P a
3 a

(ii) The volume of a rectangular pyramid = 1 a × b × h
3

(iii) The volume of an equilateral triangular pyramid = 1 × 3 a2 × h = 3 a2h
3 4 12

Worked-out examples

Example 1: Find the total surface area and volume of the following pyramids.

(i) (ii) (iii)

12 cm 15 cm 16 cm
15 cm 18 cm
12 cm 13 cm

18 cm 10 cm

18 cm 14 cm 10 cm

Solution:

(i) In the square-based pyramid,
Length of the base (a) = 18 cm

Slant height (l) = 15 cm

Height of the pyramid (h) = 12 cm

Now, the total surface area of the pyramid = a2 + 2al

1 = (18 cm)2 + 2 × 18 cm × 15 cm = 864 cm2
3
Also the volume of the pyramid = a2 × h

= 1 × 182 × 12 cm3 = 1,296 cm3
3
(ii) In the rectangular pyramid,

length of the base (a) = 14cm

breadth of the base (b) = 10 cm

slant height along length (l1) = 16 cm
slant height along breadth (l2) = 18 cm

height of the pyramid (h) = 15 cm

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Mensuration (III): Prism and Pyramid

Now, the total surface area of the pyramid = ab + al1 + bl2
= (14 × 10 + 14 × 16 + 10 × 18) cm2
= (140 + 224 + 180)cm2 = 544 cm2

Also, the volume of the pyramid = 1 a×b×h
3

= 1 × 14 × 10 × 15 cm3 = 700 cm3
3
(iii) In equilateral triangle-based pyramid,

length of the base (a) = 10 cm

slant height (l) = 13 cm

height of the pyramid (h) = 12 cm

Now, the total surface area of the pyramid = 3 a2 + 3 al
4 2

= 3 × 102 + 3 × 10 × 13 cm2
4 2

= (25 3 + 195) cm2

= (25 × 1.73 + 195) cm2 = 238.25 cm2

Also, the volume of the pyramid = 3 a2h = 3 × 102 × 12 cm3 = 173 cm3
12 12

Example 2: In the given figure, the vertical height (AB) and the A 25 cm
length of the slant edge (AC) of the square-based
pyramid are 24cm and 25cm respectively. Find the 24 cm C
volume of the pyramid. B

Solution: D P
Here, vertical height (AB) = h = 24cm

Length of the slant edge (AC) = 25 cm

In right angled ' ABC, BC = 252 – 242 = 7cm

? DC = 2 × BC = 2 × 7cm = 14cm

Now, in right angled ' DPC, DP2 + PC2 = DC2

or, a2 + a2 = 142

or, a = 7 2 cm

Again, the volume of the pyramid = 1 a2 × h
3

= 1 × (7 2 )2 × 24cm3
3

= 784 cm3

Hence, the required volume of the pyramid is 784 cm3.

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Mensuration (III): Prism and Pyramid

Example 3: Given figure is a square-based pyramid in which the 8 cm
vertical height is 8cm and the length of the base is
12 cm. Find its total surface area and volume.

Solution:
Here, length of the base of pyramid (a) = 12cm

vertical height of the pyramid (h) = 8cm

Here, PQ = 1 × RS = 1 × 12 cm = 6 cm 12 cm
2 2 O

In right angled ' OPQ, OQ = OP2 + PQ2

= 82 + 62 = 10 cm 8 cm

Now, the total surface area of the pyramid = a2 + 2al

= (12cm)2 + 2×12×10 cm2 Q P
12 cm
= 384 cm2 R S

Also, the volume of the pyramid = 1 a2h = 1 × 12 × 12 × 8 cm3 = 384 cm3
3 3

Hence, the total surface area of the pyramid is 384 cm2 and its volume is 384 cm3.

Example 4 : The solid alongside is a square-based pyramid in which 13 cm
the length of its each edge is 13 cm. Find its total surface
area.

Solution:

Here, length of the base (a) = 10 cm 10 cm
O
length of the slant edge = 13 cm

Now, EB = 1 × AB = 1 × 10 cm = 5 cm
2 2

Then, in right angled ' BEO, OE = OB2 – EB2 = 132 – 52 = 12 cm D C
? Slant height (l) = 12 cm
E 5 cm B13 cm
Now, the total surface area of the pyramid = a2 + 2al O

= (10 cm)2 + 2×10×12 cm2 A

= 340 cm2
Hence, the required total surface area of the pyramid is 340 cm2.

Example 5: In the given figure, if the length of a side of the base of
the pyramid having square base is 14 cm and the volume
of the pyramid is 1,568 cm3, find the total surface area of Q
the pyramid.

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Vedanta Excel in Mathematics - Book 10

Mensuration (III): Prism and Pyramid

Solution:

Here, the length of the base of the pyramid (a) = 14 cm O
Q 7cmP
The volume of the pyramid = 1,568 cm3
1
or, 1 3 × a2 × h = 1,568 cm3
or, 3 = 1,568 cm3
× 14 × 14 × h 24 cm

or, h = 24 cm

In the figure, slant height (OQ) = l = OP2 + PQ2

= 242 + 72 = 25 cm 14 cm

Now, the total surface area of the pyramid = a2 + 2al

= (14cm)2 + 2 × 14 × 25 cm2 = 896 cm2

Hence, the required total surface area of the pyramid is 896 cm2.

Example 6: Find the total surface area and volume of the
solid object given in the figure.

Solution: 10m 7m
Here, the length of the lower most cuboid (l1) = 2.5 m
The breadth of the lower most cuboid (b1) = 2.5 m 2.5m 2.5m
The height of the lower most cuboid (h1) = 7 m O
The height of the solid = 10 m

? The height of the uppermost pyramid (h) = 10 m – 7 m = 3 m

Also, in the square-based pyramid, slant height OQ = l = OP2 + PQ2

= 32 + 1.252 3m

= 3.25 m P 1.25m Q
Now, the surface area of the uppermost pyramid = 2al 2.5m

The surface area of the lowermost cuboid = 2 × 2.5 × 3.25 m2 = 16.25 m2
= 2(l1b1 + b1h1 + l1h1) – l1b1

= 2(2.5 × 2.5 + 2.5 × 7 + 2.5 × 7) – 2.5 × 2.5

= 82.5 – 6.25 = 76.25 m2

Then, the total surface area of the solid = 16.25 m2 + 76.25 m2 = 92.5 m2

Again, the volume of the solid = volume of pyramid + Volume of cuboid

= 1 a2 × h + l1 × b1 × h1
3

= 1 × 2.5 × 2.5 × 3 + 2.5 × 2.5 × 7
3

= 6.25 + 43.75 = 50 m3

Hence, the required total surface area of the solid is 92.5 m2 and the volume is 50 m3.

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Mensuration (III): Prism and Pyramid

Example 7: The volume of the solid given alongside is 448 cm3. Find the height of the

prism. A

Solution:

Here, the length of the base of uppermost pyramid (a) = 8 cm 5cm

The slant height of the pyramid (l) = 5 cm O B
8cm 8cm
The length (l1) = the breadth (b1) of cuboid (prism) = 8 cm
In the uppermost pyramid, its height (h) = AO = AB2 – OB2

= 52 – 42 = 3 cm
1 1
Now, the volume of the pyramid = 3 a2 × h = 3 × 8 × 8 × 3cm3 = 64 cm3

The volume of the cuboid (prism) = volume of solid – volume of pyramid

= 448 cm3 – 64 cm3 = 384 cm3

Again, the volume of the prism = l1 × b1 × h1
or, 384 = 8 × 8 × h1
or, h1 = 6 cm
Hence, the required height of the prism is 6 cm.

Example 8: The gate of a stadium has two pillars, each of height is 8 ft. with four visible
lateral faces and 6 ft. × 6 ft. bases. The top of each pillar has a combined
pyramid of height 4 feet. If the combined structures of both pillars and
pyramids are painted at the rate of Rs 75 per sq. ft., calculate the total cost
of painting.

Solution:

Here, length of the base of each pillar (l) = 6 ft. 4 ft.
breadth of the base of each pillar (b) = 6 ft. 3 ft.
height of each pillar (h) = 8 ft.
Now, the lateral surface area of each pillar = 2lh + 2bh
= 2h(l + b)

= 2 × 8(6 + 6) 8 ft.

= 192 sq. ft.

? The lateral surface area of 2 pillars = 2 × 192 sq. ft.

= 384 sq. ft

Again, the slant height of each pyramid (l1) = 42 + 32 = 5 ft. 6 ft.

Then, the lateral surface area of each pyramid = 2al1 6 ft.

= 2 × 6 × 5 = 60 sq. ft.

? The lateral surface area of 2 pyramids = 2 × 60 sq. ft. = 120 sq. ft.

Now, the total area of two combined structures = (384 + 120) sq. ft.

= 504 sq. ft.

Also, the rate of cost of painting = Rs 75 per sq. ft.

? The total cost of painting = Area × Rate = 504 × Rs 75 = Rs 37,800

Hence, the required cost of painting the pillars with pyramids is Rs 37,800.

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Mensuration (III): Prism and Pyramid

EXERCISE 7.2

General section
1. a) If the length of the base of a square base pyramid is x unit and slant height is y unit,

write the formulae to find the lateral surface area and the total surface area of the
pyramid.
b) Each side of the base of an equilateral triangular base pyramid is p unit and slant
height is q unit. Write the formulae to find the L.S.A. and T.S.A. of the pyramid.
c) The length of the base of a square base pyramid is m unit and height of pyramid is
n unit. Write the formula to find it's volume.
d) If length of each side of the base of an equilateral triangular base pyramid is k unit
and height is h unit, write the formula to find it's volume.
2. a) Find the total surface area of a square-based pyramid in which the area of the base
is 48.5 cm2 and the area of each triangular face is 24.5 cm2.
b) The area of the base of a rectangular pyramid is 85.4 cm2. If the area of each
triangular face along the length and breadth is 28.5 cm2 and 23.2 cm2 respectively,
find the total surface area of the pyramid.
c) The length of the base of a square-based pyramid is 12 cm and its slant height is
15 cm. Find its total surface area.
d) The length of the base of an equilateral triangle-based pyramid is 8 cm and its slant
height is 12 cm. Find the total surface area of the pyramid.
3. a) The height of a square-based pyramid having the area of its base 81 sq. metres is
18 metres. Find the volume of the pyramid.
b) If the length of the base of a square-based pyramid is 12 cm and the height of the
pyramid is 8 cm, find the volume of the pyramid.
c) The length and the breadth of the base of a rectangular pyramid are 10 cm and
6 cm respectively. If the height of the pyramid is 15 cm, find its volume.
d) Find the volume of an equilateral triangle-based pyramid whose length of the base
is 20 cm and height is 24 cm.
e) A pyramid of height 15 cm has the base of an equilateral triangle of side 8 cm, find
the volume of the pyramid.

Creative section - A
4. Find the total surface area and volume of these pyramids:

a) b) 8cm c)
12 cm13 cm 10cm 12cm
10 cm 6cm 8cm

10 cm 12 cm 12 cm 15 cm

20 cm

5. Find the total surface area and volume of the following square-based pyramids.

a) A A O d) 13cm
S
P b) c) C
4 cm
8cm

10cm
SR D

O B O A 16cm B 10cm
Q 6 cm R P 12cm Q

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Mensuration (III): Prism and Pyramid

e) f) g) h)

P AO A

15cm
17cm

15cm
9cm
24cm
25cm
15cm 12cm
A DP S DC RQ

OM TO R A PB S B
BC Q P

6. The uppermost part of each of the following solid is a square-based pyramid. Find the
volume of the solids.

a) b) 15cm c) 10cm d)

12cm 16cm 22cm
6cm
6cm 6cm 6cm12cm12cm 9cm 9cm 16 cm
3cm 19 cm
3cm9 cm
e) The given figure is a solid object made up of a 8cm
pyramid and a prism. Find the volume of the solid
object.

f) The adjoining figure is a solid crystal formed by two pyramids 12 cm
with a common square base. If the heights of both of the
pyramids are the same, find the volume of the crystal. 3cm

Creative section - B

7. Find the total surface area of the following combined solids made up of the pyramid
and the prism in the given figures:

a) b) 8cm 13cmd) 12 cm

14cm 18cm c)

22cm
10cm
15 cm
6cm 6cm 12cm 10cm
12cm 10cm
10 cm
P

8. a) In the given square-based pyramid, the volume of the

pyramid is 1568 cm3 and half of the length of the side of DC

the base (OM) = 7 cm. Calculate the area of triangular faces O7cm M
B
of the pyramid. A

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Mensuration (III): Prism and Pyramid

b) In the given figure, if the length of a side of the base of the 12 cm C
pyramid having square base is 12cm and the volume of the E
pyramid is 384 cm3, find the total surface area of the pyramid.
D
c) In the given figure, the total surface area of the square-based
pyramid is 96 cm2 and the side of the square base is 6 cm.
Find the volume of the pyramid.

A 6 cm B

d) The adjoining figure is a square-based pyramid. If its slant 13 cm
height is 13 cm and its total surface area is 360 square cm, find 10 cm
the volume of the pyramid.

13cm

9. a) The adjoining solid object is formed with the combination of a
pyramid and a square-based prism. If the volume of the solid
object is 900 cu. cm, find the height of the prism.

b) In the adjoining solid, a square-based pyramid is situated on 22 cm
the top of a square-based cuboid. If the volume of the solid
is 4536 cm3, find the total surface area of the solid.

18 cm 18cm

c) If the total surface area of the combined solid made up of the 4 cm
pyramid and square-based prism in the adjoining figure is 6cm
240 cm2, find the volume of the solid.

6 cm

d) The solid given alongside is the combination of a 12 cm 10 cm
square-based pyramid and a square-based prism. If the total 12cm
surface area of the solid is 864 cm2, find the volume of the
solid.

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Mensuration (III): Prism and Pyramid

10. a) The given figure is a pillar of height 5 m and 60 cm × 60 cm 40 cm
bases. The top of the pillar has a combined pyramid of height
40 cm. Calculate the cost of painting the structure at the rate of 5m
Rs 200 per sq. m.
60cm
60 cm

b) The gate of a stadium has two pillars, each of height 10 ft. with four visible lateral
faces and 3 ft. × 3 ft. bases. The top of each pillar has combined pyramid of height
2 ft. If the combined structures of both pillars and pyramid are painted at the rate
of Rs 80 per sq. ft., calculate the total cost of painting.

7.5 Cone

A cone is like a pyramid with a circular base. A pyramid is the special case of a cone

where the base is a polygon. A cone can be thought of as a pyramid with an infinite

number of faces. P

Let's take a circular piece of paper with centre O. Cut off the sector A B
APB and join the edges AO and BO. In this way a pyramid with

a circular base is formed which is called a circular pyramid or a O

cone.

A cone has a curved surface and a circular base. O is the P (vertex)
centre of the circular base, OA is the radius, OP is the l (slant height)
vertical height of the cone and PA is the slant height.
h
OA

radius

7.6 Surface area and volume of cone

A cone is formed from the sector of a circle. So its curved surface is the surface of

the sector. A

1. The curved surface area of a cone = Area of the sector = Srl O
Where r is the radius of the circular base and l is the slant height.

2. The total surface area of a cone = curved surface area + area of circular base
= Sr l + Sr2

= Sr (r + l)

3. Volume of the cone = volume of the circular pyramid

= 1 Area of the circular base × height πrl
3 πr2

= 1 Sr2 h
3

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Mensuration (III): Prism and Pyramid

Worked-out examples

Example 1: From the given right circular cone, calculate it's curved 4 cm 5 cm
surface area, total surface area and volume.

Solution:
Here, height of the cone (h) = 4 cm and slant height (l) = 5 cm.

Radius of the circular base (r) = 52 – 42 = 25 – 16= 9 = 3 cm

Now, curved surface area of the cone = Srl

= 22 × 3 cm × 5 cm = 47.14 cm2
7
22
Also, total surface area of the cone = Sr (r + l) = 7 × 3 (3 + 5) = 75.43 cm2

And, volume of the cone = 1 Sr2h = 1 u 22 u 3 cm u 3 cm u 4 cm = 37.71 cm3
3 3 7

Example 2: If the total surface area of a cone is 704 cm2 and radius of its base is 7 cm,
find the volume of the cone.

Solution:
Here, the radius of the base of the cone (r) = 7cm,

The total surface area of the cone = 704 cm2

or, Sr ( r + l ) = 704 cm2

or, 22 × 7 (7 + l) = 704 cm2
7

or, l = 25 cm

Also, the vertical height of the cone (h) = l2 – r2

= 252 – 72 = 24 cm h 25cm
7cm
Now, the volume of the cone = 1 Sr2h
3

= 1 × 22 × 7 × 7 × 24cm3
3 7

= 1,232 cm3

So, the required volume of the cone is 1,232 cm3. Q 14 cm
OR
Example 3: If the volume of the given cone is 1848 cm3, and h
the radius of its base is 14 cm, find its curved
surface area. P

Solution:
Here, the radius of the base of the cone ( r ) = 14 cm

The volume of the cone = 1848 cm3

or, 1 Sr2h = 1848 cm3
3
1 22
or, 3 × 7 × 14 × 14 × h = 1848 cm3

or, h = 9 cm

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Mensuration (III): Prism and Pyramid

Also, the slant height of the cone (l) = h2 + r2

= 92 + 142 = 16.64 cm

Now, the curved surface area of the cone = Srl = 22 × 14 × 16.64 cm2 = 732.16 cm2
7

So, the required curved surface area of the cone is 732.16 cm2.

Example 4: Calculate the total surface area and the 12cm 10cm
volume of the pencil given in the figure.
80 cm
Solution:

Here, radius of the circular base of the cone = radius of
12
cylinder = r = 2 cm = 6 cm

? Height of the cone (h1) = 102 – 62 = 100 – 36 = 64 = 8 cm

Now, the curved surface area of the cone (A1)= Srl = 22 u 6 cm u 10 cm = 188.57 cm2
7

Also, the curved surface area of the cylinder (A2)= 2Srh

= 2 u 22 u 6 cm u 80 cm = 3,017.14 cm2
7

And, area of the circular base of the cylinder (A3)= Sr2 = 22 u 6 cm u 6 cm = 113.14 cm2
7

? Total surface area of the pencil= A1 + A2 + A3
= (188.57 + 3,017.14 + 113.14) cm2 = 3,318.85 cm2

Direct method

Total surface area of the pencil = Srl + 2Srh2 + Sr2

= Sr (l + 2h2 + r)

= 22 u 6 cm (10 + 2 u 80 + 6) cm
7
132
= 7 cm u 176 cm = 3,318.857 cm2

Again, volume of the cone (V1) = 1 u 22 u 6 cm u 6 cm u 8 cm = 301.71 cm3
3 7

Also, volume of the cylinder (V2) = Sr2h2 = 22 u 6 cm u 6 cm u 80 cm = 9,051.43 cm3
7

? Volume of the pencil = V1 + V2 = 301.71 cm3 + 9,051.43 cm3 = 9,353.14 cm3

Example 5: Find the volume of the given solid. 32cm

Solution: = 12 cm 12cm
Here, the height of the cone (h1) 7cm
? The height of the cylinder (h2)
Now, volume of the cone (V1) = 32 cm – 12 cm = 20 cm
Also, volume of the cylinder (V2)
= 1 Sr2h1 = 1 u 22 u 7 cm u 7 cm u 12 cm = 616 cm3
3 3 7
22
= Sr2h2 = 7 u 7 cm u 7 cm u 20 cm = 3,080 cm3

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Mensuration (III): Prism and Pyramid

And, volume of the hemisphere (V3) = 1 u 4 Sr3
2 3

= 1 u 4 u 22 u 7 cm u 7 cm u 7cm = 718.67 cm3
2 3 7

? Volume of the solid = V1 + V2 + V3 = 616 cm3 + 3,080 cm3 + 718.67 cm3 = 4,414.67 cm3

Example 6: Find the volume of the given solid object, A 25cm P
where AP = 25 cm and OP = 24 cm. O 24cm

Solution:

Here, the radius of the cone = radius of the hemisphere (r) A'

= 252 – 242

= 625 – 576 = 49 = 7 cm

Now, the volume of the solid = volume of the cone + volume of the hemisphere

= 1 Sr2h + 1 u 4 Sr3
3 2 3

= ( 1 u 22 u 7 u 7 u 24 + 2 u 22 u 7 u 7 u 7 ) cm3
3 7 3 7

= (1,232 + 718.67) cm3 = 1,950.67 cm3

Example 7: A cone is placed on a cylinder so that the base of the cone
covers exactly the base of the cylinder as shown in the figure.
The area of the base of the cylinder is 125 cm2 and the height 3 cm
of the cylinder is 3 cm. If the volume of the whole solid is
625 cm3, find the height of the solid.

Solution:

Let, h1 be the height of the cylinder and h2 be the height of the cone.

Here, area of the base of the cylinder = Sr2 = 125 cm2

Also, height of the cylinder (h1) = 3 cm

Now, volume of the solid = 625 cm3

or, Sr2h1 + 1 Sr2h2 = 625 cm3
3

or, Sr2 (h1 + 1 h2) = 625 cm3
3

or, 125 cm2 (3 + 1 h2) = 625 cm3
3

or, 9 + h2 = 5 cm
3

or, h2 = (15 – 9) cm

or, h2 = 6 cm

? Height of the solid = h1 + h2 = 3 cm + 6 cm = 9 cm

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Mensuration (III): Prism and Pyramid

Example 8: Find the volume of the solid given in the figure.

Solution: 6cm

Here, the radius of the common circular base of cones (r) = 6 cm
2
= 3 cm.
35cm
Let the height of one cone be h cm

? The height of the other cone = (35 – h) cm

Now, volume of the cone with height (h) = 1 Sr2h = 1 u 22 u3u3uh= 66h
3 3 7 7
1
Also, volume of the cone with height (35 – h) = 3 Sr2h

= 1 u 22 u 3 u 3 u (35 – h)
3 7
66 (35 – h)
= 7

? The volume of the solid = 66h + 66 (35 – h)
7 7
66h + 2,310 – 66h 2,310
= 7 = 7 = 330 cm3

So, the required volume of the solid is 330 cm3.

EXERCISE 7.3

General section

1. a) If a is the radius of the circular base and b is the slant height of a cone, write the
formula to find it's curved surface area.

b) If x is the radius of the base and y is the slant height of a cone, write the formula to
find it's total surface area.

c) The radius of the circular base of a cone of height y is x, write the formula to find
its volume.

2. a) The slant height of a right circular cone with radius 7 cm is 15 cm. Find its curved
surface area.

b) The radius of the circular base of a cone is 14 cm and its slant height is 20cm, find
the total surface area of the cone.

c) The vertical height of a right circular cone is 21 cm and the diameter of its circular
base is 14 cm. Find its volume.

3. a) The slant height of a cone is 5 cm and the radius of its base is 3 cm, find the volume
of the cone.

b) If the vertical height of a cone with 7 cm radius is 24 cm find the curved surface
area of the cone.

c) If the vertical height and the slant height of a cone are 48 cm and 50 cm respectively,
find the total surface area of the cone.

4. a) The curved surface area of a cone is 220 cm2. If the diameter of its circular base is
14 cm, find its slant height.

b) If the total surface area of a cone is 594 cm2 and its slant height is 20 cm, find the
radius of its circular base.

c) The volume of a cone is 23.1 cm3. If the radius of its base is 2.1 cm, find the vertical
height of the cone.

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Creative section - A
5. a) Find the curved surface area of the following right circular cones.

(i) 7cm (ii) O (iii)

8 cm
12 cm 4 cm O P
5 cm B

BC 6cm
Q
b) Find the total surface area of the following cones. (iii)
(i) (ii) Q A

9cm20cm 24cm
O 12cm P 25cm

14 cm P

c) Find the volume of the following right circular cones.

(i) (ii) O 9cm P (iii) O

15cm 15cm
13cm
14cm 12cm

A PA

d) If the circumference of the circular base of a cone is 44 cm and its height is 30 cm,
find its volume.

e) The perimeter of the circular base of a cone is 88 cm. If its slant height is 20 cm,
find its total surface area.

6. Find the curved surface area, the total surface area and the volume of each of the

following right circular cones.

a) b) c) d)

25cm 48cm 8cm 35cm
12cm
50cm

14cm 25cm c) 10cm 132 cm

Creative section - B 20cm
7. Find the volume of each of the following solids. 12cm

a) b) 30cm

8cm 5cm

20cm 54cm

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Mensuration (III): Prism and Pyramid

d) e) 10cm f)
8cm
A 7cm P 35cm

O 49cm
25cm
h) i)
g)
15cm
12cm 9cm 13cm
28cm
24cm
14cm

10cm 45cm

8. Find the total surface area of each of the following solids.

a) b) c)
12cm 80cm
10cm 88cm

18cm 15.5cm
12cm
42cm 18cm

d) e) f)

24cm 3.5cm 14cm
48cm
31cm
52.8cm

9. a) The total surface area of a cone is 3696 cm2. If the diameter of its circular base is
42 cm, find the volume of the cone.

b) The total surface area of a right circular cone with slant height 25 cm is 704 cm2.
Find the volume of the cone.

c) The curved surface area of a right circular cone is 2200 cm2. If the circumference of

its circular base is 88 cm, find (i) its total surface area , (ii) its volume.

10. a) The volume of a right circular cone with radius 7 cm is 1232 cm3. Find, (i) the
curved surface area and (ii) the total surface area of the cone.

b) A right circular cone is vertically 48 cm high. If the volume of the cone is 9856 cm3,
find its curved surface area and the total surface area.

c) 150 people can be accommodated inside a conical tent. If 3 m2 space and 10 m3 air
is required for each person, find the height of the tent.

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Mensuration (III): Prism and Pyramid

11. a) A tent of height 33 m is in the form of a right circular 33m
cylinder of diameter 42 m and height 5 m surmounted by a
right circular cone of the same diameter. Find the total area 5m
of cloths required to cover the tank. Also, find the cost of
cloths at Rs 125 per sq. m.

42m

5.6m

b) A tent is of the shape of a right circular cylinder of height 4.4 m 4.4m
with a conical top of the vertical height 5.6 m. If the perimeter
of its circular base is 26.4 m, find the total surface area of the
tent.

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

b) In the given figure, the base area of the cylinder is 64 cm2 and the 4 cm
height of the cylinder is 4 cm. If the volume of the whole solid is
320 cm3, find the height of the solid.

13. a) The circumference of the circular base of a cylindrical pillar is 33 ft. and it's height
is 10 ft. The top of the pillar has a combined cone of height 7 ft. Find the cost of
painting the combined structure of the pillar at the rate of Rs 40 per sq. ft.

b) The gate of a stadium has two cylindrical pillars, each of height 5 m and the
circumference of each base is 22 m. The top of each pillar has a combined cone of
height 1.2 m. If the combined structures of both pillars and cones are painted at the
rate of Rs 100 per sq. m, find the total cost of painting the pillars.

7.7 Mensuration in household activities

The figure given alongside represents a kitchen D 12 m C
garden of a family. Mother wants to fence the garden 10 m B
to protect vegetables grown inside it. She discuss 8m
with her daughter, who is studying in grade 10, about A 15 m
the cost estimation for fencing it. She asked her
daughter to estimate the cost for (i) 2 round fencing,
(ii) 4 round fencing, and (iii) 5 round fencing.

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Mensuration (III): Prism and Pyramid

Here is the daughter's estimation:
The perimeter of the garden = AB + BC + CD + DA

= 10 m + 15 m + 12 m + 8 m
= 45 m
She found the cost of per metre of fencing wire in the market is Rs 60.
She also estimated the labour charge for this activities. The table given below shows
her cost estimation.

No. of Length of fencing Cost of fencing wire Labour Total cost
rounds wire cost
90 × Rs 60 = Rs 5,400 Rs 8,000
2 2 × 45 m = 90 m 180 × Rs 60 = Rs 10,800 Rs 2,600 Rs 14,000
225 × Rs 60 = Rs 13,500 Rs 17,000
4 4 × 45 m = 180 m Rs 3,200

5 5 × 45 m = 225 m Rs 3,500

The daughter recommended for 4 rounds of fencing and the mother approved her
idea.
Similarly, can you please help your family for the following household activities.
• Cost of estimation of your kitchen garden
• Cost of estimation of carpeting your rooms of your house
• Cost of estimation of plastering the walls of the rooms of your house
• Cost of estimation of colouring the walls the room of your house
• Cost of estimation of building compound walls in your house
• Cost of estimation of building an underground water tank

Worked-out examples

Example 1: The figure alongside is a compound of a house. C 13 m B
A
Estimate the cost of fencing the compound with 8m
17 m
5 rounds of fencing wire.

Rate of cost of per metre of wire = Rs 50 D
Labourer wages = Rs 800 per labour per day 10 m

The fencing work requires 2 days for 2 labours.

Solution:

Here, the perimeter of the compound = AB + BC + CD + DA

= 10 m + 17 m + 13 m + 8 m

= 48 m

The length of wire required for 5 rounds = 5 × 48 m = 240 m

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Mensuration (III): Prism and Pyramid

The cost of the wire = 240 × Rs 50 = Rs 12,000

The wages of 2 labourers for 1 day = 2 × Rs 800 = Rs 1,600

The wages of 2 labourers for 2 days = 2 × Rs 1,600 = Rs 3,200

? The total cost of estimation of fencing the compound = Rs 12,000 + Rs 3,200

= Rs 15,200

Example 2: The dimension of a rectangular room is 7 m long, 6 m broad and 5 m high.

a) Estimate the cost of carpeting its floor at Rs 110 per sq. m. with Rs 1,200
labourer charge.

b) Estimate the cost of colouring its walls and ceiling at Rs 150 per sq. m.

Solution:

Here, the length of the room (l) = 7 m

the breadth of the room (b) = 6 m

the height of the room (h) = 5 m

a) The area of the floor = l × b = 7 m × 6 m = 42 m2

the cost of the carpet = Area of floor × Rate

= 42 m2 × Rs 110 per m2

= Rs 4,620

The cost of carpeting the floor with labourer charge = Rs 4,620 + Rs 1,200

= Rs 5,820

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

= 2 × 5 (7 + 6)

= 10 × 13

= 130 m2

The area of the ceiling = l × b

= 7 m × 6 m = 42 m2

The total area of 4 walls and ceiling = 130 m2 + 42 m2 = 172 m2

The cost of colouring walls and ceiling = Area × Rate

= 172 m2 × Rs 150 per m2

= Rs 25,800

Example 3: The figure alongside is a

rectangular compound wall of 10 cm
a house. The dimensions of the

wall is 20 m long, 15 m broad, 4 15 m

m high and 10 cm thick. Estimate

the cost of building the walls 4m

by using bricks of dimensions

20 cm × 10 cm × 5 cm and rate

of cost of bricks is Rs 18,000 per 20 m

1000 bricks. Labourer wages is

Rs 650 per labourer per day and 2 labourers for 5 days.

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Mensuration (III): Prism and Pyramid

Solution:
Here, the outer length of the wall (l) = 20 m

the inner breadth of the wall = 15 m

? the thickness of the wall (d) = 10 cm = 0.1 m

? the outer breadth of the wall (b) = inner breadth + 2d = 15 m + 2 × 0.1 m = 15.2 m

Also, the height of the wall (h) = 4 m

Now,cross-sectional area of the wall (A) = 2d(l + b – 2d)

= 2 × 0.1(20 + 15.2 – 2 × 0.1) = 7 m2

? Volume of the wall = cross-sectional Area × height
= 7 m2 × 4 m = 28 m3

= 28 × 100 × 100 × 100 cm3

Now, volume of each brick = 20 cm × 10 cm × 5 cm

volume of wall
? Number of bricks required = volume of each brick

28 × 100 × 100 × 100 cm3
= 20 × 10 × 4 cm3
= 28,000 bricks

Again, the cost of 1,000 bricks = Rs 18,000

the cost of 1 brick Rs 18,000
the cost of 28,000 bricks = 1,000

Rs 18,000
= 1,000 × 28,000
= Rs 5,04,000

Also, labourer wages for 2 labourers per day = 2 × Rs 650

= Rs 1,300
Labourer wages for 2 labourers for 5 days = 5 × Rs 1,300

= Rs 6,500
? The total estimation of building the compound wall = Rs 5,04,000 + Rs 6,500

= Rs 5,10,500

Example 4: A man wants to build an underground water tank of capacity 10,000 litre.
If length and breadth of the tank will be 2.5 m and 2 m respectively, how
Solution: deep is the tank?
Here,
the capacity of the tank = 10,000 l
? we know that, 1,000 l = 1 m3
10,000 l = 10 m3

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Mensuration (III): Prism and Pyramid

So, the volume of the tank = 10 m3

Also, the area of the base of the tank = l × b = 2.5 m × 2 m = 5 m2

Now, volume of the tank 10 m3
depth of the tank = Area of the base = 5 m2 = 2 m

So, the tank will be 2 m deep.

EXERCISE 7.4

General section

1. a) If A is the area of the floor of a room and R be the rate of cost of carpeting the floor,
write the formula to find the total cost (T) of carpeting the floor.

b) If A be the total area of a 4 walls and ceiling of a room and T be the total cost of
colouring the walls and ceiling of the room, write the formula to find the rate of the
cost (R) of colouring.

c) If R be the rate of cost and T be the total cost of plastering the floor of a room, write
the formula to find the area (A) of the floor.

2. a) If V be the volume of a wall and v be the volume of a brick, write the formula to find
the number of bricks (N) required to build the wall.

b) If V be the volume of a wall and N be the number of bricks required to build the
wall, write the formula to find the volume of each brick.

c) If v be the volume of a brick and N be the number of bricks required to build a wall,
write the formula to find the volume of the wall.

3. If the internal volume of a water tank is 1 m3, it can hold 1,000 litre of water.

a) Find the capacity of each water tank that has internal volume

(i) 1.5 m3 (ii) 2.7 m3 (iii) 4.8 m3 (iv) 10 m3 (v) 15.5 m3

b) Find the internal volume of each water tank that has the capacity of

(i) 2,000 l (ii) 3,500 l (iii) 7,400 l (iv) 12,800 l (v) 20,000 l

4. a) The area of the floor of a room is 35 m2. Find the cost of carpeting the floor at the
rate of Rs 90 per sq. m.

b) A room is 8 m long and 5 m broad. Find the cost of carpeting its floor at Rs 75 per
sq. m.

c) A rectangular room is 9 m long, 6 m broad and 5 m high. Find the cost of colouring
its 4 walls and ceiling at the rate of Rs 150 per sq. m.

5. a) The area of square base of a water tank is 4 m2 and the height of the tank is 2.5 m.
Find the capacity of the tank in litre.

b) The internal length of a square base water tank is 2 m and its height is 1.5 m. How
many litres of water does it hold when its is full?

c) The area of the circular base of a cylindrical water tank is 6.16 m2 and it is 1.5 m
high. What is the capacity of the tank in litre?

d) The internal diameter of a cylindrical water tank is 2 m and it is 3.5 m high. How
many litres of water does it hold when it is full?

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Mensuration (III): Prism and Pyramid

Creative section A

6. a) The adjoining figure represents the kitchen 10 m 20 m

garden of a house. Estimate the cost of fencing 16 m D
C C
the garden with 7 rounds of fencing wire. Rate of B 14 m
cost per metre of wire is Rs 30. Labourer wages is

Rs 650 per labourer per day and the completion of

the fencing wire requires 3 days for 2 labourers.

b) The figure alongside is a compound of a D Q
house. The length of the diagonal AC = 40 m,
DP = 25 m and BQ = 10 m. Estimate the cost P B
of plastering the compound at the rate of A
Rs 75 per sq. m.

7. a) The dimension of a rectangular room is 10 m long, 8 m broad, and 5 m high.
(i) Estimate the cost of carpeting its floor at Rs 90 per sq.m with Rs 600 labourer
charge.
(ii) Estimate the cost of colouring its walls and ceiling at Rs 105 per sq.m.

b) A rectangular room is 8 m long, 7 m broad and 4.5 m high. Estimate the cost of
plastering its floor, walls and ceiling at the rate of Rs 110 per sq.m.

8. The adjoining figure is a rectangular 12 cm
compound wall of a house. Estimate the cost
of building the walls by using bricks of size 25 m
20 cm × 8 cm × 5 cm. The rate of the cost of
bricks is Rs 1,750 per 1,000 bricks. Labourer 5m
wages is Rs 750 per labourer per day and
3 labourers are required for 4 days.

32 m

Project Work

9. a) Measure the length and breadth of your school compound. Calculate its perimeter
and area. Then,
(i) Estimate the cost of fencing the compound according to the present cost of
materials and labourer wages.
(ii) Estimate the cost of plastering the compound according to the present cost of
materials and labourer wages.

b) Measure the length, breadth and height of your classroom. Calculate the area of
(i) floor, (ii) 4 walls and (iii) ceiling. Estimate the cost of plastering the floor, walls
and ceiling at the usual rate of cost in your locality. Also, estimate the cost of
colouring the walls and ceiling.

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Objective Questions

Tick the correct alternatives.

1. Which one of the following formulae is known as Heron’s formula to calculate the area
of the triangle?

(A) 1 b×h (B) 3 a2 (C) s(s – a) (s – b) (s – c) (D) b 4a2 – b2
2 4 4

2. In any triangle, if the base is doubled and the altitude is trebled, what is the ratio of areas
of original triangle to new triangle?

(A) 2:3 (B) 1:6 (C) 6:1 (D) 1:5

3. The radii of two cylinders are in the ratio 1:2 and heights are in the ratio 2:3. The ratio of
their volumes is:

(A) 1:6 (B) 1:9 (C) 1:3 (D) 2:9

4. In a cylinder-X, the radius of base and height are double of the cylinder-Y. If the surface
area of cylinder-Y is A, the surface area of cylinder-X is

(A) A (B) 2A (C) 4A (D) 8A

5. If a cylindrical log having equal height and radius displaces 134.75 litre of water, what is
the radius of the log?

(A) 7cm (B) 14 cm (C) 21 cm (D) 35 cm

6. How much wood should be removed from a wooden cube of each edge 21 cm for making
it the greatest sphere?

(A) 9261cm3 (B) 4851cm3 (C) 4410 cm3 (D) 1260cm3

7. The surface area of a volley ball having diameter 20 cm is..

(A) 100π cm2 (B) 200π cm2 (C) 400π cm2 (D) 1600π cm2

8. Three zinc spheres of radii 3cm, 4cm and 5 cm are melted to form a new sphere. The
radius of new sphere is:

(A) 6 cm (B) 8 cm (C) 12 cm (D) 16 cm

9. If a metallic hemisphere of volume 25π cubic centimeter is melted to form a cylindrical
wire of 1 metre long, what is the diameter of the wire?

(A) 0.25 cm (B) 0.5 cm (C) 1 cm (D) 2 cm

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10. In a conical carrot, if the slant height is 7 cm and radius is 2 6 cm, its vertical height is:

(A) 4 cm (B) 4.5cm (C) 5 cm (D) 5.2cm

11. The vertical angle of a cone having diameter of its base 10 cm is 600, what is its slant
height?

(A) 5cm (B) 10 cm (C) 10 3 cm (D) 20 3 cm

12. 50 coins each of diameter 1.4 cm and height 20 mm are piled to form a right circular
cylinder. What is the volume of the cylinder?

(A) 14.5 cm3 (B) 15.4 cm3 (C) 41.5 cm3 (D) 51. 4 cm3

13. The area of rectangular surfaces of a triangular prism having right angled on its base is
84 cm2. If the sides containing the right angle are 3cm and 4 cm, what is the height of the
prism?

(A) 7 cm (B) 5 cm (C) 12 cm (D 20 cm

14. A toy is in the form of a cone surmounted by a hemisphere of equal radius. If the height
of the toy is 7cm, the height of conical part is 4 cm then the slant height of the conical
part is

(A) 3 cm (B) 4 cm (C) 5 cm (D) 11 cm

15. Ajay constructed a cuboidal water tank having the internal length 1.5 m, breadth 1.2 m
and height 1 m, how many litres of water can it hold when it is full?

(A) 1200 l (B) 1500 l (C) 1800 l (D) 2000 l

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Unit Highest Common Factor (H.C.F.) and
Lowest Common Multiples (L.C.M.)
8

8.1 H.C.F. of algebraic expressions

Let's take any two monomial expressions 4a2 and 6a3.
Here, all the possible factors of 4a2 are a, a2, 2a, 2a2, 4a, 4a2.
Also, all the possible factors of 6a3 are a, a2, a3, 2a, 2a2, 2a3, 3a, 3a2, 3a3.
Now, the common factors of 4a2 and 6a3 are a, a2 2a and 2a2. Among these common
factors, 2a2 is the highest one.

So, the Highest Common Factor (H.C.F.) of 4a2 and 6a3 is 2a2.

Thus, to find the H.C.F. of the monomial expressions, at first we should find the
H.C.F. of the numerical coefficients. Then, the common variable with the least power
is taken as the H.C.F. of the expressions. For example,

4 is the H.C.F. of 12 and 8.

In 12x4y3 and 8x3y2 x3 is the H.C.F. of x4 and x3.

y2 is the H.C.F. of y3 and y2.
So, the H.C.F. of 12x4y3 and 8x3y2 is 4x3y2.

We can find the H.C.F. of polynomial expressions either by the process of factorisation
or by division process. In case of factorisation process, the given polynomials are to
be factorised, then a common factor or the product of common factors is obtained as
their H.C.F.

Worked-out examples

Example 1: Find the H.C.F. of x4y – x2y3 and 2x3y – x2y2 – xy3
Solution:
Here, the 1st expression = x4y – x2y3

= x2y (x2 – y2) = x2y (x + y) (x – y)
The 2nd expression = 2x3y – x2y2 – xy3

= xy (2x2 – xy – y2)

= xy (2x2 – 2xy + xy – y2) = xy [2x(x – y) + y (x – y)]

= xy (x – y) (2x + y)

? The H.C.F. = xy (x – y)

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

Example 2: Find the H.C.F. of x5y + 4x3y + 16xy, x4y – 8xy and 2x3y + 4x2y + 8xy.
Solution:
Here, the 1st expression = x5y + 4x3y + 16xy

= xy (x4 + 4x2 + 16)
= xy [(x2)2 + 42 + 4x2]
= xy [(x2 + 4)2 – 2.x2.4 + 4x2]
= xy [(x2 + 4)2 – (2x)2]
= xy (x2 + 2x + 4) (x2 – 2x + 4)

The 2nd expression = x4y – 8xy = xy (x3 – 8)
= xy (x3 – 23)
= xy (x – 2) (x2 + 2x + 4)

The 3rd expression = 2x3y + 4x2y + 8xy
= 2xy (x2 + 2x + 4)

? The H.C.F. = xy (x2 + 2x + 4)

Example 3: Find the H.C.F. of x2 – 12x – 28 + 16y – y2, x2 + 2x – y2 + 2y and
x2 – y2 + 4y – 4.

Solution:
Here, the 1st expression = x2 – 12x – 28 + 16y – y2

= x2 – 2.x.6 + 62 – 62 – 28 + 16y – y2
= (x – 6)2 – (64 – 16y + y2)
= (x – 6)2 – (82 – 2.8.y + y2)
= (x – 6)2 – (8 – y)2
= (x – 6 + 8 – y) (x – 6 – 8 + y)
= (x – y + 2) (x + y – 14)

The 2nd expression = x2 + 2x – y2 + 2y
= x2 – y2 + 2(x + y)
= (x + y) (x – y) + 2(x + y)
= (x + y) (x – y + 2)

The 3rd expression = x2 – y2 + 4y – 4
= x2 – (y2 – 2. y. 2 + 22)
= x2 – (y – 2)2
= (x + y – 2) (x – y + 2)

? The H.C.F. = (x – y + 2)

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

EXERCISE 8.1

General section

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

a) 2x2 (x + 2) (x – 2) and 4x(x + 2) (x + 3)

b) 4xy2 (x – 1) (x + 2) and 6x2y (x – 1) (x – 4)

c) 6a2b2(a – b) (2a + 3b) and 9a3b3(2a + 3b) (a + b)

d) 12a2b3 (a – 3b) (a + b – 2) and 16a3b2 (a + 3b) (a + b – 2)

e) (p + q) (p – q), (p – q) (p2 + pq + q2) and (p – q) (p + q) (p2 + q2)

2. Find the H.C.F. of the following expressions:

a) a2 – b2 and a3 + b3 b) a2 – 2a and a4 – 8a

c) x2 – 9 and 3x + 9 d) x2 – y2 and x2 + 2xy + y2

e) 4x2 – 100 and 4x + 20 f) 4x3 + 8x2 and 5x3 – 20x

g) a2 – b2 – 2a + 1 and a2 – ab – a h) x3 – 8y3 and x2 + 2xy + 4y2

i) p4 – 16 and p2 – p – 6 j) x4 + 4y4 and 2x3y + 4xy3 + 4x2y2

k) a3 – 1 and a4 + a2 + 1 l) 16x4 – 4x2 – 4x – 1, 8x3 – 1

Creative Section

3. Find the H.C.F. of the following expressions:

a) a2 – 4, a3 + 8, a2 + 5a + 6
b) x2 – 9, x3 – 27, x2 + x – 12
c) 4x2 – 9, 2x2 + x – 3, 8x3 + 27
d) 3x2 – 8x + 4, 2x2 – 5x + 2, x4 – 8x
e) 5a3 – 20a, a3 – 3a2 – 10a, a3 – a2 – 2a + 8
f) m3 – m2 – m + 1, 2m4 – 2m, 3m2 + 3m – 6
g) (a + b)2 – 4ab, a3 – b3, a2 + ab – 2b2
h) x3 – 64y3, x2 – 6xy + 8y2, x2 – 16y2
i) 4x4 + 16x3 – 20x2, 3x3 + 14x2 – 5x, x4 + 125x
j) x3 – 1, x4 + x2 + 1, x3 + 1 + 2x2 + 2x
k) 8x3 + 27y3, 16x4 + 36x2y2 + 81y4, 4x3 – 6x2y + 9xy2
l) x3y + y4, x4 + x2y2 + y4, 2ax3 – 2ax2y + 2axy2
m) a2 + 2ab + b2 – c2, b2 + 2bc + c2 – a2, c2 + 2ca + a2 – b2
n) 9x2 – 4y2 – 8yz – 4z2, 4z2 – 4y2 – 9x2 – 12xy, 9x2 + 12xz + 4z2 – 4y2
o) x2 – 18x – 19 + 20y – y2, x2 + x – y2 + y, x2 – y2 + 2y – 1

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

8.2 L.C.M. of algebraic expressions

Let's take any two monomial expressions x2 and x3.
A few multiplies of x2 are x2, x3, x4, x5, x6,.....
A few multiplies of x3 are x3, x4, x5, x6, x7, ....
Here, the common multiplies are x3, x4, x5, x6, ...
Among these common multiplies, x3 is the lowest one.
So, the Lowest Common Multiple (L.C.M.) of x2 and x3 is x3.
Thus, to find the L.C.M. of the monomial expressions, at first we should find the
L.C.M. of the numerical coefficients. Then, the common variable with the highest
power is taken as the L.C.M. of the expressions. For Example,

24 is the L.C.M. of 6 and 8.

In 6x4y2 and 8x3y3 x4 is the L.C.M. of x4 and x3.

y3 is the L.C.M. of y2 and y3.

So, the L.C.M. of 6x4y2 and 8x3y3 is 24x4y3.
In case of polynomial expressions, their L.C.M. is obtained by the process of
factorisation. By this process the product of common factors and the factors which or
not common is taken as the L.C.M. of the polynomials

Worked-out examples

Example 1: Find the L.C.M. of 6x6 + 6x4 + 6x2 and 4x6 – 4x3.

Solution:

Here, the 1st expression = 6x6 + 6x4 + 6x2

= 6x2 (x4 + x2 + 1)

= 6x2 [(x2)2 + 12 + x2]

= 6x2 [(x2 + 1)2 – 2x2 + x2]

= 6x2 [(x2 + 1)2 – x2]

= 6x2 (x2 + x + 1) (x2 – x + 1)

The 2nd expression = 4x6 – 4x3

= 4x3 (x3 – 1)

= 4x3 (x – 1) (x2 + x + 1)

? The L.C.M.= 12x3 (x2 + x + 1) (x – 1) (x2 – x + 1)

= 12x3 (x3 – 1) (x2 – x + 1)

Example 2: Find the L.C.M. of x2 + 2xy + y2 – z2, y2 + 2yz + z2 – x2 and z2 + 2xz + x2 – y2.

Solution:

Here, the 1st expression = x2 + 2xy + y2 – z2

= (x + y)2 – z2

= (x + y + z) (x + y – z)

The 2nd expression = y2 + 2yz + z2 – x2

= (y + z)2 – x2

= (x + y + z) (y + z – x)

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

The 3rd expression = z2 + 2zx + x2 – y2
= (z + x)2 – y2 = (x + y + z) (x – y + z)

? The L.C.M.=(x + y + z) (x + y – z) (y + z – x) (x – y + z)

Example 3: Find the L.C.M. of a3 – 2a2b + 2ab2 – b3, a3 + b3, a4 – b4.

Solution:
Here, the 1st expression = a3 – 2a2b + 2ab2 – b3

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

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

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

The 3rd expression = a4 – b4

= (a2)2 – (b2)2

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

? The L.C.M. = (a – b) (a + b) (a2 + b2) (a2 – ab + b2)

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

EXERCISE 8.2

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

a) 3x (x +1) (x – 1) and 2x2 (x – 1) (x + 3)
b) 4x3 (x – 3) (x + 2) and 6x2 (x + 2) (x + 3)
c) 8a2b(a – b) (a2 + ab + b2) and 12ab2 (a – b) (a + b)
d) (x + 2) (x + 3), (x + 3) (x – 2), (x – 2)(x – 3)
e) (a – 3) (a – 4), (a – 4),(a – 5), (a – 5) (a – 3)

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

a) 3x2 + 6x, 2x3 + 4x2 b) ax2 + ax, ax2 – a c) 4x5y4 + 2x4y5, 10x4y3 + 5x3y4
f) a3 – b3, a2 + ab + b2
d) x2 – xy, x3y – xy3 e) 4x2 – 2x, 8x3 – 2x i) a4b – ab4, a4b2 – a2b4
l) 6x2 – x – 1, 54x4 + 2x
g) x2 + 5x + 6, x2 – 4 h) x2 – 9, 3x3 + 81

j) x4 + x2y2 + y4, x3 – y3 k) a4 + a2b2 + b4, a3 + b3

Creative Section

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

a) a2 – 4, a3 – 8, (a + 2)2 b) (a – 3)2, a2 – 9, a3 + 27

c) a3 – 2a2 + a, a3 + a2 – 2a, a3 – 4a d) x4 – y4, x2 – y2, x3 – y3

e) 4x3 – 10x2 + 4x, 3x4 – 8x3 + 4x2, x4 – 8x f) x3 – 9x, x4 – 2x3 – 3x2, x3 – 27

g) a3 – 4a, a4 – a3 – 2a2, a3 – 8 h) a4 + a2 + 1, a3 – 1, a3 – a2 + a

i) a2 + 2ab + b2 – c2, b2 + 2bc + c2 – a2, c2 + 2ca + a2 – b2

j) x3 – 2x2y + 2xy2 – y3, x4 – y4, x3 + y3

k) x2 + 3x + 2, x2 + 5x + 6, x2 + 4x + 3

l) 2x3 + 2x2 – 12x, 6x3 – 6x2 – 72x, 4x3 – 24x2 + 32x

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Unit Simplification of Rational Expressions

9

9.1 Rational Expressions - review

x , x2 , 4a2b , a+b , etc. are a few examples of the rational expressions. However,
2 3 5 a–b

a+b is not a rational expression. Rational expressions can be expressed in the
a–b

form p , where q z 0.
q

9.2 Simplification of rational expressions

In case of addition and subtraction of rational expressions with a common
denominator, we should simply add or subtract the numerators. Then, the sum is
reduced to its lowest terms.

For example,

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

In case of addition and subtraction of rational expressions with unlike denominators,
at first, we should find the L.C.M. of the denominators. Then, the L.C.M. is divided
by each denominator and the quotient is multiplied by the corresponding numerator
as like the simplification of unlike fractions in arithmetic.

Let's learn the process of simplification from the following illustrations.

Worked-out examples

Example 1: Simplify 1 – 1
(x – y)(x – z) (z – x) (y – z)

Solution:

1 – 1 = 1 – 1
(x – y)(x – z) (z – x)(y – z) (x – y)(x – z) – (x – z)(y – z)

= 1 + 1
(x – y)(x – z) (x – z)(y – z)

= y–z+x–y
(x – y)(x – z) (y – z)

= x–z
(z – y)(x – z) (y – z)

= 1 Answer
(x – y) (y – z)

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Simplification of Rational Expressions

Example 2: Simplify (a – 2 – 3) + 2 + 1
2) (a (a – 1) (3 – a) (1 – a) (2 – a)

Solution:

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

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

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

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

Example 3: Simplify 1 +1 + 1
x2 – 5x + 6 x2 – 3x + 2 x2 – 8x + 15

Solution:

1 + 1 + 2
x2 – 5x + 6 x2 – 3x + 2 x2 – 8x + 15

= 1 + 1 + 2
x2 – 3x – 2x + 6 x2 – 2x – x + 2 x2 – 5x – 3x + 15

= 1 + 1 + 2
x(x – 3) – 2(x – 3) x(x – 2) – 1(x – 2) x(x – 5) – 3(x – 5)

= (x 1 – 2) + 1 + 2
– 3) (x (x – 2) (x – 1) (x – 5) (x – 3)

= x – 1+ x – 3 + 2
(x – 3) (x– 2) (x – 1) (x–5) (x – 3)

= 2(x – 2) + 2
(x – 3) (x– 2) (x – 1) (x–5) (x – 3)

= 2(x – 5) + 2(x – 1) = 2x – 10 + 2x – 2
(x – 3) (x– 1) (x – 5) (x – 3) (x– 1) (x – 5)

= 4(x – 3) = 4 Answer
(x – 3) (x– 1) (x – 5) (x– 1) (x – 5)

Example 4: Simplify x + y + 2xy
Solution: (x – y) (x + y) y2 – x2

x + y + 2xy
(x – y) (x + y) y2 – x2
x y 2xy
= (x – y) + (x + y) – x2 – y2

= x(x + y) + y(x – y) – 2xy
(x – y) (x + y) x2 – y2

= x2+xy + xy – y2 – 2xy = x2+2xy – y2 – 2xy = x2 – y2 = 1 Answer
x2 – y2 x2 – y2 x2 – y2 x2 – y2

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Simplification of Rational Expressions

Example 5 : Simplify 1 – 1 – 2p
1 – p + p2 1 + p + p2 1 – p2 + p4

Solution:

1 – 1 p2 – 1 + 1 p2 – 2p = 1 + p2 + p – (1 + p2 – p) – 2p
p+ p+ 1 – p2 + p4 (1 + p2 – p) (1 + p2 + p) 1 + p4 – p2

= 1 + p2 + p – 1 – p2 + p – 2p
(1 + p2)2 – p2 1 + p4 – p2

= 2p – 2p
1 + 2p2 + p4 – p2 1 + p4 – p2

= 1 + 2p p2 – 2p
p4 + 1 + p4 – p2

= 2p(1 + p4 – p2) – 2p(1 + p4 + p2)
(1 + p4 + p2) (1 + p4 – p2)

= 2p + 2p5 – 2p3 – 2p – 2p5 – 2p3
(1 + p4)2 – (p2)2

= –4p3 = –4p3 Answer
1 + 2p4 + p8 – p4 1 + p4 + p8

Example 6 : Simplify y–2 + y2 y +2 4 – 16
y2 – 2y + 4 + 2y + y4 + 4y2 + 16

Solution:

y–2 + y2 y +2 4 – y4 + 16 + 16
y2 – 2y + 4 + 2y + 4y2

= (y – 2) (y2 + 2y + 4) + (y + 2) (y2 – 2y + 4) – 16
(y2 – 2y + 4) (y2 + 2y + 4) y4 + 4y2 + 16

= y3 – 23 + y3 + 23 – 16
[(y2 + 4) – 2y] [(y2 + 4) + 2y] y4 + 4y2 + 16

= (y2 + 2y3 (2y)2 – 16
4)2 – y4 + 4y2 + 16

= 2y3 – 16
y4 + 8y2 + 16 – 4y2 y4 + 4y2 + 16

= 2y3 – 16
y4 + 4y2 + 16 y4 + 4y2 + 16

2y3 – 16 2(y3 – 8)
= y4 + 4y2 +16 = (y2 – 2y +4) (y2 + 2y + 4)

= 2(y – 2) (y2 + 2y + 4) = 2(y – 2) Answer
(y2 – 2y +4) (y2 + 2y + 4) y2 – 2y +4

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Simplification of Rational Expressions

Example 7 : Simplify x2 – (y – z)2 + y2 – (x – z)2 + z2 – (x – y)2
(x + z)2 – y2 (x + y)2 – z2 (y + z)2 – x2

Solution:

x2 – (y – z)2 + y2 – (x – z)2 + z2 – (x – y)2
(x + z)2 – y2 (x + y)2 – z2 (y + z)2 – x2

= (x + y – z) (x – y + z) + (y + x – z) (y – x + z) + (z + x – y) (z – x + y)
(x + y + z) (x – y + z) (x + y + z) (x + y – z) (x + y + z) (y + z – x)

= x+y–z + y–x+z + z+x–y
x+y+z x+y+z x+y+z

= x+y–z+y–x+z+z+x–y
x+y+z

= x+y+z = 1 Answer
x+y+z

Example 8 : Simplify a4 + a4 – 1 1 – 1
a2 + 1 a2 – 1 a2 + a2 – 1

Solution:

a4 + a4 – 1 – 1 = a4 – 1 + a4 – 1
a2 + 1 a2 – 1 a2 + 1 a2 – 1 a2 + 1 a2 + 1 a2 – 1 a2 – 1

= a4 – 1 + a4 – 1
a2 + 1 a2 – 1

= (a2 + 1) (a2 – 1) + (a2 + 1) (a2 – 1)
a2 + 1 a2 – 1

= a2 – 1 + a2 + 1 = 2a2 Answer

Example 9 : Simplify 2 + 1 1 + 3a + 1 a
1+a a– 1 – a2 + a3

Solution:

2 + 1 + 3a + a = 2 – 1 + 3a + 1 a
1+a a–1 1 – a2 1 + a3 1+a 1–a 1 – a2 + a3

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

= 1 + a
(1 + a) (1 – a) (1 + a) (1 – a + a2)

= 1 – a + a2 + a – a2
(1 + a) (1 – a) (1 – a + a2)

= (1 – 1 a3) Answer
a) (1+

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Simplification of Rational Expressions

Example 10: Simplify 1 – 2 1 + 1 – 2
x–1 2x – x+1 2x + 1

Solution:

1 – 2 1 + 1 – 1 = 1 + 1 – 2 – 2 1
x–1 2x – x+1 x+1 x–1 x+1 2x – 1 2x +

= x +1 + x – 1 – 2(2x + 1) + 2(2x – 1)
(x – 1) (x + 1) (2x – 1) (2x + 1)

= 2x – 4x + 2 + 4x – 2
x2 – 1 4x2 – 1

2x – 8x
= x2 – 1 4x2 – 1

= 2x(4x2 – 1) – 8x(x2 – 1)
(x2 – 1) (4x2 – 1)

= 8x3 – 2x – 8x3 + 8x = 6x Answer
(x2 – 1) (4x2 – 1) (x2 – 1) (4x2 – 1)

Example 11: Simplify x–1 – x+1 – 2x + 16x
x–2 x+2 x2 + 4 x4 + 16

Solution:

x–1 – x+1 – 2x 4 + 16x
x–2 x+2 x2 + x4 + 16

= (x – 1) (x + 2) – (x + 1) (x – 2) – 2x + 16x
(x – 2) (x + 2) x2 + 4 x4 + 16

= x2 + x – 2 – x2 + x + 2 – 2x + 16x
x2 – 4 x2 + 4 x4 + 16

= 2x – 2x + 16x
x4 – 4 x4 + 4 x4 + 16

= 2x(x2 + 4) – 2x(x2 – 4) + 16x
(x2 – 4) (x2 + 4) x4 + 16

= 2x3 + 8x – 2x3 + 8x + 16x
x4 – 16 x4 + 16

= 16x 16x
x4 – 16 + x4 + 16

16x(x4 + 16) + 16x(x4 – 16)
= (x4 – 16) (x4 + 16)

= 16x5 + 256x + 16x5 – 256x = 32x5 Answer
x8 – 256 x8 – 256

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Simplification of Rational Expressions

EXERCISE 9.1

General section

1. Simplify:

a) a+ 1 + a2 – 1 b) p+ 2 – p– 2
a–1 a+1 p–2 p+2

c) 4x – x+ 1 d) x– 4 – x– 6
x2 – 1 x–1 x+6 x+4

e) a2 + ab + b2 + a2 – ab + b2 f) x–2 – x+1
a+ b a–b x2 – 1 x2 – 2x + 1

g) a+2 2 + 3 h) x + y
a2 + a – a2 – 1 x2 + 2xy + y2 x2 – y2

i) 1 + 1 j) 1 + 1
(x – y) (x – z) (x – z) (y – z) (a – b) (b – c) (c – b) (a – c)

Creative section - A

2. Simplify:

a) (x – 1 – 3) + (x – 1 – 1) – (x 2 1)
2) (x 2) (x – 3) (x –

b) (x 1 + (x + 3 – x) + (x 2 4)
– 3) (x + 2) 2) (4 – 3) (x –

c) 2 (a – 3) + a–1 + a–2
(a – 4) (a – 5) (3 – a) (a – 4) (5 – a) (a – 3)

d) (2x – x–1 + 2) + (x + 3 – 1) – 1
1) (x 2) (x (1 – x) (1 – 2x)

3. Simplify:

a) x2 – 1 + 6 – x2 – 2 + 3 – 1
5x 4x x2 –3x + 2

b) 1 – 2 – 1
a2 – 5a + 6 a2 – 4a + 3 a2 – 3a + 2

c) x2 x–1 2 + x2 x–2 6 + x–5
– 3x + – 5x + x2 – 8x + 15

d) a2 2a – 6 20 – a2 a– 1 12 – a2 a– 2 15
– 9a + – 7a + – 8a +

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Simplification of Rational Expressions

4. Simplify:

a) x + x – 4x b) 2 – 2 b + 4a
x+2 x–2 x2 – 4 a+b a– a2 – b2

c) x+y – x–y + 4xy d) a+b – a–b – 2ab
x–y x+y x2 + y2 a–b a+b a2 – b2

e) x+1 + x–1 – 2x2 f) x+y + x–y – 2 (x2 – y2)
x–1 x+1 x2 – 1 x–y x+y x2 + y2

5. Simplify:

a) 1 + 1 + x2 – 1 – 1 x2 – 2x
x x+ 1 + x2 + x4

b) x+3 + x2 x–3 9 – 54
x2 + 3x + 9 – 3x + x4 + 9x2 + 81

c) x2 x– y y2 + x2 x+y y2 – 2y3
– xy + + xy + x4 + x2y2 + y4

d) a+2 – a–2 – 2a2
1 + a + a2 1 – a + a2 1+ a2 + a4

e) 4a2 2a + b b2 + 2a – b – 2b3
+ 2ab + 4a2 – 2ab + b2 16a4 + 4a2b2 + b4

3x – 1 3x + 1 54x3
f) 9x2 – 3x + 1 – 9x2 + 3x + 1 + 81x4 + 9x2 + 1

6. Simplify:

a) a2 – (b – c)2 + b2 – (a – c)2 + c2 – (a – b)2
(a + c)2 – b2 (a + b)2 – c2 (b + c)2 – a2

b) (a – b)2 – c2 + (b – c)2 – a2 + (c – a)2 – b2
a2 – (b + c)2 b2 – (c + a)2 c2 – (a + b)2

7. Simplify:

a) 1 x) – (1 1 x) + (1 x
(1 – + – x)

b) 2 y+ 3 – 5 x– y
x+ x– x–y
y

c) 1 –1 a) + 8 2 a
8 (1 – a ) 8 (1 + (1 – a)

d) 1 – 1) + 8 ( 1 + 2x
8( x x + 1) 8 (x – 1)

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Simplification of Rational Expressions

Creative section - B

8. Simplify:

a) 1 – 1 + 1 – 1 b) 1 – 1 + 1 – 1
x–5 x–3 x+5 x+3 a–3 a–1 a+3 a+1

c) 1 – 2 + 1 – 2 d) a3 + a3 – 1 + a 1 1
x–a 2x + a x+a 2x – a a–1 a+1 a–1 +

e) x3 – x2 – x x 1 + x 1 1 f) 1 – 3 + 3 – 1
x–1 x+1 – + a–1 a a+1 a+2

9. Simplify:

a) x 3 1 – x – x 2 1 – 5 1
– x2 + 1 + x2 –

b) 2 – a – 1 – 3
a–1 a2 + 1 a+1 a2 – 1

c) 4 – 2x – 2 + 6
x–1 x2 + 1 x+1 1 – x2

d) x4 + x4 – 1 – 1
x2 + 1 x2 – 1 x2 + 1 x2 – 1

10. Simplify:

a) 1 + 1 1 a + 2 + 4 b) 1 – 1 – 2 – 4
1–a + 1 + a2 1 + a4 a–1 a+1 a2 + 1 a4 + 1

c) 1 + 1 – 2x – 4x d) b + a b b + 2ab + 4a3b
x+1 x–1 x2 + 1 x4 + 1 a–b + a2 + b2 a4 + b4

e) 1 + 1 + 2x + 4x3 f) b – b + 6b2 – 8b4
x–y x+y x2 + y2 x4 + y4 a+b a–b a2 – b2 a4 – b4

g) a + a + 2a + 16a3 h) 1–x – 1+x – 4x – 8x
1 – 2a 1 + 2a 1 + 4a2 16a4 – 1 1+x 1–x 1 + x2 1 + x4

i) 1+x – 1–x + 4x + 8x3 j) 1+a + 1–a – 1 + a2 – 1 – a2
1–x 1+x 1 + x2 1 – x4 1–a 1+a 1 – a2 1 + a2

k) x+y – x–y + 4xy – 8x3y l) 1 – a + a2 + a4
x–y x+y x2 + y2 x4 + y4 a+1 a2 – 1 a4 – 1 a8 – 1

m) 1 + 1 + 2x 1 – x + 4x3 n) 1 + a b b + 2ab – a + 4a3b
x–1 x2 + x+1 x4 + 1 – a2 + b2 a+b a4 + b4

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Unit Indices

10

10.1 Indices - review

Let's multiply an algebraic term x four times by itself. It is written as x × x × x × x and
the product is shown in the abbreviated form which is x4. In this case, x is called the
base and 4 is called the index or exponent.

Thus, the index refers to the power to which a number is raised. In the example x4,
the base x is raised to the power 4. Indices is the plural form of index.

10.2 Laws of indices

There are certain verified rules which are used in the operations of indices of
algebraic terms. Product rule, quotient rule, power rule, etc. are the examples of
such rules. The verifications of these rules have been already discussed in our earlier
grades. The table given below shows the summary of these rules.

Laws of indices at a glance

(i) Product law am u an = am + n, where 'm' and 'n' are positive
integers

(ii) Quotient law am ÷ an = am – n when m > n
(iii) Power law of indices am ÷ an = 1 when m < n

an – m

(am)n = am u n, (ab)m = ambm, am = am
b bm

(iv) Law of negative index a– m = 1 or am = 1
am a–m

(v) Law of zero index aq = 1, bq = 1, xq = 1 and so on
(vi) Root law of indices
n am m

= an

Worked-out examples

Example 1: Evaluate 8 – 1 81 – 1 32 – 1
3 4 5
27 × 16 ÷
243

Solution:

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Indices

11 1
27 3 16 4 243 5
= 8 × 81 ÷ 32

= 3 3 × 1 × 2 4 × 1 ÷ 3 5 × 1
2 3 3 4 5

2

= 3 × 2 × 2 = 2 Answer
2 3 3 3

Example 2: Prove that 11x+2 + 11x = 1
122 × 11x
Solution:

L.H.S. = 11x+2 + 11x
122 × 11x
11x × 112 + 11x
= 122 × 11x

= 11x (121 + 1)
122 × 11x

= 11x × 122 = 1 R.H.S. Proved
122 × 11x

Example 3: Simplify 7a+ 1 + 9 × 7a
7a + 2 – 45 × 7a
Solution:

Here, 7a+ 1 + 9 × 7a = 7a × 7 + 9 × 7a
7a + 2 – 45 × 7a 7a × 72 – 45 × 7a

= 7a (7 + 9) = 16 = 4 Answer
7a (49 – 45) 4

Example 4: Simplify a a2 + ab + b2 b b2 + bc + c2 xc c2 + ca + a2
Solution: xa
xx ×b xx ×c

a a2 + ab + b2 b b2 + bc + c2 xc c2 + ca + a2
xa
xx ×b xx ×c

= x × x × x(a – b) (a2 + ab + b2) (c – a) (c2 + ca + a2)
(b – c) (b2 + bc + c2)

= xa3 – b3 × xb3 – c3 × xc3 – a3

= xa3 – b3 + b3 – c3 + c3 – a3 = x° = 1 Answer

Example 5: Simplify x x+y–z y y+z–x az z + x – y
Solution: ax
aa ×y aa ×z

x x+y–z y y+z–x az z + x – y y2 + yz – xy z2 + zx – yz
ax yz + z2 – zx zx + x2 – xy
aa ×y aa ×z x2 + xy – zx

= aa ×× aa ×× aaxy + y2 – yz

x2 + y2 + z2
x2 + y2 +z2
x2 + xy – zx + y2 + yz – xy + z2 + zx – yz

= aa = aa = 1 Answerxy + y2 – yz + yz + z2 – zx + zx + x2 – xy

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Indices

bc a

Example 6: Simplify xc × xa × xb
Solution: bc c ca a ab b

xb xc xa

b c a b c a

bc xc × ca xa × ab xb = xc ÷ bc × xa ÷ ca × xb ÷ ab
÷ bc ÷ ca ÷ ab
c a b c a b

xb xc xa xb xc xa

b × 1 × c × 1 a × 1 11 1
bc × ca ab x c2 × xa2 × x b2
= xc xa ×xb = =1 Answer
× 1 × 1 × 1 1 1 1
c bc a ca b ab x b2 × x c2 × xa2

xb xc ×xa

11 1

Example 7: Simplify m + (mn2)3 + (m2n)3 × 1 – n3
Solution: 1
m–n
m3

11 1 12 21 11

m + (mn2)3 + (m2n)3 × 1– n3 = m + m3 n3 + m3 n3 m3 – n3
1 ×1
m–n m–n m3
m3
12 2 11
11

m3 m3 + n3 + m3n3 m3 – n3
= m–n × 1
m3

11 12 11 12

m3 – n3 m3 + m3n3 + n3
=
m–n

13 13
m3 – n3
= m–n = m – n = 1 Answer
m – n

x+ 1 a 1 –x b
y y
Example 8: Simplify ×
Solution:
y+ 1 a 1 –y b
x x
×

x+ 1 a 1 –x b xy + 1 a 1 – xy b
y y y y
× ×
1 – xy b
1 a 1 b = xy + 1 a x
x x x
y+ × –y ×

= (xy + 1)a (1 – xy)b xa + b xa + b x a+b
× (xy + 1)a (1 – xy)b = ya + b = y
ya + b Answer

1 – a2 a 1 –a b–a a a+b
b2 b b
Example 9: Show that =
Solution:
1 – b2 b 1 +b a–b
a2 a

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Indices

1 a 1 –a b–a 1 a 1 a 1 – a b–a
b2 b +b a–b b b b
– a2 1 +a –a
a
L.H.S. = 1 b = 1 b 1 b 1 a–b
a2 a a a
– b2 +b –b +b

1 a 1 b–a+a 1 + ab a 1 – ab b
b b b b
1 +a –a
a 1 + ab a 1 – ab b
= b+a–b 1 b = a a
a
+b –b

= (1 + ab)a (1 – ab)b × aa + b
ba + b ab)a (1
(1 + – ab)b

= a a+b Proved.
b

Example 10: 111
Simplify 1 + ax – y + az – y + 1 + ay – z + ax – z + 1 + az – x + ay – x
Solution:

111
1 + ax – y + az – y + 1 + ay – z + ax – z + 1 + az – x + ay – x

= 1 az + 1 ax + 1 ay
ax ay ay az az ax
1+ ay + 1 + az + 1+ ax +

= 1 + 1 + 1
ay + ax + az az + ay + ax ax + az + ay
ay az ax

ay az ax ax + ay + az
= ax + ay + az + ax + ay + az + ax + ay + az = ax + ay + az = 1 Answer

Example 11: If a + b + c = 0, show that

Solution: 111
1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a = 1

Here, a + b + c = 0, ? b + c = – a
111

1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a

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

xb xc 1
= xb + xa + b + 1 + xc + xb + c + 1 + 1 + xc + xb + c

xb.xc xc 1
= xc (xb + xa + b + 1) + xc + xb + c + 1 + 1 + xc + xb + c

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Indices

xb + c xc 1
= xb + c + xa + b + c +xc + xb + c + xc+ 1 + xb + c + xc + 1

xb + c xc 1
= xb + c + x0+ xc + xb + c + xc+1 + xb + c + xc + 1

xb + c xc 1 xb + c + xc + 1
= xb + c + xc+1 + xb + c + xc+1 + xb + c + xc + 1 = xb + c + xc + 1 = 1 Proved.

22

Example 12: If x2 + 2 = 33 + 3 3, show that 3x(x2 + 3) = 8.

Solution:

22

Here, x2 + 2 = 33 + 3 3

12 12

or, x2 = 33 + 3 3 – 2

12 12 1 1
or, x2 = 33 + 3 3 – 2.33.3 3

1 12

or, x2 = 33 – 3 3

11

or, x = 33 – 3 3

1 13

or, x3 = 33 – 3 3

13 13 11 1 1

or, x3 = 33 – 3 3 – 3.33.3 3 33 – 3 3

or, x3 = 3– 1 – 3x
8
3

or, x3 + 3x = 3
or, 3x3 + 9x = 8

or, 3x (x2 + 3) = 8 Proved.

EXERCISE 10.1

General section

1. Evaluate:

8 – 1 4 – 1 125 – 1 2 25 – 1
27 3 2 3 2
a) ÷ b) ÷
9 64 4

– 1 – 1 – 1 –1
3 4 5
c) 8 × 81 ÷ 32 d) 729
27 16
243 3
64

– 1 2 1 1 1
3 2 3
125 25 – 125 3 8 –
64 ÷
e) 64 f) 16 27

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Indices

2. Simplify:

a) 2x × 3 – 2x b) 5m + 2 – 5m c) 4m + 4m + 1
2x+2 – 2x 5m+1 + 5m 4m+2 – 4m
+1

d) 6n + 2 – 6n e) 2x +4 – 2x f) 5n + 2 – 2.5n
6n + 1 + 6n 5.2x 23.5n

g) 5n + 2 – 10 × 5n h) 33a + 2 – 33a + 1 i) 132x +1 + 5 × 169x
3 × 5n 6 × 27a 9 × 169x

j) 112x + 1 – 6 × 121x k) 7a + 1 + 9 × 7a l) 9n + 2 + 10 × 9n
5× 121x 7a + 2 – 45 × 7a 9n + 1 × 11 – 8 × 9n

m) 11n + 2 – 55.11n – 1 n) 273n – 1 (243)– 4n 2n
11n × 116 5
o) (243) 5 . 32n + 1
9n + 1 33n – 5 9n + 1 × 32(n – 2)

3. Simplify: b) 3n a3 × an n – 1 2
e) a6b–2c4 ÷ 4 a4b–4c8
2 h) 3 3x7y11z–1 × 3 72x–1yz4 c) (125a3 ÷ 27b–3) 3

a) (64x3 ÷ 27a–3) 3 f) 3 27a12b9 ÷ 4 16a16b2

3 2

d) (81a4 ÷ 16b–4) 4 i) 3 (x + y)–8 .(x + y)3

g) 3 9x–2y7 × 3 3x5y–1

1

j) 3 (a + b)–7 .(a + b)3

4. Simplify: b) ax + y x – y × ay + z y – z × az – x z + x
a) xb – c × xc – a × xa – b
ya b zb a xa b
c) xa + b × xa – b × xc – 3a d) za . xb . ya
xc – a
f) 1 + 1
e) 1 + 1 1 – ax – y 1 – ay – x
1 + xa – b 1 + xb – a
h) a2 + b2 + c2
g) x–1y–1 + y–1z–1 + z–1x–1 a–2b–2 + b–2c–2 + c–2a–2
x+y+z

Creative section - A

5. Simplify:

xa + b a – b xc + a c – a xb + c b – c xa a – b xb b – c xc c + a
a) xc × xb × xa b) x–b × x–c × xa

xl l2 + lm + m2 xm m2 + mn + n2 xn n2 + nl + l2
c) xm × xn × xl

d) (xa ÷ xb)a2 + ab + b2 × (xb ÷ xc)b2 + bc + c2 × (xc ÷ xa)c2 + ca + a2

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Indices

(aax + y)2 (ay+z)2 (az + x)2 xa2 + b2 a–b xb2 +bc b–c xc2 +ac c –
(ax.ay.az)4 x–ab x–c2 x–a2
e) f) × ×

xb b + c – a xc c + a – b xa a + b – c xb + c a – b xa + b b – c xa + b c – a
g) xc × xa × xb h) xc – a × xa – c × xb – c

xl + m n – l xm + n l – m xn + l m – n xa a – b xb b – c xc c – a
i) xl – m × × xn – l j) x.x–b × x.x–c × x.x–a
xm – n

111 b 1 c 1 a 1
xb xc xa ab bc ca ab
xc bc xa ac xb xc xa xb
k) l) c × a × b
× ×

xb xc xa

m) 1 1 1 1 . 1 1 111
y–x z–y
ax–y x–z . ay–z az–x n) xm2 – n2 m+ n × xn2 – p2 n+ p × xp2 – m2 p+ m

11 1

o) ax .ay x – y az y+z × az z+x p) p + (pq2)3 + (p2q)3 × 1– q3
ay ax p–q 1

p3

6. Simplify: b) xy ax × yz ay × zx az
a) xa +b a2 – b2 × b +c xb2 – c2 × c +a xc2 – a2 ay az ax

yzx 1 111
ab 1 1
c) yz az × zx ax × xy ay d) xa × bc xb × ca xc
z x y 1 1 1

ay az ax xb xc xa

Creative section - B

7. Simplify:

a+ 1 x 1 –a y 1 –x a x+ 1 a
b b y y
a) × b) ×

b+ 1 x 1 –b y y+ 1 a 1 –y a
a a x x
× ×

a2 – 1 a a– 1 b–a 1+ x x 1– y y
b2 b y x x–y
× x–y ×

c) 1 b 1 a–b d) y x x y
a2 a x x–y y x–y
b2 – × b+ +1 × –1

y+ 1 x+y x– 1 x+y
x y
e) ×

x2 – 1 x y2 – 1 y
y2 x2
×

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Indices

8. Simplify:

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

b) 1 + ax 1 + ax – z + 1 + ay 1 + ay – z + 1 + az 1 + az – y

–y –x –x

y y2 – yz + z2 z z2 – zx + x2
a a ax x2 – xy + y2
9. a) If x3 + y3 + z3 = 1, show that, a–y a–z a–x = a2

y y–z
a a ax x – y z z–x

b) If x2 + y2 + z2 = xy + yz + zx, show that ay . az . ax =1

c) If pqr = 1, show that 1 + 1 q–1 + 1 + 1 r–1 + 1 + 1 p–1 = 1.
p+ q+ r+

d) If abc + 1 = 0, prove that 1 + 1 + 1 = 1
1 – a – b–1 1 – b – c–1 1 – c – a–1

22

e) If x2 + 2 = 23 + 2 3, show that 2x (x2 + 3) = 3

12

f) If x = 33 + 33 , prove that x3 – 9x – 12 = 0
111

g) If a + b + c = 0, show that 1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a = 1

10.3 Exponential equation

Let's take an equation 2x = 8. In this equation, the unknown variable 'x' is the base
and 2 is its coefficient.

Now, take another equation 2x = 8. In this case, the unknown variable 'x' is the
exponent of the base 2. Such an equation in which the unknown variable appears as
an exponent of a base is known as exponential equation. The following axioms are
useful while solving the exponential equations:

(i) If ax = ab, then x = b (ii) If ax = 1, then ax = a° and x = 0

Thus, while solving an exponential equation, we should simplify the equation till
the equation will be obtained in the forms ax = ab or ax = 1.

Worked-out examples

Example 1: Solve 2x + 2 + 2x + 3 = 1
2
Solution:
2x + 2 + 2x + 3 = 1
Here, 2 Check

or, 2x × 2 2 + 2x × 23 × 2–1 = 1 2x + 2 + 2x + 3 = 1
2
or, 2x (22 + 22) = 1 or, 2–3 + 2 + 2– 3 + 3 = 1
or, 2x = 1
8 or, 1 2 1 =1
+
or, 2x = 1 22
23
or, 1 = 1
or, 2x = 2–3

? x =–3

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