This document discusses mensuration and defines perimeter and area of closed figures. It provides examples of calculating the perimeter of rectangles, squares, equilateral triangles, and other shapes. It also defines area as the amount of surface enclosed by a closed figure and provides examples of calculating the area of rectangles and squares by counting the number of squares covered on a grid. The key learning objectives are to interpret the perimeter of closed figures and illustrate the area of closed figures.

2. What is Mensuration?
Learning Objective

3. • To interpret perimeter of closed figure.
• To illustrate area of closed figure .
Learning Outcomes

4. What do you know
about perimeter of
closed figure ?
Learning Objective

5. Look at the given figures :

6. Look at the given figures :

7. Look at the given figures :

8. Look at the given figures :

9. Look at the given figures :
Can you make them
wire or a string.
S
S
S

10.  From the point S in each case and
move along the line segments then you
again reach the point S.

11.  From the point S in each case and move
along the line segments then you again
reach the point S.
 Make a complete round of the shape .

12.  From the point S in each case and move
along the line segments then you again
reach the point S.
 Make a complete round of the shape .
 The distance covered is equal to the
length of wire used to draw the figure.

13.  From the point S in each case and move
along the line segments then you again
reach the point S.
 Make a complete round of the shape .
 The distance covered is equal to the
length of wire used to draw the figure.
 This distance is known as the perimeter
of the closed figure.

14. Perimeter is the distance covered
along the boundary forming a
closed figure when you go round
the figure once.

15. 1. Find the perimeter of the following figures:
Perimeter = AB + BC + CD + DA

16. 1. Find the perimeter of the following figures:
Perimeter = AB + BC + CD + DA
= _____+_____+_____+_____
= ______
40 cm 10cm 40cm 10cm
100 cm

17. 1. Find the perimeter of the following figures:
Perimeter = AB + BC +
CD + DE + EF + FG + GH
+HI + IJ + JK + KL + LA

18. 1. Find the perimeter of the following figures:
Perimeter = AB + BC
+ CD + DE + EF + FG +
GH +HI + IJ + JK + KL
+ LA
=1cm + 3cm + 3cm +
1cm + 3cm + 3cm +
1cm + 3cm + 3cm +
1cm + 3cm + 3cm
= 28cm

19. 1. Find the perimeter of the following figures:
Perimeter = AB + BC + CD
+ DE + EF + FA

20. 1. Find the perimeter of the following figures:
Perimeter = AB + BC
+ CD + DE + EF + FA
=100cm + 120cm +
90cm + 45cm +
60cm + 80cm
= 495 cm

21. Perimeter of a rectangle :
A B
C
D

22. Perimeter of a rectangle :
A B
C
D
= AB + BC + CD + DA
l
l
b
b

23. Perimeter of a rectangle :
A B
C
D
= AB + BC + CD + DA
= + + +
l
l
b
b
l
b
l b

24. Perimeter of a rectangle :
A B
C
D
= AB + BC + CD + DA
= l + b + l + b
= = 2l + 2b
l
l
b b

25. Perimeter of a rectangle :
A B
C
D
= AB + BC + CD + DA
= l + b + l + b
= 2l + 2b
= 2 ( l + b )
l
l
b b

26. Perimeter of a rectangle :
A B
C
D
l
l
b
b
Perimeter = 2 ( length + breadth )

27. Example 1 : Find the perimeter of a rectangle
whose length and breadth are 150 cm and 1 m
respectively.

28. Example 1 : Find the perimeter of a rectangle
whose length and breadth are 150 cm and 1 m
respectively.

29. Example 1 : Find the perimeter of a rectangle
whose length and breadth are 150 cm and 1 m
respectively.
Length = 150 cm
Breadth = 1m = 100 cm

30. Example 1 : Find the perimeter of a rectangle
whose length and breadth are 150 cm and 1 m
respectively.
Length = 150 cm
Breadth = 1m = 100 cm
Perimeter of the rectangle
= 2 × (length + breadth)

31. Example 1 : Find the perimeter of a rectangle
whose length and breadth are 150 cm and 1 m
respectively.
Length = 150 cm
Breadth = 1m = 100 cm
Perimeter of the rectangle
= 2 × (length + breadth)
= 2 × (150 cm + 100 cm)
= 2 × (250 cm)
= 500 cm
= 5 m

32. Example 2 : A farmer has a rectangular field of
length and breadth 240 m and 180 m
respectively. He wants to fence it with 3 rounds
of rope as shown in figure 10.4. What is the total
length of rope he must use?

33. Example 2 : A farmer has a rectangular field of
length and breadth 240 m and 180 m
respectively. He wants to fence it with 3 rounds
of rope as shown in figure 10.4. What is the total
length of rope he must use?

34. Sol:
The farmer has to cover three
times the perimeter of that field.

35. Sol:
The farmer has to cover three
times the perimeter of that field.
Therefore, total length of rope
required is thrice its perimeter.
Perimeter of the field
= 2 × (length + breadth)
= 2 × ( 240 m + 180 m)
= 2 × 420 m = 840 m

36. Sol:
The farmer has to cover three
times the perimeter of that field.
Therefore, total length of rope
required is thrice its perimeter.
Perimeter of the field
= 2 × (length + breadth)
= 2 × ( 240 m + 180 m)
= 2 × 420 m = 840 m
Total length of rope required
= 3 × 840 m
= 2520 m

37. Example 3 : Find the cost of fencing a rectangular
park of length 250 m and breadth 175 m at the
rate of ₹12 per metre.

38. Example 3 : Find the cost of fencing a rectangular
park of length 250 m and breadth 175 m at the
rate of ₹12 per metre.
Sol: Length of the rectangular park = 250 m
Breadth of the rectangular park = 175 m

39. Example 3 : Find the cost of fencing a rectangular
park of length 250 m and breadth 175 m at the
rate of ₹12 per metre.
Sol: Length of the rectangular park = 250 m
Breadth of the rectangular park = 175 m
To calculate the cost of fencing we require perimeter.
Perimeter of the rectangle =
2 × (length + breadth)

40. Example 3 : Find the cost of fencing a rectangular
park of length 250 m and breadth 175 m at the
rate of ₹12 per metre.
Sol: Length of the rectangular park = 250 m
Breadth of the rectangular park = 175 m
To calculate the cost of fencing we require
perimeter.
Perimeter of the rectangle =
2 × (length + breadth)
= 2 × (250 m + 175 m) = 2 × (425 m) =
850 m

41. Example 3 : Find the cost of fencing a rectangular
park of length 250 m and breadth 175 m at the
rate of ₹12 per metre.
Sol: Length of the rectangular park = 250 m
Breadth of the rectangular park = 175 m
To calculate the cost of fencing we require
perimeter.
Perimeter of the rectangle =
2 × (length + breadth)
= 2 × (250 m + 175 m) = 2 × (425 m) = 850 m
Cost of fencing 1m of park = ₹ 12
Therefore, the total cost of fencing the park
= ₹ 12 × 850 = ₹ 10200

42. Perimeter of a square:
S
S
S
S
A B
C
D

43. Perimeter of a square:
S
S
S
S
A B
C
D
Perimeter = AB + BC + CD + DA

44. Perimeter of a square:
S
S
S
S
A B
C
D
Perimeter = AB + BC + CD + DA
= S + S + S + S
= 4 x S

45. Perimeter of a square:
S
S
S
S
A B
C
D
Perimeter = AB + BC + CD + DA
= S + S + S + S
= 4 x S
= 4 x sides

46. Perimeter of a square:
S
S
S
S
A B
C
D
Perimeter = AB + BC + CD + DA
= S + S + S + S
= 4 x S
= 4 x sides
Or
= 4 x length of a side

47. Perimeter of an equilateral triangle A
B
C
S S
S

48. Perimeter of an equilateral triangle A
B
C
S S
S
Perimeter = AB + BC + CA
= S + S + S

49. Perimeter of an equilateral triangle A
B
C
S S
S
Perimeter = AB + BC + CA
= S + S + S
= 3 x S
Or

50. Perimeter of an equilateral triangle A
B
C
S S
S
Perimeter = AB + BC + CA
= S + S + S
= 3 x S
Or
= 3 x length of a side

51. Example 4 : Pinky runs around a square field of side
75 m, Bob runs around a rectangular field with
length 160 m and breadth 105 m. Who covers more
distance and by how much?
75 m
160 m
105 m

52. 75 m
Sol:
Length of the square field =
75m
Perimeter of the square field

53. 75 m
Sol:
Length of the square field =
75m
Perimeter of the square field
= 4 x length of a
side
= 4 x 75m
= 300 m

54. 105 m
160 m
Length of the field = 160m
Breadth of the field =
105m
Perimeter of the
rectangular field
= 2 x (length + breadth )

55. 105 m
160 m
Length of the field = 160m
Breadth of the field = 105m
Perimeter of the
rectangular field
= 2 x (length + breadth )
= 2 x ( 160 + 105 )
= 2 x (265)
= 530cm

56. Therefore , pinky runs= 300m
and
Bob runs = 530m
Hence , Bob covers more distance than pinky and
by 230m .

57. Example 5 : A piece of string is 30 cm long. What
will be the length of each side if the string is used
to form :
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?

58. Sol: total length of the string = 30 m
total length of the string = perimeter of the
closed figure
(a)Square
perimeter of the square = total length of the
string

59. Sol: total length of the string = 30 m
total length of the string = perimeter of the closed
figure
(a)Square
perimeter of the square = total length of the string
4 x length of a side = 30 m

60. Sol: total length of the string = 30 m
total length of the string = perimeter of the
closed figure
(a)Square
perimeter of the square = total length of the
string
4 x length of a side = 30 m
length of a side
= 30 ÷ 4
= 7.5 m
Each side of the square = 7.5 m

61. Sol: total length of the string = 30 m
total length of the string = perimeter of the closed
figure
(b) Equilateral triangle

62. Sol: total length of the string = 30 m
total length of the string = perimeter of the closed
figure
(b) Equilateral triangle
perimeter of Equilateral triangle = total length of
the string

63. Sol: total length of the string = 30 m
total length of the string = perimeter of the closed
figure
(b) Equilateral triangle
perimeter of Equilateral triangle = total length of
the string
3 x length of a side = 30 m
length of a side = 30 ÷ 3
Therefore , length of each side of the equilateral
triangle
= 10m

64. Sol: total length of the string = 30 m
total length of the string = perimeter of the closed
figure
(a) Regular hexagon

65. Sol: total length of the string = 30 m
total length of the string = perimeter of the closed
figure
(a) Regular hexagon
perimeter of regular hexagon = total length of the
string

66. Sol: total length of the string = 30 m
total length of the string = perimeter of the closed
figure
(a) Regular hexagon
perimeter of regular hexagon = total length of the
string
6 x length of a side = 30m
length of a side = 30 ÷ 3
= 10m

67. • To interpret perimeter of closed figure.
• To illustrate area of closed figure .
Learning Outcomes
How confident do you feel?

68. • To interpret perimeter of closed figure.
• To illustrate area of closed figure .
Learning Outcomes
How confident do you feel?

69. What do you know
about area of closed
figure ?
Learning Objective

70. Example : Look at the closed figures. Can you tell which one
occupies more region?

71. What is Area ?

72. What is Area ?
The amount of surface
enclosed by a closed figure
is called its area.

73. It is difficult to tell
just by looking at
these figures

74. It is difficult to tell
just by looking at
these figures
So, what do you do?

75. Place them on a squared paper or graph
paper where every square measures 1 cm
× 1 cm.

76. Place them on a squared paper or
graph paper where every square
measures 1 cm × 1 cm.
The area of one full square is
taken as 1 sq unit.

77. Place them on a squared paper or graph paper
where every square measures 1 cm × 1 cm.
The area of one full square is
taken as 1 sq unit.
 Ignore portions of the area that
are less than half a square.

78. Place them on a squared paper or graph
paper where every square measures 1 cm × 1
cm.
The area of one full square is
taken as 1 sq unit.
 Ignore portions of the area that
are less than half a square.
If more than half of a square is in a
region, just count it as one square.

81. Area of a rectangle
Example : What will be the area of a rectangle
whose length is 5 cm and breadth is 3 cm?

82. Area of a rectangle
 Draw the rectangle on a graph paper having
1 cm × 1 cm sq
Example : What will be the area of a rectangle
whose length is 5 cm and breadth is 3 cm?

83. Area of a rectangle
 Draw the rectangle on a graph paper
having 1 cm × 1 cm sq
Example : What will be the area of a rectangle
whose length is 5 cm and breadth is 3 cm?
The rectangle covers 15 squares
completely.

84. Area of a rectangle
 Draw the rectangle on a graph paper having 1
cm × 1 cm sq
Example : What will be the area of a rectangle
whose length is 5 cm and breadth is 3 cm?
The rectangle covers 15 squares
completely.
The area of the rectangle = 15 sq
cm

85. Area of a rectangle
 Draw the rectangle on a graph paper having 1
cm × 1 cm sq
Example : What will be the area of a rectangle whose
length is 5 cm and breadth is 3 cm?
The rectangle covers 15 squares
completely.
The area of the rectangle = 15 sq cm
which can be written as 5 × 3 sq cm
i.e. (length × breadth).

86. Area of a rectangle = (length × breadth)

87. Area of a square
Let us now consider a square of side 4 cm

88. Area of a square
Let us now consider a square of side 4 cm
What will be its area?

89. Area of a square
Let us now consider a square of side 4 cm
What will be its area?
It covers 16 squares
i.e. the area of the square =
16 sq cm
= 4 × 4 sq cm

90. Area of the square = side × side

91. Example 6 : Find the area of a rectangle whose
length and breadth are 12 cm and 4 cm
respectively.

92. Example 6 : Find the area of a rectangle whose
length and breadth are 12 cm and 4 cm respectively.
Solution : Length of the rectangle = 12 cm
Breadth of the rectangle = 4 cm

93. Example 6 : Find the area of a rectangle whose
length and breadth are 12 cm and 4 cm
respectively.
Solution : Length of the rectangle = 12 cm
Breadth of the rectangle = 4 cm
Area of the rectangle = length × breadth

94. Example 6 : Find the area of a rectangle whose
length and breadth are 12 cm and 4 cm
respectively.
Solution : Length of the rectangle = 12 cm
Breadth of the rectangle = 4 cm
Area of the rectangle = length × breadth
= 12 cm × 4 cm = 48 sq cm.

95. Example 7 : The area of a rectangular piece of
cardboard is 36 sq cm and its length is 9 cm. What
is the width of the cardboard?

96. Example 7 : The area of a rectangular piece of
cardboard is 36 sq cm and its length is 9 cm. What
is the width of the cardboard?
Solution : Area of the rectangle = 36 sq cm
Length = 9 cm
Width = ?

97. Example 7 : The area of a rectangular piece of
cardboard is 36 sq cm and its length is 9 cm. What is
the width of the cardboard?
Solution : Area of the rectangle = 36 sq cm
Length = 9 cm
Width = ?
Area of a rectangle = length × width

98. Example 7 : The area of a rectangular piece of
cardboard is 36 sq cm and its length is 9 cm. What
is the width of the cardboard?
Solution : Area of the rectangle = 36 sq cm
Length = 9 cm
Width = ?
Area of a rectangle = length × width
So, width =
𝑨𝒓𝒆𝒂
𝒍𝒆𝒏𝒈𝒕𝒉
=
𝟑𝟔
𝟒
= 4cm

99. Example 7 : Bob wants to cover the floor of a room
3 m wide and 4 m long by squared tiles. If each
square tile is of side 0.5 m, then find the number of
tiles required to cover the floor of the room.

100. Solution :
Total area of tiles must be equal to the area of the
floor of the room.
Length of the room = 4 m
Breadth of the room = 3 m

101. Solution :
Total area of tiles must be equal to the area of the
floor of the room.
Length of the room = 4 m
Breadth of the room = 3 m
Area of the floor = length × breadth

102. Solution :
Total area of tiles must be equal to the area of the
floor of the room.
Length of the room = 4 m
Breadth of the room = 3 m
Area of the floor = length × breadth
= 4 m × 3 m = 12 sq m

103. Solution :
Total area of tiles must be equal to the area of the
floor of the room.
Length of the room = 4 m
Breadth of the room = 3 m
Area of the floor = length × breadth
= 4 m × 3 m = 12 sq m
Area of one square tile = side × side
= 0.5 m × 0.5 m
= 0.25 sq m

104. Number of tiles required =
Area of the floor
Area of one tile
=
𝟏𝟐
𝟎.𝟐𝟓
=
𝟏𝟐𝟎𝟎
𝟐𝟓
= 48tiles.

105. Example 8: Find the area in square metre of a
piece of cloth 1m 25 cm wide and 2 m long.
Solution :

106. Example 8: Find the area in square metre of a piece
of cloth 1m 25 cm wide and 2 m long.
Solution :
Length of the cloth = 2 m
Breadth of the cloth= 1 m 25 cm

107. Example 8: Find the area in square metre of a piece
of cloth 1m 25 cm wide and 2 m long.
Solution :
Length of the cloth = 2 m
Breadth of the cloth= 1 m 25 cm
= 1 m + 0. 25 m = 1.25 m (since 25 cm = 0.25m)

108. Example 8: Find the area in square metre of a piece
of cloth 1m 25 cm wide and 2 m long.
Solution :
Length of the cloth = 2 m
Breadth of the cloth= 1 m 25 cm
= 1 m + 0. 25 m = 1.25 m (since 25 cm = 0.25m)
Area of the cloth = length of the cloth × breadth of
the cloth
= 2 m × 1.25 m
= 2.50 sq m

109. Example 9 : By splitting the following figures into
rectangles, find their areas (The measures are
given in centimetres).

110. Example 9 : By splitting the following figures into
rectangles, find their areas (The measures are
given in centimetres).
1 1
2
1
1
2
2 I
II
III

111. Area of fig I : l x b
= 2 x 1
= 2 sq cm

112. Area of fig I : l x b
= 2 x 1
= 2 sq cm
Area of fig II : l x b
= 2 x 1
=2 sq cm

113. Area of fig I : l x b
= 2 x 1
= 2 sq cm
Area of fig II : l x b
= 2 x 1
=2 sq cm
Area of fig III : l x b
= 5 x 1
= 5 sq cm

114. Area of fig I : l x b
= 2 x 1
= 2 sq cm
Area of fig II : l x b
= 2 x 1
=2 sq cm
Area of fig III : l x b
= 5 x 1
= 5 sq cm
Total area of the given fig
= Area of fig I + Area of fig II +
Area of fig III
= 2 sq cm + 2 sq cm + 5 sq cm
= 9 sq cm

115. • To interpret perimeter of closed figure.
• To illustrate area of closed figure .
Learning Outcomes
How confident do you feel?

116. • To interpret perimeter of closed figure.
• To illustrate area of closed figure .
Learning Outcomes
How confident do you feel?