What is Mensuration

Mensuration

1. What is Mensuration? Learning Objective

2. To interpret perimeter of closed figure. To illustrate area of closed figure . Learning Outcomes

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

4. Look at the given figures :

5. Look at the given figures :

6. Look at the given figures :

7. Look at the given figures :

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

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

10. 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 .

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 . The distance covered is equal to the length of wire used to draw the figure.

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. This distance is known as the perimeter of the closed figure.

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

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

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

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

17. 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

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

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

20. Perimeter of a rectangle : A B C D

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

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

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

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

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

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

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. Length = 150 cm Breadth = 1m = 100 cm

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 Perimeter of the rectangle = 2 × (length + breadth)

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) = 2 × (150 cm + 100 cm) = 2 × (250 cm) = 500 cm = 5 m

31. 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?

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

34. 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

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 Total length of rope required = 3 × 840 m = 2520 m

36. 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.

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. Sol: Length of the rectangular park = 250 m Breadth of the rectangular park = 175 m

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 To calculate the cost of fencing we require perimeter. Perimeter of the rectangle = 2 × (length + breadth)

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) = 2 × (250 m + 175 m) = 2 × (425 m) = 850 m

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 Cost of fencing 1m of park = ₹ 12 Therefore, the total cost of fencing the park = ₹ 12 × 850 = ₹ 10200

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

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

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

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 = 4 x sides

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 Or = 4 x length of a side

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

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

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

49. 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

50. 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

51. 75 m Sol: Length of the square field = 75m Perimeter of the square field

52. 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

53. 105 m 160 m Length of the field = 160m Breadth of the field = 105m Perimeter of the rectangular field = 2 x (length + breadth )

54. \\ 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

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

56. 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 square? an equilateral triangle? a regular hexagon?

57. Sol: total length of the string = 30 m total length of the string = perimeter of the closed figure Square perimeter of the square = total length of the string

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

59. Sol: total length of the string = 30 m total length of the string = perimeter of the closed figure 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  

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

61. 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

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 3 x length of a side = 30 m length of a side = 30 3 Therefore , length of each side of the equilateral triangle = 10m  

63. Sol: total length of the string = 30 m total length of the string = perimeter of the closed figure Regular hexagon

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

65. Sol: total length of the string = 30 m total length of the string = perimeter of the closed figure 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  

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

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

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

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

70. What is Area ?

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

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

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

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

75. 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.

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. Ignore portions of the area that are less than half a square.

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. If more than half of a square is in a region, just count it as one square.

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

79. 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?

80. 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.

81. 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

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? 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).

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

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

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

86. 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

87. Area of the square = side × side

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

89. 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

90. 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

91. 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.

92. 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?

93. 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 = ?

94. 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

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? Solution : Area of the rectangle = 36 sq cm Length = 9 cm Width = ? Area of a rectangle = length × width So, width = = = 4cm  

96. 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.

97. 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

98. 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

99. 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

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 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

101. Number of tiles required = = = = 48tiles .  

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

103. 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

104. 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)

105. 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

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

107. 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

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

109. 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

110. 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

111. 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

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

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