## NCERT Solutions for class 7 Maths Chapter 11: Perimeter and Area## Exercise 11.1
Length of the rectangular land = 500m Breadth of the rectangular land = 300m Area of the rectangular land = 500m × 300m = 150000 m
= 150000 m Cost of the land = Area × Cost of the land per 1m Cost of the land = 150000 m Cost of the land = ₹ 1,500,000,000
Perimeter of a square park = 4 × Side of the square park 320 m = 4 × Side of the square park Side of the square park = 320/4 = 80 m Area of the square park = Side × Side or (Side) Area of the square park = 80 m × 80 m Area of the square park = 6400 m
Area of the rectangular land = Length × Breadth 440 m Breadth = 440 m Breadth = 20 m Perimeter of the rectangular land = 2 × (Length + Breadth) Perimeter of the rectangular land = 2 × (22 + 20) m Perimeter of the rectangular land = 2 × 42 m Perimeter of the rectangular land = 84 m
100 cm = 2 × (35 + Breadth) cm (35 + Breadth) = 100/2 (35 + Breadth) = 50 Breadth = 50 - 35 Breadth = 15 cm Area of the rectangular sheet = Length × Breadth Area of the rectangular sheet = 35 cm × 15 cm Area of the rectangular sheet = 525 cm
Area of the rectangular park = Length × Breadth Area of the rectangular park = 90 m × Breadth Side of the square park = 60 m Area of the square park = Side × Side Area of the square park = 60 m × 60 m = 3600 m Area of the square park = Area of the rectangular park 3600 m Breadth of the rectangular park = 3600 m Breadth of the rectangular park = 40 m
Breadth = 22 cm Perimeter of the rectangle = 2 × (Length + Breadth) Perimeter of the rectangle = 2 × (40 + 22) cm Perimeter of the rectangle = 2 × 62 cm Perimeter of the rectangle = 124 cm ## Note: When a wire is bent in the form of any shape, its perimeter is always considered. Perimeter is the sum of all sides of a given polygon.Perimeter of a square = 4 × Side The same wire when bent in the form of a square will have the same perimeter. 124 cm = 4 × Side Side = 124 cm/4 Side = 31 cm The side of the square is 31 cm. Area of the rectangle = Length × Breadth Area of the rectangle = 40 cm × 22 cm Area of the rectangle = 880 cm Area of the square = Side × Side Area of the square = 31 cm × 31 cm = 961 cm Area of square > Area of the rectangle Hence, square encloses more area.
Length =? Breadth = 30 cm Perimeter of the rectangle = 2 × (Length + Breadth) 130 cm = 2 × (Length + 30) cm (Length + 30) cm = 130/2 (Length + 30) cm = 65 cm Length = 65 cm - 30 cm Length = 35 cm Area of the rectangle = Length × Breadth Area of the rectangle = 30 cm × 35 cm Area of the rectangle = 1050 cm
Breadth of the door = 1 m Area of the door = Length × Breadth Area of the door = 2 m × 1 m Area of the door = 2 m Length of the wall = 4.5 m Breadth of the wall = 3.6 m Area of the wall = Length × Breadth Area of the wall = 4.5 m × 3.6 m Area of the wall = 16.2 m Area of whitewashing the wall = Area of the wall - Area of the door Area of whitewashing the wall = 16.2 m Cost of whitewashing the wall = Area of whitewashing the wall × Rate per m Cost of whitewashing the wall = 14. 2 m Cost of whitewashing the wall = ₹ 284 ## Exercise 11.2
Base of the parallelogram = 7 cm Height of the parallelogram = 4 cm Area of the parallelogram = 7 cm × 4 cm Area of the parallelogram = 28 cm
Base of the parallelogram = 5 cm Height of the parallelogram = 3 cm Area of the parallelogram = 5 cm × 3 cm Area of the parallelogram = 15 cm
Base of the parallelogram = 2.5 cm Height of the parallelogram = 3.5 cm Area of the parallelogram = 2.5 cm × 3.5 cm Area of the parallelogram = 8.75 cm
Base of the parallelogram = 5 cm Height of the parallelogram = 4.8 cm Area of the parallelogram = 5 cm × 4.8 cm Area of the parallelogram = 24 cm
Base of the parallelogram = 2 cm Height of the parallelogram = 4.4 cm Area of the parallelogram = 2 cm × 4.4 cm Area of the parallelogram = 8.8 cm
Base of the triangle = 4 cm Height of the triangle = 3 cm Area of triangle = ½ × 4 cm × 3 cm Area of triangle = ½ × 12 cm Area of triangle = 6 cm
Base of the triangle = 5 cm Height of the triangle = 3.2 cm Area of triangle = ½ × 5 cm × 3.2 cm Area of triangle = ½ × 16 cm Area of triangle = 8 cm
Base of the triangle = 3 cm Height of the triangle = 4 cm Area of triangle = ½ × 3 cm × 4 cm Area of triangle = ½ × 12 cm Area of triangle = 6 cm
Base of the triangle = 3 cm Height of the triangle = 2 cm Area of triangle = ½ × 3 cm × 2 cm Area of triangle = ½ × 6 cm Area of triangle = 3 cm
Area of the parallelogram = Base × Height Base of the parallelogram = 20 cm Height of the parallelogram =? Area of the parallelogram = 246 cm Area of the parallelogram = Base × Height 246 cm Height = 246 cm Height = 12.3 cm
Area of the parallelogram = Base × Height Base of the parallelogram =? Height of the parallelogram = 15 cm Area of the parallelogram = 154.5 cm Area of the parallelogram = Base × Height 154.5 cm Base = 154.5 cm Height = 10.3 cm
Area of the parallelogram = Base × Height Base of the parallelogram =? Height of the parallelogram = 8.4 cm Area of the parallelogram = 48.72 cm Area of the parallelogram = Base × Height 48.72 cm Base = 48.72 cm Height = 5.8 cm
Area of the parallelogram = Base × Height Base of the parallelogram = 15.6 cm Height of the parallelogram =? Area of the parallelogram = 16.38 cm Area of the parallelogram = Base × Height 16.38 cm Height = 16.38 cm Height = 1.05 cm
Area of triangle = ½ × Base × Height Base of the triangle = 15 cm Height of the triangle =? Area of triangle = 87 cm 87 cm Height = (87 cm Height = 174 cm Height = 11.6 cm
Area of triangle = ½ × Base × Height Base of the triangle =? Height of the triangle = 31.4 mm Area of triangle = 1256 mm 1256 mm Base = (1256 mm Base = 2512 mm Base = 80 mm
Area of triangle = ½ × Base × Height Base of the triangle = 22 cm Height of the triangle =? Area of triangle = 170.5 cm 170.5 cm Height = (170.5 cm Height = 341 cm Height = 15.5 cm
Base of the parallelogram = SR = 12 cm Height of the parallelogram = QM = 7.6 cm Area of the parallelogram = 12 cm × 7.6 cm Area of the parallelogram = 91.2 cm
Area of the parallelogram = Base × Height Base of the parallelogram = PS = 8 cm Height of the parallelogram = QN =? Area of the parallelogram = 8 cm × Height 91.2 cm QN = 91.2 cm QN = 11.4 cm
Base of the parallelogram = AB and AD Let's first find the length of the height DL. Base of the parallelogram = AB = 35 cm Height of the parallelogram = DL =? Area of the parallelogram = 35 cm × Height 1470 cm DL = 1470 cm
Base of the parallelogram = AD = 49 cm Height of the parallelogram = BM =? Area of the parallelogram = 49 cm × Height 1470 cm BM = 1470 cm
Area of triangle = ½ × Base × Height Base of the triangle = AB = 12 cm Height of the triangle = AB = 5 cm Area of triangle = ½ × 12 cm × 5 cm Area of triangle = ½ × 60 cm Area of triangle = 30 cm Let's find the height AD. Height of the triangle = AD Base of the triangle = BC = 13 cm Area of triangle = 30 cm 30 cm AD = (30 cm AD = 60 cm AD = 60/13 cm
Area of triangle = ½ × Base × Height Base of the triangle = BC = 9 cm Height of the triangle = AD = 6 cm Area of triangle = ½ × 9 cm × 6 cm Area of triangle = ½ × 54 cm Area of triangle = 27 cm Let's find the height CE. Height of the triangle = CE Base of the triangle = AB = 7.5 cm Area of triangle = 27 cm 27 cm CE = (27 cm CE = 54 cm CE = 7.2 cm ## Exercise 11.3
Where, π = 22/7 r = Radius of the circle r = 14 cm Circumference of the circle = 2 × π × r Circumference of the circle = 2 × 22/7 × 14 Circumference of the circle = 88 cm
Where, π = 22/7 r = Radius of the circle r = 28 mm Circumference of the circle = 2 × π × r Circumference of the circle = 2 × 22/7 × 28 Circumference of the circle = 176 mm
Where, π = 22/7 r = Radius of the circle r = 21 cm Circumference of the circle = 2 × π × r Circumference of the circle = 2 × 22/7 × 21 Circumference of the circle = 132 cm
Where, π = 22/7 r = Radius of the circle r = 14 mm Area of the circle = π × r × r Area of the circle = 22/7 × 14 mm × 14 mm Area of the circle = 616 mm
Where, π = 22/7 r = Radius of the circle D = Diameter of the circle D = 2r 49 m = 2r r = 49/2 m Area of the circle = π × r × r Area of the circle = 22/7 × 49/2 m × 49/2 m Area of the circle = (22 × 49 × 49)/ (7 × 2 × 2) Area of the circle = 1886.5 m
Where, π = 22/7 r = Radius of the circle r = 5 cm Area of the circle = π × r × r Area of the circle = 22/7 × 5 cm × 5 cm Area of the circle = 550/7 cm
Where, π = 22/7 r = Radius of the circle r =? Circumference of the circle = 2 × π × r 154 m = 2 × 22/7 × r r = (154 m × 7)/ (2 × 22) r = 49/2 m r = 24.5 m Area of the circle = πr Area of the circle = π × r × r Area of the circle = 22/7 × 49/2 m × 49/2 m Area of the circle = (22 × 49 × 49)/ (7 × 2 × 2) Area of the circle = 1886.5 m
D = 2r 21 = 2r r = 21/2 m Length of the rope to fence the circular garden = Circumference of the garden Circumference of the circle = 2πr Circumference of the circle = 2 × π × r Circumference of the circle = 2 × 22/7 × 21/2 Circumference of the circle = 66 m Length of the rope to fence 1 round = 66 m Length of the rope to fence 2 rounds = 2 × 66 = 132 m Cost of the rope = Cost per meter × Total length of the rope Cost of the rope = ₹ 4 × 132 m Cost of the rope = ₹ 528
Area of the circle = π × r × r Area of the circle with radius 4 cm = 3.14 × 4 cm × 4 cm Area of the circle with radius 4 cm = 50.26 cm Area of the circle with radius 3 cm = 3.14 × 3 cm × 3 cm Area of the circle with radius 3 cm = 28.27 cm Remaining area of the sheet = Area of the sheet with radius 4 cm - Area of the sheet with radius 3 cm Remaining area of the sheet = 50.26 cm Remaining area of the sheet = 21.98 cm
d = 2r 1.5 = 2r r = 1.5/2 m Length of the lace to fence the circular table = Circumference of the table Circumference of the circle = π d Circumference of the circle = π × d Circumference of the circle = 3.14 × 1.5 Circumference of the circle = 4.71 m Length of the lace required to cover the edge of the table = 4.71 m Cost of the lace = Cost per meter × Total length of the lace Cost of the lace = ₹ 15 × 4.71 m Cost of the lace = ₹ 70.65
Circumference of the circle = 2πr Circumference of the semi-circle = 2πr/2 = πr Diameter = 10 cm D = 2r 2r = 10 r = 10/2 = 5 cm Radius of the circle = 5 cm Circumference of the semi-circle = π × 5 Circumference of the semi-circle = 3.141 × 5 Circumference of the semi-circle = 15.70 cm Perimeter of the semi-circle including its diameter = Circumference of the semi-circle + Diameter = 15.70 cm + 10 cm = 25.70 cm
2r = 1.6 m r = 1.6/2 r = 0.8 m Area of the circular table-top = πr = π × r × r Area of the circular table-top = 3.14 × 0.8 m × 0.8 m Area of the circular table-top = 2.0096 m Cost of polishing the circular table-top = Area of the table-top × Cost per meter square Cost of polishing the circular table-top = 2.0096 m Cost of polishing the circular table-top = = ₹ 30.14
Perimeter of the circle = 44 cm 44 cm = 2πr Where, r = radius of the circle π = 22/7 44 cm = 2 × 22/7 × r r = (44 × 7)/ (2 × 22)
Thus, the radius of the circle is 7 cm. Area of the circle = πr = π × r × r = 22/7 × 7 × 7
Length of the wire bent into the shape of a square = Perimeter of the square Perimeter of the square = 44 cm 44 cm = 4 × (Side of a square) Side of a square = 44/4 cm Side of a square = 11 cm Area of the square = Side × Side Area of the square = 11 cm × 11cm
Area of the circle > Area of the square 154 cm Thus,
Area of the card sheet = πr = π × r × r = 22/7 × 14 × 14
Radius of smaller circle = 3.5 cm Area of the smaller circle = πr = π × r × r = 22/7 × 3.5 cm × 3.5 cm
Area of rectangle = Length × breadth Length of the rectangle = 3 cm Breadth of the rectangle = 1 cm Area of the rectangle = 3 cm × 1 cm
Area of the remaining sheet = Area of the card sheet - Area of the first circle - Area of the second circle - Area of the rectangle Area of the remaining sheet = 616 cm Area of the remaining sheet = 536 cm
Area of the square = Side × Side Area of the square = 6 cm × 6 cm Area of the square = Radius of the circle = 2 cm Area of the circle = πr = π × r × r Area of the circle = 3.14 × 2 cm × 2 cm Area of the circle = 12.56 cm Area of the leftover aluminium sheet = Area of the square - Area of the cut out circle Area of the leftover aluminium sheet = 36 cm Area of the leftover aluminium sheet = 23.44 cm
Circumference of the circle = 2πr 2πr = 31.4 cm 2 × 3.14 × r = 31.4 cm r = (31.4)/ (2 × 3.14) r = 10/2 r = 5 cm Area of the circle = πr = π × r × r Area of the circle = 3.14 × 5 cm × 5 cm Area of the circle = 78.5 cm
Radius of the flower bed = 66/2 = 33 m The width of the surrounded path = 4 m Area of the path = Area of the flower bed with path width - Area of the flower bed without the path width Diameter of the flower bed with path width = 4 + 66 + 4 = 74 m Radius of the flower bed with path width = 74/2 = 37 m Area of the circle = πr = π × r × r Area of the path = Area of the flower bed with path width - Area of the flower bed without the path width Area of the path = 3.14 × 37 × 37 - 3.14 × 33 × 33 Area of the path = 3.14 × (37 × 37 - 33 × 33) Area of the path = 3.14 × (1369 - 1089) Area of the path = 3.14 × 280 Area of the path =
Radius of the circle that a sprinkler can cover = 12 m Area of the circle = πr = π × r × r Area of the circle = 3.14 × 12 × 12 Area of the circle = 452.16 m Area of the circle that a sprinkler can cover is greater than the area of the circular flower garden. Hence, the sprinkler can easily water the entire garden. The answer is '
Radius of the outer circle = 19 m Width of the outer circle = 10 m Radius of the inner circle = Radius of the outer circle - Width of the outer circle Radius of the inner circle = 19 m - 10 m = 9 m Circumference of the inner circle = 2× 3.14 × 9 Circumference of the Circumference of the outer circle = 2× 3.14 × 19 Circumference of the
28 cm = 28/100 = 0.28 m Radius of the wheel = 0.28 m Circumference of the wheel = 2πr Circumference of the wheel = 2× 22/7 × 0.28 Circumference of the wheel = 1.76 m The number of times a wheel can rotate = 352 m/ Circumference of the wheel The number of times a wheel can rotate = 352 m/1.76 m The number of times a wheel can rotate = 200 times
In 1 hour, the minute hand covers the full 12 clock hours, i.e., the full circle. The length of the minute hand moved in hour = Circumference of the clock Circumference of the clock = 2πr Circumference of the clock = 2× 3.14 × 15 Circumference of the clock = 94.2 cm Thus, the length of the minute hand moved in hour is 94.2 cm. ## Exercise 11.4
Breadth of the garden = 75 m Area of the rectangular garden = Length × Breadth Area of the rectangular garden = 90 m × 75 m Area of the rectangular garden = 6750 m Path to be built outside the garden = 5 m Length of the garden with the width = 90 + 5 + 5 = 100 m Breadth of the garden with the width = 75 + 5 + 5 = 85 m Area of the rectangular garden = Length × Breadth Area of the rectangular garden = 100 m × 85 m Area of the rectangular garden = 8500 m Area of the path = Area of the rectangular garden with outside width - Area of the inside rectangular garden Area of the path = 8500 m Area of the path = 1750 m Area of the garden = 6750 m 1 square meter = 0.0001 hectare = 6750 m = 0.675 hectare
Breadth of the park = 65 m Area of the rectangular park = Length × Breadth Area of the rectangular park = 125 m × 65 m Area of the rectangular park = 8125 m Path to be built outside the park = 3 m Length of the park with the width = 125 + 3 + 3 = 131 m Breadth of the park with the width = 65 + 3 + 3 = 71 m Area of the rectangular park = Length × Breadth Area of the rectangular park = 131 m × 71 m Area of the rectangular park = 9301 m Area of the path = Area of the rectangular park with outside width - Area of the inside rectangular park Area of the path = 9301 m Area of the path = 1176 m
Width of the cardboard = 5 cm Area of the cardboard = Length × Width Area of the cardboard = 8 cm × 5 cm Area of the cardboard = 40 cm Margin along the sides of the cardboard = 1.5 cm Length of the painting = 8 cm - 1.5 cm - 1.5 cm = 5 cm Width of the painting = 5 cm - 1.5 cm - 1.5 cm = 2 cm Area of the painting = Length × Width Area = 5 cm × 2 cm Area = 10 cm Area of the margin = Area of the cardboard- Area of the painting Area of the margin = 40 cm Area of the margin = 30 cm
Width of the room = 4m Area of the room = Length × Width Area of the room = 5.5 m × 4 m Area of the room = 22 m Width of the verandah = 2.25 m Length of the room with verandah = 5.5 m + 2.25 m + 2.25 m = 10 m Width of the room with verandah = 4 m + 2.25 m + 2.25 m = 8.5 m Area of the room with verandah = Length × Width = 10 m × 8.5 m = 85 m Area of the verandah = Area of the room with verandah - Area of the room Area of the verandah = 85 m Area of the verandah = 63 m
Cost of cementing the floor of the verandah = 63 m Cost of cementing the floor of the verandah = ₹ 12600 ## Note: When the width is added outside any area, it is added twice to the length and the breadth of the given area. Similarly, if the width lies inside any area, it is subtracted twice from the length and the breadth.
Find:
Area of the garden = Side × Side Area of the garden = 30 m × 30 m Area of the garden = 900 m Width of the path built inside the square garden = 1 m Side of the square built inside the garden = 30 m - 1 m - 1 m = 28 m Area of the inside garden = Side × Side Area of the inside garden = 28 m × 28 m Area of the inside garden = 784 m Area of the path = Area of the garden - Area of the inside garden Area of the path = 900 m Area of the path = 116 m
Cost of planting trees in the remaining portion of the garden = 784 m Cost of planting trees in the remaining portion of the garden = ₹ 31360
Breadth of the rectangular park = 300 m The width of the cross road = 10 m Area of the rectangular park = Length × Width Area of the rectangular park = 700 m × 300 m Area of the rectangular park = 210000 m Length of one side of the cross road = 300 m Width = 10 m Area = Length × Width Area = 300 m × 10 m Area = 3000 m Length of the other side of the cross road = 10 m Width = 700 m Area = Length × Width Area = 10 m × 700 m Area = 7000 m Area of the centre region (common area of the intersection of two crossroads) = 10 m × 10 m Area = 100 m Total area of the two cross roads = 3000 m Total area of the two cross roads = 1 square meter = 0.0001 hectare = 9900 m Total area of the two cross roads = Area of the park excluding the cross road = Total area of the rectangular park - Area of the cross roads Area of the park excluding the cross road = 210000 m Area of the park excluding the cross road = 200100 m 1 square meter = 0.0001 hectare = 200100 m Area of the park excluding the cross road =
Breadth of the rectangular field = 60 m The width of the cross road = 3 m Area of the rectangular park = Length × Width Area of the rectangular park = 90 m × 60 m Area of the rectangular park = 5400 m Length of one side of the cross road = 60 m Width = 3 m Area = Length × Width Area = 60 m × 3 m Area = 180 m Length of the other side of the cross road = 3 m Width = 90 m Area = Length × Width Area = 3 m × 90 m Area = 270 m Area of the centre region (common area of the intersection of two crossroads) = 3 m × 3 m Area = 9 m Total area of the two cross roads = 180 m Total area of the two cross roads =
Cost of constructing the roads = 441 m Cost of constructing the roads = ₹ 48,510
Side of the square = 4 cm The length of the cord required to cover a circular pipe = Perimeter of the circle The length of the cord required to cover a square = Perimeter of the square Perimeter of the circular pipe = 2 π r = Circumference of the circle Perimeter of the circular pipe = 2 × 3.14 × 4 Perimeter of the circular pipe = 25.12 cm So, length of the cord required to cover a circular pipe is 25.12 cm. Perimeter of the square = 4 × Side Perimeter of the square = 4 × 4 Perimeter of the square = 16 cm Perimeter of the square < Perimeter of the circle Chord left = Length of the cord required to cover the circle - Length of the cord required to cover the square Chord left = 25.12 cm - 16 cm Chord left = 9. 12 cm
Length of the rectangle = 10 m Width of the rectangle = 5 m Area of the rectangular lawn = Length × Width Area of the rectangular lawn = 10 m × 5 m Area of the rectangular lawn = 50 m
Area of the circular flower bed = π r Area of the circular flower bed = 3.14 × 2 × 2 Area of the circular flower bed = 12.56 m
Area of the rectangular lawn = 50 m Area of the circular flower bed = 12.56 m Area of the lawn excluding the area of the flower bed = 50 m Area of the whole land = 37.44 m
Circumference of the flower bed = 2 π r Circumference of the flower bed = 2 × 3.14 × 2 Circumference of the flower bed =12.56 m
Length of the rectangle = 18 cm Width of the rectangle = 10 cm Area of the rectangle = Length × Width Area of the rectangle = 18 cm × 10 cm Area of the rectangle ABCD = 180 cm Area of the triangle AEF = ½ × Base × Height Base of the triangle AEF = 10 cm Height of the triangle AEF = 6 cm Area of the triangle AEF = ½ × 10 cm × 6 cm Area of the triangle AEF = 30 cm Area of the triangle BEC = ½ × Base × Height Base of the triangle BEC = 8 cm Height of the triangle BEC =10 cm Area of the triangle BEC = ½ × 8 cm × 10 cm Area of the triangle BEC = 40 cm Area of the shaded portion = Area of the rectangle ABCD - Area of the triangle AEF - Area of the triangle BEC Area of the shaded portion = 180 cm Area of the shaded portion = 110 cm
Length of the rectangle = 20 cm Width of the rectangle = 20 cm Area of the rectangle = Length × Width Area of the rectangle = 20 cm × 20 cm Area of the rectangle PQRS = 400 cm Area of the triangle PQT = ½ × Base × Height Base of the triangle PQT = 20 cm Height of the triangle PQT =10 cm Area of the triangle PQT = ½ × 20 cm × 10 cm Area of the triangle PQT = 100 cm Base of the triangle TUS = 10 cm Height of the triangle TUS =10 cm Area of the triangle TUS = ½ × 10 cm × 10 cm Area of the triangle TUS = 50 cm Base of the triangle QUR = 10 cm Height of the triangle QUR =20 cm Area of the triangle QUR = ½ × 10 cm × 20 cm Area of the triangle QUR = 100 cm Area of the shaded portion = Area of the rectangle PQRS - Area of the triangle PQT - Area of the triangle TUS - Area of the triangle QUR Area of the shaded portion = 400 cm Area of the shaded portion = 150 cm
Area of the triangle ABC = ½ × Base × Height of the perpendicular BM Base of the triangle ABC = 22 cm Height of the triangle ABC =3 cm Area of the triangle ABC = ½ × 22 cm × 3 cm Area of the triangle ABC = 33 cm Area of the triangle ACD = ½ × Base × Height of the perpendicular DN Base of the triangle ACD = 22 cm Height of the triangle ACD =3 cm Area of the triangle ACD = ½ × 22 cm × 3 cm Area of the triangle ACD = 33 cm Area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle ACD Area of the quadrilateral ABCD = 33 cm Area of the quadrilateral ABCD = 66 cm |