Selina Solutions Concise Maths Class 7 Chapter 20 Mensuration (Perimeter and Area of Plane Figures) has important formulas and shortcut methods explained in simple language. The motto of providing chapter wise solutions is to boost the exam preparation of students. For a better academic score, Selina Solutions Concise Maths Class 7 Chapter 20 Mensuration (Perimeter and Area of Plane Figures), PDF links are provided below with a free download option.
Chapter 20 deals with the method of finding the perimeter and area of figures like rectangle, square and triangle. The important formulas and the method of solving problems are given in a stepwise manner, for a better conceptual knowledge among students.
Selina Solutions Concise Maths Class 7 Chapter 20: Mensuration (Perimeter and Area of Plane) Download PDF
Exercises of Selina Solutions Concise Maths Class 7 Chapter 20 – Mensuration (Perimeter and Area of Plane)
Access Selina Solutions Concise Maths Class 7 Chapter 20: Mensuration (Perimeter and Area of Plane)
Exercise 20A page: 220
1. The length and the breadth of a rectangular plot are 135 m and 65 m. Find, its perimeter and the cost of fencing it at the rate of ₹ 60 per m.
Solution:
It is given that
Length of a rectangular plot = 135 m
Breadth of a rectangular plot = 65 m
We know that
Perimeter of a rectangular plot = 2 (length + breadth)
Substituting the values
= 2 (135 + 65)
= 2 (200)
= 400 m
Here the cost of fencing = ₹ 60 per m
So the cost of fencing 400 m = 60 × 400 = ₹ 24,000
2. The length and breadth of a rectangular field are in the ratio 7 : 4. If its perimeter is 440 m, find its length and breadth.
Also, find the cost of fencing it @ ₹150 per m.
Solution:
It is given that
Perimeter of rectangular field = 440 m
Consider 7x as the length and 4x as the breadth of rectangular field
So we get
2 (l + b) = Perimeter
Substituting the values
2 (7x + 4x) = 440
By further calculation
2 (11x) = 440
22x = 440
So we get
x = 440/22 = 11 m
Here
Length = 7x = 7 × 11 = 77m
Breadth = 4x = 4 × 11 = 44m
We know that
Cost of fencing = ₹150 per m
So the cost of fencing 440 m = 150 × 440 = ₹ 66,000
3. The length of a rectangular field is 30 m and its diagonal is 34 m. Find the breadth of the field and its perimeter.
Solution:
It is given that
Length of a rectangular field = 30 m
Diagonal of a rectangular field = 34 m
Consider the breadth of rectangular field = b m
Using the Pythagoras theorem
AC2 = AB2 + BC2
Substituting the values
342 = 302 + b2
By further calculation
1156 = 900 + b2
So we get
b2 = 1156 – 900 = 256
b = √256 = 16 m
We know that
Perimeter = 2 (l + b)
Substituting the values
= 2 (30 + 16)
= 2 × 46
= 92 m
4. The diagonal of a square is 12 √2 cm. Find its perimeter.
Solution:
It is given that
Diagonal of a square = 12 √2 cm
We know that diagonal = side × √2
Here the side = 12 cm
So the perimeter = 4 × 12 = 48 cm
5. Find the perimeter of a rectangle whose length = 22.5 m and breadth = 16 dm.
Solution:
It is given that
Length = 22.5 m
Breadth = 16 dm = 1.6 m
We know that
Perimeter of a rectangle = 2 (l + b)
Substituting the values
= 2 (22.5 + 1.6)
So we get
= 2 (24.1)
= 48.2 m
6. Find the perimeter of a rectangle with length = 24 cm and diagonal = 25 cm.
Solution:
It is given that
Length = 24 cm
Diagonal = 25 cm
Consider the breadth of a rectangle = b m
Using Pythagoras theorem in triangle ABC
AC2 = AB2 + BC2
Substituting the values
252 = 242 + b2
625 = 576 + b2
By further calculation
b2 = 625 – 576 = 49
b = √49 = 7 cm
Here the perimeter of rectangle = 2 (l + b)
Substituting the values
= 2 (24 + 7)
So we get
= 2 (31)
= 62 cm
7. The length and breadth of rectangular piece of land are in the ratio of 5 : 3. If the total cost of fencing it at the rate of ₹48 per metre is ₹19,200, find its length and breadth.
Solution:
It is given that
Length and breadth of rectangular piece of land are in the ratio = 5 : 3
Cost of fencing = ₹19,200
Rate = ₹48 per metre
We know that
Perimeter of rectangular piece of land = 19,200/48 = 400 m
Consider length = 5x
Breadth = 3x
So the perimeter = 2 (l + b)
Substituting the values
400 = 2 (5x + 3x)
By further calculation
400 = 2 (8x)
400 = 16x
So we get
x = 400/16 = 25
Length = 5x = 5 × 25 = 125 m
Breadth = 3x = 3 × 25 = 75 m
8. A wire is in the shape of square of side 20 cm. If the wire is bent into a rectangle of length 24 cm, find its breadth.
Solution:
It is given that
Side of square = 20 cm
So the perimeter of square = 4 × 20 = 80 cm
Perimeter of rectangle = 80 cm
We know that
Length of rectangle = 24 cm
So the perimeter of rectangle = 2 (l + b)
Substituting the values
80 = 2 (24 + b)
By further calculation
40 = 24 + b
b = 40 – 24 = 16 m
9. If P = perimeter of a rectangle, l= its length and b = its breadth; find:
(i) P, if l = 38 cm and b = 27 cm
(ii) b, if P = 88 cm and l = 24 cm
(iii) l, if P = 96 m and b = 28 m
Solution:
(i) It is given that
l = 38 cm
b = 27 cm
We know that
Perimeter = 2 (l + b)
Substituting the values
= 2 (38 + 27)
= 2 (65)
= 130 cm
(ii) It is given that
P = 88 cm
l = 24 cm
Consider b as the breadth
We know that
P = 2 (l + b)
It can be written as
b = P/2 – l
Substituting the values
b = 88/2 – 24
b = 44 – 24
b = 20 cm
(iii) It is given that
P = 96 m
B = 28 m
Consider l as the length
We know that
P = 2 (l + b)
It can be written as
l = P/2 – b
Substituting the values
l = 96/2 – 28
l = 48 – 42
l = 20 m
10. The cost of fencing a square field at the rate of ₹75 per meter is ₹67,500. Find the perimeter and the side of the square field.
Solution:
Cost of fencing = ₹67,500
So the length of fence = 67,500/75 = 900 m
We know that the perimeter of square field = length of fence = 900 m
Here
Perimeter of a square = 4 × Length of its side
Substituting the values
Length of the side of a square = Perimeter/ 4
So we get
= 900/4
= 225 m
11. The length and the breadth of a rectangle are 36 cm and 28 cm. If its perimeter is equal to the perimeter of a square, find the side of the square.
Solution:
It is given that
Length of a rectangle = 36 cm
Breadth of a rectangle = 28 cm
We know that
Perimeter = 2 (l + b)
Substituting the values
= 2 (36 + 28)
= 2 (64)
= 128 cm
It is given that
Perimeter of a square = Perimeter of a rectangle = 128 cm
So the side of square = perimeter/4
Substituting the value
= 128/ 4
= 32 cm
12. The radius of a circle is 21 cm. Find the circumference (Take π = 3 1/7).
Solution:
It is given that
Radius of a circle = 21 cm
We know that π = 22/7
So the circumference of a circle = 2 πr
Substituting the values
= 2 × 22/7 × 21
So we get
= 2 × 22 × 3
= 132 cm
13. The circumference of a circle is 440 cm. Find its radius and diameter. (Take π = 22/7).
Solution:
It is given that
Circumference of a circle = 440 cm
So the radius = C/ 2π
Substituting the values
= (440 × 7)/ (2 × 22)
So we get
= 3088/44
= 70 cm
Diameter of the circle = 2 × radius
So we get
= 2 × 70
= 140 cm
14. The diameter of a circular field is 56 m. Find its circumference and cost of fencing it at the rate of ₹80 per m. (Take π = 22/7).
Solution:
It is given that
Diameter of a circular field = 56 m
So the radius = 56/2 = 28 m
We know that
Circumference of the circle = 2 πr
Substituting the values
= 2 × 22/7 × 28
So we get
= 2 × 22 × 4
= 176 m
Here the cost of fencing 176 m = 176 × 80 = ₹ 14,080
15. The radii of two circles are 20 cm and 13 cm. Find the difference between their circumferences. (Take π = 22/7).
Solution:
It is given that
Radius of first circle = 20 cm
We know that
Circumference of the circle = 2 πr
Substituting the values
= 2 × 22/7 × 20
So we get
= 880/7
= 122.8 cm
Similarly
Radius of the second circle = 13 cm
We know that
Circumference of the circle = 2 πr
Substituting the values
= 2 × 22/7 × 13
So we get
= 572/7
= 81.7
So the difference of circumference of two circles = 122.8 – 81.7 = 41.1 cm
Exercise 20B page: 228
1. Find the area of a rectangle whose length and breadth are 25 cm and 16 cm.
Solution:
It is given that
Length of a rectangle = 25 cm
Breadth of a rectangle = 16 cm
We know that
Area of a rectangle = l × b
Substituting the values
= 25 × 16
= 400 cm2
2. The diagonal of a rectangular board is 1 m and its length is 96 cm. Find the area of the board.
Solution:
It is given that
Length of rectangular board = 96 cm
Diagonal of rectangular board = 1 m = 100 cm
Using the Pythagoras theorem in right angled triangle ABC
AC2 = AB2 + BC2
Substituting the values
1002 = 962 + BC2
By further calculation
10000 = 9216 + BC2
So we get
BC2 = 10000 – 9216 = 784
BC = √784 = 28 cm
Here the area of rectangular board = l × b
Substituting the values
= 96 × 28
= 2688 cm2
3. The sides of a rectangular park are in the ratio 4 : 3. If its area is 1728 m2, find:
(i) its perimeter
(ii) cost of fencing it at the rate of ₹40 per meter.
Solution:
It is given that
Ratio in the sides of a rectangular park = 4 : 3
Area = 1728 m2
Consider 4x as the length and 3x as the breadth
We know that
Area = l × b
Substituting the values
1728 = 4x × 3x
By further calculation
12x2 = 1728
x2= 1728/12 = 144
x = √144 = 12
Here we get
Length = 4x = 4 × 12 = 48 m
Breadth = 3x = 3 × 12 = 36 m
(i) We know that
Perimeter = 2 (l + b)
Substituting the values
= 2 (48 + 36)
So we get
= 2 (84)
= 168 m
(ii) It is given that
Rate of fencing = ₹40 per meter
So the total cost of fencing = 168 × 40 = ₹ 6720
4. A floor is 40 m long and 15 m broad. It is covered with tiles, each measuring 60 cm by 50 cm. Find the number of tiles required to cover the floor.
Solution:
Given below are the dimensions of floor
Length = 40 m
Breadth = 15 m
We know that
Area = l × b = 40 × 15 = 600 m2
Here the length of one tile = 60 cm = 6/10 m
Breadth of one tile = 50 cm = 5/10 m
So the area of one tile = 6/10 × 5/10 = 30/100 = 3/10 m2
We know that
Number of tiles = total area of floor/ area of one tile
Substituting the values
= 600/ 3/10
= (600 × 10)/ 3
= 2000
5. The length and breadth of a rectangular piece of land are in the ratio 5 : 3. If the total cost of fencing it at the rate of ₹24 per meter is ₹9600, find its:
(i) length and breadth
(ii) area
(iii) cost of levelling at the rate of ₹60 per m2.
Solution:
It is given that
Ratio in length and breadth of a rectangular piece of land = 5 : 3
Cost of fencing = ₹9600
Rate = ₹24 per meter
We know that
Perimeter = total cost of fencing/ rate per meter
Substituting the values
= 9600 /24
= 400 m
Consider 5x as the length and 3x as the breadth
Here perimeter = 2 (l + b)
Substituting the values
400 = 2 (5x + 3x)
By further calculation
400 = 2 (8x)
So we get
400 = 16x
x = 400/16 = 25
(i) Length of land = 5x = 5 × 25 = 125 m
Breadth of land = 3x = 3 × 25 = 75 m
(ii) Area = l × b
Substituting the values
= 125 × 75
= 9375 m2
(iii) Cost of levelling at the rate of ₹60 per m2 = 60 × 9375 = ₹ 5,62,500
6. Find the area of the square whose perimeter is 56 cm.
Solution:
It is given that
Perimeter of the square = 56 cm
We know that
4 × side = 56 cm
So we get
Side = 56/4 = 14 cm
Here
Area of the square = side2
Substituting the values
Area of the square = 142 = 196 cm2
7. A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m2, find the area of the lawn.
Solution:
It is given that
Area of the path = 165 m2
Width of the path = 2.5 m
Consider x m as the side of square lawn
So the outer side = x + 2 × 2.5
We get
Outer side = (x + 5) m
Here the area of path = (x + 5)2 – x2
Substituting the values
x2 + 10x + 25 – x2 = 165
By further calculation
10x = 165 – 25 = 140
So we get
x = 140/10 = 14 m
Here the side of lawn = 14 m
Area of the lawn = 142 = 196 m2
8. For each figure, given below, find the area of shaded region: (All measurements are in cm)
Solution:
(i) It is given that
Outer length = 20 cm
Outer breadth = 16 cm
Outer area = l × b
Substituting the values
= 20 × 16
= 320 cm2
Similarly
Inner length = 15 cm
Inner breadth = 10 cm
So the inner area = l × b
Substituting the values
= 15 × 10
= 150 cm2
Here the area of shaded region = area of whole region – area of unshaded region
Substituting the values
= 320 – 150
= 170 cm2
(ii) It is given that
Outer length = 30 cm
Outer breadth = 20 cm
Outer area = l × b
Substituting the values
= 30 × 20
= 600 cm2
Similarly
Inner length = 12 cm
Inner breadth = 12 cm
So the inner area = l × b
Substituting the values
= 12 × 12
= 144 cm2
Here the area of shaded region = area of outer figure – area of inner figure
Substituting the values
= 600 – 144
= 456 cm2
(iii) Here the area shaded portion = area of outer region – area of unshaded region
Substituting the values
= 40 × 40 – 32 × 15
So we get
= 1600 – 480
= 1120 cm2
(iv) Here the area of shaded region = area of outer region – area of inner region
Substituting the values
= 40 × 40 – 15 × 15
So we get
= 1600 – 225
= 1375 cm2
(v) Area of shaded portion = 2 × 20 + 2 × 8 + 2 × (12 + 2)
By further calculation
= 40 + 16 + 28
= 84 cm2
9. One side of a parallelogram is 20 cm and its distance from the opposite side is 16 cm. Find the area of the parallelogram.
Solution:
We know that
Area of parallelogram = base × height
Here
Area of parallelogram = AB × DE
Substituting the values
= 20 × 16
= 320 cm2
10. The base of a parallelogram is thrice it height. If its area is 768 cm2, find the base and the height of the parallelogram.
Solution:
It is given that
Area of parallelogram = 768 cm2
Consider x as the height and 3x as the base of parallelogram
So we get
Area = base × height
Substituting the values
768 = 3x × x
By further calculation
768 = 3x2
x2 = 768/3 = 256 cm
x = √ 256 = 16 cm
Height = x = 16 cm
Base = 3x = 3 × 16 = 48 cm
11. Find the area of the rhombus, if its diagonals are 30 cm and 24 cm.
Solution:
It is given that
Diagonal d1 = 30 cm
Other diagonal d2 = 24 cm
Consider AC and BD as the diagonals of rhombus
We know that
Area = ½ × product of diagonals
So we get
= ½ × AC × BD
= ½ × d1 × d2
Substituting the values
= ½ × 30 × 24
= 15 × 24
= 360 cm2
Hence, the area of the rhombus is 360 cm2.
12. If the area of a rhombus is 112 cm2 and one of its diagonals is 14 cm, find its other diagonal.
Solution:
It is given that
Area of a rhombus = 112 cm2
One diagonal = 14 cm
Consider x cm as the second diagonal
We know that
Area = product of diagonal/2
Substituting the values
112 = (14 × x)/ 2
By further calculation
x = (112 × 2)/ 14
So we get
x = 224/14 = 16 cm
Hence, the other diagonal of the rhombus is 16 cm.
13. One side of a parallelogram is 18 cm and its area is 153 cm2. Find the distance of the given side from its opposite side.
Solution:
It is given that
Area of parallelogram = 153 cm2
One side = 18 cm
So the distance between AB and DC = area/ base
Substituting the values
= 153/ 18
= 17/2
= 8.5 cm
14. The adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 6 cm, find the distance between the shorter sides.
Solution:
It is given that
AB = DC = 15 cm
BC = AD = 10 cm
Here the distance between longer sides AB and DC = 6 cm
So perpendicular DL = 6 cm
DM is perpendicular to BC
We know that
Area of parallelogram = base × altitude
So we get
= AB × DL
Substituting the values
= 15 × 6
= 90 cm2
Consider DM = x cm
Similarly
Area of parallelogram ABCD = BC × DM
Substituting the values
= 10 × x
= 10 x cm2
By equating both
10 x = 90
x = 90/10 = 9 cm
15. The area of a rhombus is 84 cm2 and its perimeter is 56 cm. Find its height.
Solution:
It is given that
Area of a rhombus = 84 cm2
Perimeter of a rhombus = 56 cm
So the side of a rhombus = 56/4 = 14 cm
Here the height = area/ base
Substituting the values
= 84/14
= 6 cm
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