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Maths Wiz Class 7 Solutions Chapter 14 Perimeter and Area
Welcome to NCTB Solutions. Here with this post we are going to help 14th class students for the Solutions of Maths Wiz Class 7 Math Book, Chapter 14, Perimeter and Area. Here students can easily find step by step solutions of all the problems for Perimeter and Area, Exercise 14A, 14B and 14C Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 14 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Perimeter and Area Exercise 14A Solution :
Question no – (1)
Solution :
(a) Perimeter PQRST
= 7 + 3 + 5 + 6 + 2
= 23
(b) Perimeter ABCDEF
= 10 + 8 + 6 + 4 + 7 + 15
= 50
(C) Perimeter LMNOPQRSTUVW
= 1 + 2 + 3 + 1 + 3 + 2 + 1 + 2 + 3 + 1 + 3 + 2
= 24
(d) Perimeter of ABCDEFGHIJKLMNOPQRST
= 1 + 3 + 2 + 3 + 4 + 1 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4
= 52
Question no – (2)
Solution :
Let, length of the side = a
Perimeter of polygon = 53 m
According to question –
= 7 + a + 5 + 9 + 4 + 16
= 53
Or, 41 + a
= 53
Or, a = 53 – 41
= 12 m
Therefore, the length of the side a will be 12 m.
Question no – (3)
Solution :
As per the question,
(a) equilateral triangle with side 8 cm
∴ Perimeter of triangle,
= 8 + 8 + 8
= 24 cm
Therefore, the perimeter of an equilateral triangle will be 24 cm.
(b) Square with Side = 10 cm
As we know, Perimeter of a square = 4 × (side length)
∴ Perimeter of square,
= (4 × 10) cm
= 40 cm
Thus, the perimeter of the square will be 40 cm.
(c) Regular pentagon of sides 6.2 cm.
Perimeter of pentagon,
= (6.2 × 8)
= 49.6 cm
Therefore, the perimeter of pentagon will be 49.6 cm.
(d) Regular hexagon with side 9.5 cm
∴ Perimeter of hexagon
= 9.5 + 9.5 + 9.5 + 9.5 + 9.5 + 9.5
= 55 cm
Therefore, the perimeter of hexagon will be 55 cm.
(e) Isosceles triangle with equal sides 5 cm each and third side 7 cm.
∴ Perimeter of Isosceles triangle,
= 5 + 5 + 7
= 17 cm
Therefore, the perimeter of Isosceles triangle will be 17 cm.
Question no – (4)
Solution :
(a) a square, if the perimeter is 36.8 cm
Let, length of square = L
According to question,
L + L + L + L
= 36.8 CM
OR, 4L = 36.8
OR, L = 36/4
= 9.2 cm
Therefore, the length of one side of the square will be 9.2 cm.
(b) an equilateral triangle, if its perimeter is 22.5 cm
Let length of triangle = L
Acceding to question,
L + L + L= 22.5
OR, 3L = 22.5
OR, L = 22.5/3
= L = 7.5 cm
Hence, the length of one side of an equilateral triangle will be 7.5 cm.
(c) a regular octagon, if its perimeter is 42.4 cm
Let, length of octagon = L
∴ According to question,
L + L + L + L + L + L+ L + L = 42.4
or 8L = 42.4
or, L = 42.4/8
= L = 5.3 cm
Thus, the length of one side of a regular octagon is 5.3 cm.
(d) a rhombus, if its perimeter is 25.2 cm
Let, length of octagon = L
∴ According to question,
L + L + L + L + L + L+ L + L + L + L+ L + L
= 25.2
OR, 12L = 25.2
Or, L = 25.2/12
= 6.3 cm
Therefore, the length of one side of the rhombus will be 6.3 cm.
Question no – (5)
Solution :
As per the question the string is 48 cm long,
Now, (a) A squares
= length = 48/4
= 12 cm
(b) An equilateral triangle
= length = 48/3
= 16 cm
(c) A regular hexagon
= length = 48/6
= 8 cm
(d) A regular decagon
= length = 48/10
= 4.8 cm
Question no – (6)
Solution :
Let, equal sides = L
∴ According to question,
L + L + 14 = 50
OR, 2L = 50 – 14
= 36
OR, L = 36/2
= 18 cm
Therefore, the length of each of the equal sides will be 18 cm.
Question no – (7)
Solution :
According to the given question,
(a) Perimeter,
= 8 + 8 + 5 + 5
= 26 cm
(b) perimeter,
= 107 + 10.7 + 7.3 + 7.3
= 36 cm
(c) perimeter,
= 120 + 120 + 90 + 90
= 420 mm
= 42 cm
(d) perimeter,
= 85 + 85 + 70 + 70
= 310 cm
Question no – (8)
Solution :
(a) Let, length = L
∴ According to question,
L + L + 5 + 5
= 32
Or, 2L + 10
= 32
Or, 2L = 32 – 10
= 22
Or, L = 22/2
= 11 cm
Thus, length of the rectangle will be 11 cm.
(b) Let, breadth = B
∴ According to question,
24 + 24 + B +B = 80
or, 48 + 2B = 80
or, 2B = 80 – 48 = 32
B = 32/2
= B = 16 cm.
Therefore, breadth of a rectangle will be 16 cm.
Question no – (9)
Solution :
As per the given question,
Breadth = 15 cm
Length = 3 × 15
= 45 cm
∴ Perimeter,
= 45 + 45 + 15 + 15
= 120 cm
Therefore, the perimeter of the rectangle 120 cm.
Question no – (10)
Solution :
As per the given question,
L : B = 5 : 3
∴ According to question,
5x + 5x + 3x + 3x
= 112
Or, 10x + 6x
= 112
Or, 16x = 112
Or, x = 112/16
= 7
∴ Length,
= 5 × 7
= 35 cm
∴ Breadth
= 3 × 7
= 21 cm.
Question no – (11)
Solution :
First, Perimeter of squares,
= 80 + 80 + 80 + 80
= 320 m²
Second, Perimeter of rectangle,
= 65 + 65 + 55 + 55
= 130 + 110
= 240 m²
∴ 320 m² > 240 m²
Therefore, Prem runs greater distance.
Question no – (12)
Solution :
Perimeter of rectangle,
= 850 + 850 + 70 + 70
= 3400 m.
5 row weirs length,
= 3100 × 5
= 15.500 m.
Cost of fencing,
= 15,500 × 80
= 12,40,000 Rs.
Question no – (13)
Solution :
Perimeter of rectangular park,
= 100 + 100 + 80 + 80
= 360 m
He covered total are at 6 round,
= 360 × 6
= 21 60 m
Speed of Mridul,
= 9 km/hour
= 9000/3600
= 2.5 m/s
He covered the distance,
= 2160/2.5
= 865 second
= 14 minute 25 second.
Question no – (14)
Solution :
Length of garden = 150
Breadth = 120
∴ Perimeter = 2 × (length + breadth)
= 2 × (150 + 120)
= 2 × 270
= 540
= 2700/540 means 2700 divided by 540 = 5
Therefore, Prashant take 5 rounds of the garden.
Question no – (15)
Solution :
Perimeter of packet,
= 35 + 35 + 25 + 25
= 120
Length of string,
= 120 + 8
= 128 cm
Therefore, the length of the string is 128 cm.
Perimeter and Area Exercise 14B Solution :
Question no – (1)
Solution :
(a) Given, Length is = 8 cm
Breadth is = 4 cm
Area = ?
As we know, Area = Length × Breadth
∴ Now the Area,
= 8 × 4 = 32 cm²
Thus, area of the rectangle will be 32 cm²
(b) Length is = 12 m
Breadth is = 6 m
Area = ?
As we know, Area = Length × Breadth
∴ Now the Area,
= 12 × 6
= 72 m²
Thus, the area of rectangle will be 72 m²
(c) Length is = 8 cm
Breadth is = 4 cm
Area = ?
As we know, Area = Length × Breadth
∴ Now the Area,
= 10 × 7.5
= 75 cm²
Hence, area of the rectangle will be 75 cm²
(d) Given, Length is = 7 mm
Breadth is = 14 mm
Area = ?
As we know, Area = Length × Breadth
∴ Now the Area,
= 7 × 14
= 98 mm²
Therefore, area of the rectangle will be 98 mm²
Question no – (2)
Solution :
(a) Each side = 3 cm
Area = ?
As we know, Area of square = Side × Side
∴ Now the Area of square,
= 3 × 3
= 9 cm²
Therefore, area of the square will be 9 cm.
(b) Each side = 10 cm
Area = ?
As we know, Area of square = Side × Side
∴ Now the Area of square,
= 10 × 10
= 100 m²
Therefore, area of the square will be 100 m²
(c) Each side = 5 mm
Area = ?
As we know, Area of square = Side × Side
∴ Now Area of square,
= 5 × 5
= 25 mm²
Therefore, area of the square will be 25 mm²
(d) Each side = 1.6 cm
Area = ?
As we know, Area of square = Side × Side
∴ Area of square,
= 1.6 × 1.6
= 2.56 cm²
Therefore, area of the square will be 2.56 cm²
Perimeter and Area Exercise 14C Solution :
Question no – (1)
Solution :
(a) Given, Length is = 17 cm
Breadth is = 10 cm
Area = ?
As we know, Area of rectangle = Length × Breadth
∴ Now Area of rectangle,
= 17 cm × 10 cm
= 170 cm2
Therefore, area of the rectangle is 170 cm2
(b) Given, Length is = 5 m
Breadth is = 8 m
Area = ?
As we know, Area = Length × Breadth
∴ Area of the wall,
= 5 m × 8 m
= 40 m2
Therefore, area of the rectangle will be 40 m2
(c) Given, Length is = 200 m
Breadth is = 120 m
Area = ?
As we know, Area = Length × Breadth
∴ Area of the field,
= 200 m × 120 m
= 24000 m2
Therefore, area of the field will be 24000 m2
(d) Each side = 1.5 m.
Area = ?
As we know, Area of square = Side × Side
∴ Now Area of square,
= 1.5 m × 1.5 m
= 2.25 m2
Hence, area of the square board will be 2.25 m2
(e) Each side = 6.3 km
Area = ?
As we know, Area of square = Side × Side
∴ Area of square,
= 6.3 km × 6.3 km
= 69.69 km2
Therefore, area of the square lawn is 69.69 km2
Question no – (2)
Solution :
(a) Area = 90 sq.cm
length = 15 cm
∴ Breadth,
= 90/15
= 6 cm
∴ Perimeter,
= 15 + 15 + 6 + 6
= 42 cm
Therefore, the breadth will be 6 cm and perimeter will be 42 cm.
(b) Given, Area = 117 sq.cm;
length = 13 cm
∴ Breadth,
= 117/13
= 9 cm
∴ Perimeter,
= 13 + 13 + 9 + 9
= 44 cm
Hence, the breadth will be 9 cm and perimeter will be 44 cm.
Question no – (3)
Solution :
(a) Area = 42 sq.cm;
breadth = 6 cm
∴ Length,
= 42/6
= 7 cm
∴ Perimeter,
= 7 + 7 + 6 + 6
= 26 cm
Hence, the length will be 7 cm and perimeter will be 26 cm.
(b) Area = 480 sq.cm;
breadth = 20 cm
∴ Length,
= 480/20
= 24 cm
∴ Perimeter,
= 24 + 24 + 20 + 20
= 88 cm
Hence, the length will be 24 cm and perimeter will be 88 cm.
Question no – (4)
Solution :
(a) Given Area = 49 cm2
∴ 49 cm2 = (7 × 7) = L × B
∴ Perimeter,
= 7 + 7 + 7 + 7
= 28 cm
Thus, the perimeter of the square is 28 cm.
(b) area = 144 cm²
∴ 144 cm²
= (12 × 12) cm
= L × B
∴ Perimeter
= 12 + 12 + 12 + 12
= 48 cm
Hence, the perimeter of the square will be 48 cm.
(c) Area = 256 cm²
∴ 256 cm² = (16 ×16) = L × B
∴ Perimeter
= 16 + 16 + 16 + 16
= 64 cm
Therefore, the perimeter of the square Will be 64 cm.
(d) Area = 900 cm²
∴ 900 cm² = (30 × 30) = L × B
∴ Perimeter
= 30 + 30 + 30 + 30
= 120 cm
Therefore, the perimeter of the square will be 120 cm.
Question no – (5)
Solution :
Total floor area of room
= 7 × 7
= 49 m²
Area of carpet
= 5 × 4
= 20 m2
∴ Uncovered of carpet room,
= 49 – 20
= 29 m2
Hence, the area of the floor left uncovered is 29 m2
Question no – (6)
Solution :
Area = 10 × 8
= 80 m²
Length = 10 + 2 = 12 m
Breadth = 8 + 10 = 10 m ….(Lawn with path)
∴ Lawn with path’s area,
= 12 × 10
= 120 m²
∴ The area of the path,
= 120 – 80
= 40 m²
Therefore, the area of the path will be 40 m²
Question no – (7)
Solution :
Picture area,
= 30 × 24
= 720 cm²
∴ Length of card with picture,
= 30 + (3 + 3)
= 36 cm
∴ Breadth of card with picture,
= 24 + (3 + 3)
= 30 cm
∴ Area of card with picture,
= 36 × 30
= 1080 cm²
∴ Only margin are picture,
= 1080 – 720
= 360 cm²
Hence, the area of the margin will be 360 cm²
Question no – (8)
Solution :
Courtyard area,
= 40 × 24
= 960 m²
∴ With path courtyard length,
= 40 – 6
= 34 m
∴ With path courtyard breadth,
= 24 – 6
= 18 m
∴ With path courtyard area,
= 34 × 18
= 612 m²
∴ Paved path’s area,
= 960 – 612
= 348 m²
Question no – (9)
Solution :
As we know that, 1 hectares = 10000 m²
6 hectares,
= 6 × 10000
= 60000 m²
Landing area,
= 40 × 15
= 600 m²
Number of garden,
= 60000/600
= 100
Therefore, 100 gardens can be made in 6 hectares of land.
Question no – (10)
Solution :
Brick’s area,
= 22 × 15
= 330 cm²
Court’s area,
= 30 × 22
= 660 m²
= 66000 cm²
Number of brick’s,
= 66000/330
= 200 × 100
= 20000
Therefore, 20000 bricks will be required.
Question no – (11)
Solution :
Floor area = 9 m × 6 m
= 54 m2
= 5400 cm²
∴ Tiles area
= 25 × 30
= 750 cm²
∴ No of tiles
= 5400/750
= 7.2 cm
∴ Cost of tiles
= 7.2 × 200
= 1440 Rs.
Hence, the cost of covering the floor of the room will be 1440 Rs.
Question no – (12)
Solution :
Given, Dimensions : 7 m long, 5 m wide and 3m high
∴ 4 walls area of room,
= 2 × 3 (7 + 5)
= 72 m2
Therefore, the total area of the four walls of room is 72 m2
Question no – (13)
Solution :
Area of total room
= 2 × 4 (8 + 6)
= 8 × 14
= 112 m²
∴ Area of paper,
= 0.8 × 5
= 4 m²
∴ Paper required,
= 112/4
= 28
∴ Cost of paper required,
= 28 × 120
= 3360
Thus, the cost of papering the room will be 3360 Rs.
Question no – (14)
Solution :
Area of room
= 5 × 4.8
= 24 m²
∴ Area of carpet,
= 1.2 × 5
= 6 m²
∴ No of carpet covered room,
= 24/6
= 4
∴ Length of 4 carpet,
= 4 × 5
= 20 m
∴ Cost of 20 m carpet,
= 150 × 20
= 3000 Rs.
Therefore, the cost to carpet the room will be 3000 Rs.
Question no – (15)
Solution :
Figure – (a)
Area = (5 × 4) + (7 × 20
= 20 + 14
= 34 m2
Figure – (b)
Area = (10 × 3) + (14 × 3) + (18 × 3)
= 126 cm2
Figure – (c)
Area = (9 × 4) + (2 × 6) + (12 × 5)
= 108 cm2
Question no – (16)
Solution :
Figure – (a)
Area = (15 × 6) – (7 × 2)
= 90 – 14
= 76 cm2
Figure – (b)
Area = (20 × 16) – {20 – (4 + 4)} × {16 – (3 + 5)}
= 320 – [(20 – 8) × (16 – 8)]
= 320 – (12 × 8)
= 320 – 96
= 224 cm2
Figure – (c)
Area = {(6 + 4 + 6) × (5 + 3 + 5)} – {4 × {6 × 5}]
= (16 × 13) – (4 × 30)
= 208 – 120
= 88 cm2
Figure – (d)
Area = [(5 + 2 + 4 + 2 + 5) × (4 + 3 + 4)] – [(5 × 4) + (4 × 4) + (5 × 4) + {4 × (18 – 6)]
= (18 × 11) – (20 + 16 + 20 + 48)
= 198 – 104
= 94 cm2
Question no – (17)
Solution :
Area of 4 flower beds
= (5 × 4) × 4
= 20 × 4
= 80 m²
∴ Total area
= (5 + 2 + 5) × (4 + 2 + 4)
= 12 × 10
= 120 m2
Total grass area,
= 120 – 80
= 40 m²
If wide of gravel is 3 m
∴ Then, Total area with gravel path
= (12 + 6) × (10 + 6)
= 18 × 16
= 288 m²
∴ Area of gravel path
= 288 – 120
= 168 m²
Therefore, the area of the gravel path will be 168 m²
Next Chapter Solution :
👉 Chapter 15 👈
