Number Line Prime Class 7 Solutions Chapter 21 Perimeter and Area
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Number Line Prime Class 7 Math Book, Chapter 21, Perimeter and Area. Here students can easily find step by step solutions of all the problems for Perimeter and Area, Exercise 21A, 21B, 21C, 21D and 21E 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 21 solutions. Here in this post all the solutions are based on latest Syllabus.
Perimeter and Area Exercise 21A Solution :
Question no – (1)
Solution :
(a) Side = 4 cm
∴ Perimeter of square – 4 × 4
= 16 cm
(b) Length = 5 cm
Breadth = 4 cm
∴ Perimeter of rectangle = 2 (length + breadth)
= 2 (5 + 4)
= 2 × 9
= 18 cm
Question no – (2)
Solution :
l = 40 cm, b = 10 cm
∴ The length of the tape,
= 2(40 + 10)
= 20 × 50
= 100 cm
Question no – (3)
Solution :
l = 7m, b = 5m
∴ Perimeter of rectangle
= 2(7 + 5)
= 24 m
∴ Each side is to be fenced with 4 rows
∴ The length of the wire needed,
= 4 × 24 m
= 96 m
Question no – (4)
Solution :
(a) Perimeter of rectangle = Sum of the sides
= (3 + 4 + 5) cm
= 12 cm
(b) Perimeter of a equilateral triangle
∴ Perimeter of a square,
= 4 × Side
= 4 × 9 cm
= 36 cm
(c) Perimeter of isosceles triangle,
= 8 + 8 + 6 cm
= 22 cm
Question no – (5)
Solution :
The string is 36 cm long
(a) The length for a square,
= 36/4 cm
= 9 cm
(b) The length for equilateral triangle,
= 36/3 cm
= 12 cm
Perimeter and Area Exercise 21B Solution :
Question no – (1)
Solution :
As we know that,
Area of a rectangle = length × breadth
(a) l = 4 cm b = 3 cm
∴ Area – (4 × 3)cm
= 12 cm
(b) l = 3 m, b = 2 m
∴ Area = (3 × 2) m
= 6 m
Question no – (2)
Solution :
∴ Area of a square – (side) 2
(a) Given, 10 cm
∴ S = 10 cm
∴ Area = (10)2
= 100 cm2
(b) Given, 6 m
∴ S = 6 m
∴ Area = (6)2
= 36 m2
Question no – (3)
Solution :
Given, l = 5 m, b = 4 m
Area of the floor
= 2(5 + 4)
= 2 × 9
= 18 m2
Area of the carpet
= (3)2 m2
= 9 m2
∴ Area of the floor that is not carpeted
= (18 – 9) m2
= 9 m2
Question no – (4)
Solution :
∴ Area of the floor,
= (2 × 1.5) m2
= 3 m2 × 10000
= 30000 m2
∴ Area of the tiles ,
= (25 × 25)
= 625 cm2
∴ There are = 48/12
= 4 dozen tiles
∴ Cost of the tiles
= (420 × 4)
= 1680
Question no – (5)
Solution :
∴ Area of 2 walls
= 10.2/10 × 4.6/10
= 46.92 m2
∴ Area of the 2 walls
= 4.6 × 8.5
= 39.10 m2
∴ Total area
= (46.92 + 39.10)
= 8602 m2
∴ Area of the door
= (1.5 × 1.6) m2
= 3.9 m2
Area of the window
= (2 × 1.7) m2
= 3.4 m2
∴ Area of wall we need to tiling,
= [86.02 – (3.9 + 3.4)]
= 86.02 – 7.3
= 78.72 m2
∴ Area of tiling the walls,
= (78.72 × 13.50)
= 1062.72
Question no – (6)
Solution :
Given, L = 750 m, b = 250m
∴ Area of the playground,
= (750 × 250)
= 187500
∴ Cost of levelling at 16 rupees per 100 sqm
= 187500 × 16/100
= 30000 Rs
Perimeter and Area Exercise 21C Solution :
Question no – (1)
| Sl.No. | Base | Height | Area of
Parallelogram |
| (a) | 10cm | 3.6cm | |
| (b) | 9 cm | 136.8 cm2 | |
| (c) | 7.8 cm | 50.7 cm2 | |
| (d) | 7.5 m | 528 cm |
Solution :
∴ Area of parallelogram = base × height
(a) b = 10 cm, H = 3.6 cm
∴ Area = 10 × 3.6/10 cm2
= 36 m2
(b) A = 136.8 cm2, h = 9 cm
∴ b = A/h = 136.8/9
= 15.9 cm
(c) b = 7.8, A = 50.7 cm2
∴ h = 50.7/7.8
= 6.5 cm
(d) b = 7.5 m, h = 528, cm = 5.28 cm
∴ A = 7.5 × 5.28
= 39.60 m2
Question no – (2)
Solution :
(a) b= 8.4 cm, h = 3cm
∴ Area = 8.4 × 3 = 25.2 cm2
(b) Base = 12 cm, h = 608 cm
∴ Area = 12 × 6.8 = 81.6 cm2
Question no – (3)
Solution :
b = 5 cm, h = 3 cm
∴ Area = 5 × 3
= 15 cm2
Question no – (4)
Solution :
b = 12 m, h = 5 m
∴ Area = 12 × 5
= 60 m2
∴ Cost rate = 60 × 6.50/100
= 390
Question no – (5)
Solution :
b = 12 cm, A = 108 cm2
∴ The altitude = 108/12
= 9 cm
And b = 8 cm, A = 108 cm2
∴ The altitude = 108/8
= 13.5 cm
Perimeter and Area Exercise 21D Solution :
Question no – (1)
| S.NO. | base | Height | Area of Triangle |
| (a) | 20.75 cm | 257.20 cm2 | |
| (b) | 6.4 m | 225 cm | |
| (c) | 38 cm | 494 cm2 | |
| (d) | 15 m | 285 m2 |
Solution :
As we know that,
Area of a triangle = 1/2 × base × height
(a) b = 20.75 cm, A = 257.20 cm2
∴ A = 1/2 (b × h)
∴ b × h = 2A
∴ hb = 2A/b
∴ hb = 2 × 257.20/20.75
= 24.79
(b) b = 6.4 m, h = 225, cm = 2.25 m
∴ A = 1/2 (6.4 × 2.25)
= 1/2 × 14.4
= 7.2 m2
(c) b = 38 cm, A = 494 cm2
∴ h = 2A/b
∴ h = 2 × 494/38
∴ h = 26 cm
(d) b = 15 m, A = 285 m2
∴ h = 2A/b
∴ h = 285/15
∴ h = 19 m
Question no – (2)
Solution :
(a) b = 9.5 cm, h = 7.3 cm
∴ A = 1/2 × 9.5 × 7.3
∴ A = 34.675 cm2
(b) b = 4.9 cm, h = 10.2 cm
∴ A = 1/2 × 4.9 × 10.2
∴ A = 24.99 cm2
Question no – (3)
Solution :
A = 70 cm2, one-side – 14 cm
∴ Other side = area/one-side
= 70/14
= 5 cm
Question no – (4)
Solution :
∴ Area of the rectangle = l × b
= 91 × 24
= 2184 cm2
∴ Area of a triangular = 1/2 × 12 × 13
= 78 cm2
∴ There are – 2184/78
= 28 triangular flower beds.
Question no – (5)
Solution :
∴ Area of rectangular
= 25 × 16
= 400 cm2
∴ Area of triangular
= 1/2 × 14 × 10
= 70 cm2
∴ Area of the remaining piece,
= (400 – 70)
= 330 cm2
Perimeter and Area Exercise 21E Solution :
Question no – (1)
Solution :
Circumference of a circle – 2πr
(a) Given, 14 cm
∴ r = 14 cm
∴ Circumference,
= 2 × π × 14
= 2 × 22/7 × 14
= 88 cm
(b) Given, 35 cm
∴ r = 35 cm
∴ c = 2 × 22/7 × 75
∴ c = 220 cm
Question no – (2)
Solution :
∴ Circumference = π × diameter.
∴ Diameter = Circumference/ π
(a) Given, 132 cm
∴ c = 132 cm
∴ d = 132/ π
∴ d = 132 × 7/22
∴ d = 42 cm
(b) Given, 11 cm
∴ d = 11 × 7/22
∴ d = 3.5 cm
Question no – (3)
Solution :
∴ Here, circumference of circle = Perimeter of the circle
∴ Square = 4 × 15.5
= 62 cm
∴ Diameter of the circle,
= 62 × 7/22
= 217/11
= 19.7 cm
Question no – (4)
Solution :
∴ d = 3 m
(a) Length of the metal boundary = π × 3
= 22/7 × 3
= 66/7
= 9.42 m
(b) The boundary costs,
= 9.42 × 25
= 235.5
Question no – (5)
Solution :
(a) Given, 530.66 m2
∴ A = 530.66m2
∴ A = πr2
∴ r2 = A/π = 530.66/3.14
∴ r2 = √169
= 13 cm
(b) Given, 706.5 cm2
∴ A = 706.5 cm2
∴ r2 = 706.5/3.14
∴ r = √225
= 15 cm
Question no – (6)
Solution :
∴ r = 21 m
∴ Area of the portion that is
∴ Grazed by the goat,
= 1/2 × πr2
= 1/2 × 22/7 × 21 × 21
= 693 m2
Therefore, the area of the field will be 693 m2
Next Chapter Solution :
