NCERT Solutions for Class 6 Maths Chapter 10
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1, Exercise 10.2 and Exercise 10.3 in English Medium as well as Hindi Medium updated for new academic session 2020-2021 based on new NCERT Books.
Solutions of Prashnavali 10.1, Prashnavali 10.2 and Prashnavali 10.3 in Hindi Medium PDF file format to free download. Download links for NCERT Books 2020-2021 and Offline NCERT Solutions Apps 2020-21 are also available to free download. Solutions of each question with suitable answers and explanation are given on the website. Download NCERT Solution Offline Apps for Class 6 all subjects.Page Contents
NCERT Solutions for Class 6 Maths Chapter 10
| Class: | 6 |
| Subject: | Maths – गणित |
| Chapter 10: | Mensuration |
Download Class 6 Maths Chapter 10 Solutions in PDF Format
Download NCERT Solutions for Class 6 Maths Chapter 10 Mensuration in PDF format updated for new session 2020-21 using the latest NCERT Books 2020-21. For any difficulty, please notify us for help. We are here to help you free of cost.
Class 6 Maths Chapter 10 Solutions in English Medium
Class 6 Maths Chapter 10 Solutions in Hindi Medium
Class 6 Maths Exercise 10.1 & 10.2 Solutions in Video
Class 6 Maths Exercise 10.3 Solutions in Video
About Class 6 Maths Chapter 10
In 6 Maths Chapter 10 Mensuration, we will study about the perimeter and area related to various 2-D figures. This chapter is important as per further classes also. We should learn the formulae of perimeter and areas of various figures. Some the the formulae are given below:
Perimeter of a rectangle = 2 × (length + breadth)
Perimeter of a square = 4 × length of its side
Perimeter of an equilateral triangle = 3 × length of a side
Perimeter is the distance covered along the boundary forming a closed figure when you go round the figure once.
Area of a rectangle = length × breadth
Area of a square = side × side
To calculate the area of a figure using a squared paper, the following conventions are adopted:
Ignore portions of the area that are less than half a square.
If more than half a square is in a region. Count it as one square.
If exactly half the square is counted, take its area as 1/2 sq units.
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Important Questions on 6 Maths Chapter 10
= Perimeter of rectangle
= 2 (length + breadth)
= 2 (40 + 10)
= 2 x 50
= 100 cm
= 1 m
Thus, the total length of tape required is 100 cm or 1 m.
= 2 m 25 cm
= 2.25 m
Breadth of table top
= 1 m 50 cm
= 1.50 m
Perimeter of table top
= 2 x (length + breadth)
= 2 x (2.25 + 1.50)
= 2 x 3.75
= 7.50 m
Thus, the perimeter of table top is 7.5 m.
फोटो फ्रेम का परिमाप
= 2 x (लंबाई + चौड़ाई)
= 2 (32 + 21)
= 2 x 53 cm
= 106 cm
लकड़ी की पट्टी की लंबाई = फोटो फ्रेम का परिमाप = 106 cm
अतः, आवश्यक लकड़ी की पट्टी की लंबाई 106 cm है।
अतः, आवश्यक तार की लंबाई
= 4 x आयताकार भूखंड का परिमाप
आयताकार भूखंड का परिमाप
= 2 x (लंबाई + चौड़ाई)
= 2 x (0.7 + 0.5)
= 2 x 1.2
= 2.4 km
= 2.4 x 1000 m
= 2400 m
अतः, आवश्यक तार की लंबाई
= 4 x 2400
= 9600 m
= 9.6 km
= Sum of all three sides
= 10 cm + 14 cm + 15 cm
= 39 cm
Thus, the perimeter of triangle is 39 cm.
= 6 x length of one side
= 6 x 8 m
= 48 m
Thus, the perimeter of hexagon is 48 m.
20 = 4 x भुजा
भुजा = 20/4 = 5 cm
अतः, वर्ग की भुजा की लंबाई 5 cm है।
5 x भुजा = 100 cm
भुजा = 100/5 = 20 cm
अतः, सम पंचभुज की प्रत्येक भुजा की लंबाई 20 cm है।
= 250 m
Perimeter of square
= 4 x side
= 4 x 250
= 1000 m
Since, cost of fencing of per meter = ₹ 20
Therefore, the cost of fencing of 1000 meters
= 20 x 1000
= ₹20,000
पार्क की चौड़ाई = 125 m
पार्क का परिमाप
= 2 x (लंबाई + चौड़ाई)
= 2 x (175 + 125)
= 2 x 300 m
= 600 m
प्रति मीटर बाड़ लगाने का व्यय = ₹ 12
अतः, चारों ओर बाड़ लगाने का कुल व्यय
= 12 x 600
= ₹ 7,200
= वर्गाकार पार्क का परिमाप
= 4 x भुजा
= 4 x 75 = 300 m
अतः, स्वीटी द्वारा तय दूरी 300 m है।
अब, बुलबुल द्वारा तय दूरी
= आयताकार पार्क का परिमाप
= 2 x (लंबाई + चौड़ाई)
= 2 x (60 + 45)
= 2 x 105 = 210 m
अतः, बुलबुल द्वारा तय दूरी 210 m है।
इसप्रकार, बुलबुल कम दूरी तय करती है।
= 2 m
Breadth of table
= 1 m 50 cm
= 1.50 m
Area of table
= length x breadth
= 2 m x 1.50 m
= 3 m^2
= 4 m
कमरे की चौड़ाई
= 3 m 50 cm
= 3.50 m
आवश्यक गलीचे का क्षेत्रफल
= लंबाई x चौड़ाई
= 4 x 3.50
= 14 वर्ग मीटर
चौड़ाई = 4 m
फर्श का क्षेत्रफल
= लंबाई x चौड़ाई
= 5 m x 4 m
= 20 m^2
अब, वर्गाकार गलीचे की भुजा = 3 m
गलीचे का क्षेत्रफल
= भुजा x भुजा
= 3 x 3
= 9 m^2
फर्श के उस भाग का क्षेत्रफल, जिस पर गलीचा नहीं बिछा है
= 20 m^2 – 9 m^2
= 11 m^2
Area of square bed
= side x side
= 1 m x 1 m
= 1 m^2
Area of 5 square beds
= 1 x 5
= 5 m^2
Now, Length of land = 5 m
Breadth of land = 4 m
Area of land
= length x breadth
= 5 m x 4 m
= 20 m^2
Area of remaining part
= Area of land – Area of 5 flower beds
= 20 m^2 – 5 m^2
= 15 m^2
