NCERT Solutions For Class 6 Maths Chapter 10 : Mensuration
NCERT Solutions For Class 6 Maths Chapter 10 Mensuration is an important resource for students to prepare the topic of Mensuration. In exams, several questions are asked from this section, making it one of the most crucial chapters of Class 6 Maths subject. Practising the NCERT Solutions for Class 6 Maths will help the students in understanding the concepts in a better way.
Mathematics is a high scoring subject. Securing high marks in this subject will largely improve the overall aggregate for students. To help students achieve the maximum marks in examinations, individual subject experts at BYJU’S have created the NCERT Solutions following the latest CBSE guidelines. Students can access these solutions in both online and offline modes at any time and anywhere.Chapter 10 MeasurementExercise 10.1
NCERT Solutions for Class 6 Chapter 10: Mensuration Download PDF
Mensuration chapter solutions are prepared by our experts as per the CBSE syllabus, to develop a strong conceptual base among students. These solved questions will help students to resolve their difficulties while solving the Mensuration problems present in the NCERT textbook and also while solving sample papers and previous year question papers.
Access NCERT Solutions for Class 6 Chapter 10: Mensuration
Exercise 10.1 Page no: 212
1. Find the perimeter of each of the following figures:
Solutions:
(a) Perimeter = Sum of all the sides
= 1 + 2 + 4 + 5
= 12 cm
(b) Perimeter = Sum of all the sides
= 23 + 35 + 35 + 40
= 133 cm
(c) Perimeter = Sum of all the sides
= 15 + 15 + 15 + 15
= 60 cm
(d) Perimeter = Sum of all the sides
= 4 + 4 + 4 + 4 + 4
=20 cm
(e) Perimeter = Sum of all the sides
= 1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4
= 15 cm
(f) Perimeter = Sum of all the sides
= 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3
= 52 cm
2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all around with tape. What is the length of the tape required?
Solutions:
Length of required tape = Perimeter of rectangle
= 2 (Length + Breadth)
= 2 (40 + 10)
= 2 (50)
= 100 cm
∴ Required length of tape is 100 cm
3. A table top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table top?
Solutions:
Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 (Length + Breadth)
= 2 (2.25 + 1.50)
= 2 (3.75)
= 2 × 3.75
= 7.5 m
∴ The perimeter of the table top is 7.5 m
4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Solutions:
Required length of wooden strip = Perimeter of photograph
= 2 (Length + Breadth)
= 2 (32 + 21)
= 2 (53)
= 2 × 53
= 106 cm
∴ Required length of the wooden strip is 106 cm
5. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solutions:
Perimeter of the field = 2 (Length + Breadth)
= 2 (0.7 + 0.5)
= 2 (1.2)
= 2 × 1.2
= 2.4 km
Each side is to be fenced with 4 rows = 4 × 2.4
= 9.6 km
∴ Total length of the required wire is 9.6 km
6. Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm
(b) An equilateral triangle of side 9 cm
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Solutions:
(a) Perimeter of triangle = 3 + 4 + 5
= 12 cm
(b) Perimeter of an equilateral triangle = 3 × side
= 3 × 9
= 27 cm
(c) Perimeter of isosceles triangle = 8 + 8 + 6
= 22 cm
7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solutions:
Perimeter of triangle = 10 + 14 + 15
= 39 cm
∴ The perimeter of triangle is 39 cm
8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Solutions:
Perimeter of hexagon = 6 × 8
= 48 m
∴ Perimeter of regular hexagon is 48 m
9. Find the side of the square whose perimeter is 20 m.
Solutions:
Perimeter of square = 4 × side
20 = 4 × side
Side = 20 / 4
Side = 5 m
∴ The side of the square is 5 m
10. The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solutions:
Perimeter of regular pentagon = 100 cm
5 × side = 100 cm
Side = 100 / 5
Side = 20 cm
∴ Side of the pentagon is 20 cm
11. A piece of strings is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Solutions:
(a) Perimeter of square = 30 cm
4 × side = 30
Side = 30 / 4
Side = 7.5 cm
(b) Perimeter of an equilateral triangle = 30 cm
3 × side = 30
Side = 30 / 3
Side = 10 cm
(c) Perimeter of a regular hexagon = 30 cm
6 × side = 30
Side = 30 / 6
Side = 5 cm
12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Solutions:
Let x cm be the third side
Perimeter of triangle = 36 cm
12 + 14 + x = 36
26 + x = 36
x = 36 – 26
x = 10 cm
∴ The third side is 10 cm
13. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Solutions:
Side of square = 250 m
Perimeter of square = 4 × side
= 4 × 250
= 1000 m
Cost of fencing = ₹ 20 per m
Cost of fencing for 1000 m = ₹ 20 × 1000
= ₹ 20,000
14. Find the cost of fencing a rectangular park of length 175 cm and breadth 125 m at the rate of ₹ 12 per metre.
Solutions:
Length = 175 cm
Breadth = 125 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (175 + 125)
= 2 (300)
= 2 × 300
= 600 m
Cost of fencing = 12 × 600
= 7200
∴ Cost of fencing is ₹ 7,200
15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Solutions:
Perimeter of square = 4 × side
= 4 × 75
= 300 m
∴ Distance covered by Sweety is 300 m
Perimeter of rectangular park = 2 (Length + Breadth)
= 2 (60 + 45)
= 2 (105)
= 2 × 105
= 210 m
∴ Distance covered by Bulbul is 210 m
Hence, Bulbul covers less distance than Sweety.
16. What is the perimeter of each of the each of the following figures? What do you infer from the the answers?
Solutions:
(a) Perimeter of square = 4 × side
= 4 × 25
= 100 cm
(b) Perimeter of rectangle = 2 (40 + 10)
= 2 × 50
= 100 cm
(c) Perimeter of rectangle = 2 (Length + Breadth)
= 2 (30 + 20)
= 2 (50)
= 2 × 50
= 100 cm
(d) Perimeter of triangle = 30 + 30 + 40
= 100 cm
All the figures have same perimeter.
17. Avneet buys 9 square paving slabs, each with a side of 1 / 2 m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [fig 10.7(i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig 10.7 (ii)]?
(c) Which has greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e they cannot be broken.)
Solutions:
(a) Side of square = 3 × side
= 3 × 1 / 2
= 3 / 2 m
Perimeter of Square = 4 × 3 / 2
= 2 × 3
= 6 m
(b) Perimeter = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1
= 10 m
(c) The arrangement in the form of cross has greater perimeter
(d) Perimeters greater than 10 m cannot be determined.
Exercise 10.2 page no: 216
1. Find the areas of the following figures by counting square:
(a) The figure contains only 9 fully filled squares. Hence, the area of this figure will be 9 square units.
(b) The figure contains only 5 fully filled squares. Hence, the area of this figure will be 5 square units.
(c) The figure contains 2 fully filled squares and 4 half filled squares. Hence, the area of this figure will be 4 square units.
(d) The figure contains only 8 fully filled squares. Hence, the area of this figure will be 8 square units.
(e) The figure contains only 10 fully filled squares. Hence, the area of this figure will be 10 square units.
(f) The figure contains only 2 fully filled squares and 4 half filled squares. Hence, the area of this figure will be 4 square units.
(g) The figure contains 4 fully filled squares and 4 half filled squares. Hence, the area of this figure will be 6 square units.
(h) The figure contains 5 fully filled squares. Hence, the area of this figure will be 5 square units.
(i) The figure contains 9 fully filled squares. Hence, the area of this figure will be 9 square units.
(j) The figure contains 2 fully filled squares and 4 half filled squares. Hence, the area of this figure will be 4 square units.
(k) The figure contains 4 fully filled squares and 2 half filled squares. Hence, the area of this figure will be 5 square units.
(l) From the given figure, we observe
| Covered Area | Number | Area estimate (square units) |
| Fully filled squares | 2 | 2 |
| Half filled squares | – | – |
| More than half filled squares | 6 | 6 |
| Less than half filled squares | 6 | 0 |
Therefore total area = 2 + 6
= 8 square units.
(m) From the given figure, we observe
| Covered Area | Number | Area estimate (square units) |
| Fully filled squares | 5 | 5 |
| Half filled squares | – | – |
| More than half filled squares | 9 | 9 |
| Less than half filled squares | 12 | 0 |
Therefore total area = 5 + 9
= 14 square units
(n) From the given figure, we observe
| Covered Area | Number | Area estimate (square units) |
| Fully filled squares | 8 | 8 |
| Half filled squares | – | – |
| More than half filled squares | 10 | 10 |
| Less than half filled squares | 9 | 0 |
Therefore total area = 8 + 10 = 18 square units
Exercise 10.3 page no: 219
1. Find the area of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Solutions:
We know that
Area of rectangle = Length × Breadth
(a) l = 3 cm and b = 4 cm
Area = l × b = 3 × 4
= 12 cm2
(b) l = 12 m and b = 21 m
Area = l × b = 12 × 21
= 252 m2
(c) l = 2 km and b = 3 km
Area = l × b = 2 × 3
= 6 km2
(d) l = 2 m and b = 70 cm = 0.70 m
Area = l × b = 2 × 0.70
= 1.40 m2
2. Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
Solutions:
(a) Area of square = side2
= 102
= 100 cm2
(b) Area of square = side2
= 142
= 196 cm2
(c) Area of square = side2
= 52
=25 cm2
3. The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
Solutions:
(a) Area of rectangle = l × b
= 9 × 6
= 54 m2
(b) Area of rectangle = l × b
= 17 × 3
= 51 m2
(c) Area of rectangle = l × b
= 4 × 14
= 56 m2
Area of rectangle 56 m2 i.e (c) is the largest area and area of rectangle 51 m2 i.e (b) is the smallest area
4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Solutions:
Area of rectangle = length × width
300 = 50 × width
width = 300 / 50
width = 6 m
∴ The width of the garden is 6 m
5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m.?
Solutions:
Area of land = length × breadth
= 500 × 200
= 1,00,000 m2
∴ Cost of tiling 1,00,000 sq m of land = (8 × 1,00,000) / 100
= ₹ 8000
6. A table top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solutions:
Given
l = 2m
b = 1m 50 cm = 1.50 m
Area = l × b = 2 × 1.50
= 3 m2
7. A room is 4 m long and 3 m 50 cm wide. Howe many square metres of carpet is needed to cover the floor of the room?
Solutions:
Given
l = 4m
b = 3 m 50 cm = 3.50 m
Area = l × b = 4 × 3.50
=14 m2
8. A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Solutions:
Area of floor = l × b = 5 × 4
= 20 m2
Area of square carpet = 3 × 3
= 9 m2
Area of floor that is not carpeted = 20 – 9
= 11 m2
∴ Area of the floor that is not carpeted is 11 m2
9. Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Solutions:
Area of flower square bed = 1 × 1
= 1 m2
Area of 5 square bed = 1 × 5
= 5 m2
Area of land = 5 × 4
= 20 m2
Remaining part of the land = Area of land – Area of 5 square bed
= 20 – 5
= 15 m2
∴ Remaining part of the land is 15 m2
10. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).
Solutions:
(a)
Area of yellow region = 3 × 3
= 9 cm2
Area of orange region = 1× 2
= 2 cm2
Area of grey region = 3 × 3
= 9 cm2
Area of brown region = 2 × 4
= 8 cm2
Total area = 9 + 2 + 9 + 8
= 28 cm2
∴ Total area is 28 cm2
(b)
Area of brown region = 3 × 1
= 3 cm2
Area of orange region = 3 × 1
= 3 cm2
Area of grey region = 3 × 1
= 3 cm2
Total area = 3 + 3 + 3
= 9 cm2
∴ Total area is 9 cm2
11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)
Solutions:
(a)
Total area of the figure = 12 × 2 + 8 × 2
= 40 cm2
(b)
There are 5 squares. Each side is 7 cm
Area of 5 squares = 5 × 72
= 245 cm2
(c)
Area of grey rectangle = 2 × 1
= 2 cm2
Area of brown rectangle = 2 × 1
= 2 cm2
Area of orange rectangle = 5 × 1
= 5 cm2
Total area = 2 + 2 + 5
= 9 cm2
12. How many tiles whose length and breadth are 12 cm and 5 cm, respectively will be needed to fit in a rectangular region whose length and breadth are respectively?
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm
Solutions:
(a) Area of rectangle = 100 × 144
= 14400 cm
Area of one tile = 5 × 12
= 60 cm2
Number of tiles = (Area of rectangle) / (Area of one tile)
= 14400 / 60
= 240
Hence, 240 tiles are needed
(b) Area of rectangle = 70 × 36
= 2520 cm2
Area of one tile = 5 × 12
= 60 cm2
Number of tiles = (Area of rectangle) / (Area of one tile)
= 2520 / 60
= 42
Hence, 42 tiles are needed.
Frequently Asked Questions on NCERT Solutions for Class 6 Maths Chapter 10
How to calculate an area of irregular shapes covered in the Chapter 10 of NCERT Solutions for Class 6 Maths?
Area of an irregular figure can be calculated :
Step 1: Divide the irregular shape into regular shapes that you can recognize (eg. triangles, rectangles, circles and squares)
Step 2: Find the area of these individual shapes and add them. Sum will be the area of the irregular figure.
Explain the perimeter of irregular shapes as per the Chapter 10 of NCERT Solutions for Class 6 Maths.
Irregular shapes are the shapes which do not have all sides and angles equal. The perimeter of irregular shapes is equal to total length covered by the shape. In the figure given, perimeter is the sum of all sides.
What is mensuration covered in the Chapter 10 of NCERT Solutions for Class 6 Maths.
Mensuration is a branch of mathematics which is a topic in Geometry. It is a study of various geometrical shapes, their length, breadth, volume, and area for 2D as well as 3D shapes. Mensuration is the branch of mathematics that deals with the measurement of length, area or volume of various geometric shapes.