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# NCERT Solutions Class 6 Maths Chapter 10 Mensuration | class 6th maths

NCERT Solutiontions Class 6 Maths Chapter 10 Mensuration class 6th maths. Here We learn what is in class 6 maths chapter 10 and how to solve questions with easiest method. In this chapter we solve the question of NCERT class 6 maths chapter 10 exercise 10.1, maths, class 6 maths chapter 10 exercise 10.2, class 6 maths Solutiontions exercise 10.3, class 6 maths chapter 10 exercise 10.4 and ncert Solutiontions for class 6 maths chapter 10 exercise 10.5. class 6th maths
class 6th maths ncert Solutiontions are part of NCERT Solutiontions Class 6 Maths Chapter 10 Mensuration Solutiontion PDF. ncert class 6 maths chapter 10 Mensuration with formula and Solutiontion.

Here we solve class 6th maths NCERT Solutiontions Class 6 Maths Chapter 10 Mensuration concepts all questions with easy method with expert Solutiontions. It help students in their study, home work and preparing for exam. Soon we provide NCERT Solutiontions Class 6 Maths Chapter 10 Mensuration question and answers. Soon we provided ncert Solutiontions for class 6th maths chapter 10 Mensuration in free PDF here. ncert Solutiontions for class 6 maths chapter 10 pdf will be provide soon. class 6th maths chapter 10 NCERT Solutiontion and ncert Solutiontions for class 6 maths chapter 10 pdf download book PDF.

# NCERT Solutiontions Class 6 Maths Chapter 10 Mensuration Class 6th maths

## NCERT Solutiontions Class 6 Maths Chapter 10 Mensuration class 6th maths Ex 10.1  (adsbygoogle = window.adsbygoogle || []).push({});

Exercise 10.1

1. Find the perimeter of each of the following figures :
Solutions :
Perimeter : Perimeter is the sum of all side of a figure.
(a) Perimeter of figure (a) = 4 + 2 + 1 + 5 = 12 cm
(b) Perimeter of figure (b) = 23 + 35 + 40 + 35 = 133 cm
(c) Perimeter of figure (c) = 15 + 15 + 15 + 15 = 60 cm
(d) Perimeter of figure (d) = 4 + 4 + 4 + 4 + 4 = 20 cm
(e) Perimeter of figure (e) = 1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4 = 15 cm
(f) Perimeter of figure (f) = 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 = 52 cm

2. The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Solution :
Rectangular box length = 40 cm
Rectangular box breadth = 10 cm
Perimeter of rectangular box = 2 × (Length + Breadth)
= 2 × (40 + 10)
= 2 × 50
= 100 cm
So, the length of the tape required is 100 cm or 1 meter.

3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Solution :
A table-top measures are = 2 m 25 cm by 1 m 50 cm
Perimeter of table = Perimeter of rectangle (because table is a rectangle shape)
Perimeter of table = 2 × (Length + Breadth)
= 2 × (2.25 + 1.50)
= 2 × 3.75
= 7.5 meter
So, the perimeter of the table-top is 7.5 meters.

4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Solution :
Length of photograph = 32 cm
Breadth of photograph = 21 cm
the wooden strip perimeter = 2 × (Length + Breadth)
= 2 × (32 + 21)
= 2 × 53
= 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?
Solution :
Length of rectangular piece of land = 0.7 km
Breadth of rectangular piece of land = 0.5 km
Perimeter of piece of land = Perimeter of rectangle
Perimeter of piece of land = 2 × (Length + Breadth)
= 2 × (0.7 + 0.5)
= 2 × 1.2
= 2.4 km
∵ one row length = 2.4 km
∴ four rows length = 4 × 2.4 = 9.6 km
So, the length of the wire needed 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.
Solution :
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
∵ All sides of tringle is different so this is a scalene triangle.
Perimeter of a scalene triangle =  Sum of all three sides
= 3 + 4 + 5 = 12 cm

(b) An equilateral triangle of side 9 cm.
Perimeter of an equilateral triangle = 3 × side
= 3 × 9
= 36 cm

(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Perimeter of an isosceles triangle = (2 × equal side) + third side
= (2 × 8) + 6
= 16 + 6
= 22 cm

7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution :
∵ All sides of tringle is different so this is a scalene triangle.
Perimeter of a scalene triangle =  Sum of all three sides
= 10 + 14 + 15
= 39 cm

8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution :
Perimeter of a regular hexagon = Sum of all sides
= 6 × 8
= 48 meter

9. Find the side of the square whose perimeter is 20 m.
Solution :
Perimeter of square = 20 meter
4 × side = 20
Side = $$\displaystyle \frac{{20}}{4}$$
Side = 5 m
So, the side of the square is 5 meter.

10. The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution :
The perimeter of a regular pentagon is = 100 cm
5 × side = 100
Side = $$\displaystyle \frac{{100}}{5}$$
Side = 20 cm
So, its each side is 20 cm long.

11. A piece of string 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?
Solution :
(a) a square?
If a square is made from a thread, then the perimeter of that square is = 30 cm
4 × side = 30
Side = $$\displaystyle \frac{{30}}{4}$$
Side = 7.5 cm
Therefore, if a square is made with string, then the side of that square will be 7.5 cm.

(b) an equilateral triangle?
If an equilateral triangle is made from the string, then the perimeter of that equilateral triangle is = 30 cm
3 × side = 30
Side = $$\displaystyle \frac{{30}}{3}$$
Side = 10 cm
Therefore, if an equilateral triangle is made from the string, then the side of that equilateral triangle will be 10 cm.

(c) a regular hexagon?
If a regular hexagon is made with string, then the perimeter of that regular hexagon = 30 cm
6 × side = 30
Side = $$\displaystyle \frac{{30}}{6}$$
Side = 5 cm
Therefore, if a regular hexagon is made with string, then the side of that regular hexagon will be 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?
Solution :
Let the third side of the triangle be = x cm
Perimeter of triangle = 36 cm
sum of three sides of a triangle = 36
So, 12 + 14 + x + = 36
26 + x = 36
x = 36 − 26
x = 10 cm

13. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Solution :
Side of a square park is = 250 m
Perimeter of a square park is = 4 × Side
= 4 × 250
= 1000 m
∵ Fencing cost on one meter area = ₹20
∴ Fencing cost on 1000 meters area = 20 × 1000
= ₹20000
Hence, the cost of fencing a square garden of side 250 m will be ₹ 20000.

14. Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹ 12 per metre.
Solution :
Length of a rectangular park = 175 m
Breadth of a rectangular park = 125 m
Perimeter of a rectangular park = 2 × (Length + Breadth)
Perimeter of a rectangular park = 2 × (175 + 125)
= 2 × 300
= 600 m
∵ one meter fencing cost = ₹20
∴ cost of fencing 600 meters = 20 × 600
= ₹12000
Hence, the cost of fencing all around will be ₹ 12000.

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?
Solution :
Side of a square park is = 75 m
Perimeter of square park = 4 × side
= 4 × 75
= 300 m

Length of rectangular park is = 60 m
Breadth of rectangular park is = 45 m
Perimeter of rectangular park = 2 × (Length + Breadth)
= 2 × (60 + 45)
= 2 × 105
= 210 m

So, Sweety cover less distance.

16. What is the perimeter of each of the following figures? What do you infer from the answers?
Solution :
Perimeter of figure (a) = Perimeter of square
Perimeter of square  = 4 × Side
= 4 × 25
= 100 cm

Perimeter of figure (b) = Perimeter of rectangle
Length of rectangle = 30 cm
Breadth of rectangle = 20 cm
Perimeter of rectangle = 2 × (Length + Breadth)
= 2 × (30 + 20)
= 2 × 50
= 100 cm

Perimeter of figure (c) = Perimeter of rectangle
Length of rectangle = 40 cm
Breadth of rectangle = 10 cm
Perimeter of rectangle = 2 × (Length + Breadth)
= 2 × (40 + 10)
= 2 × 50
= 100 cm

Perimeter of figure (d) = Perimeter of isosceles triangle
Length of equal side of an isosceles triangle = 30 cm
Length of third side of isosceles triangle = 40 cm
Perimeter of isosceles triangle = (2 × equal side) + third side
Perimeter of equilateral triangle = (2 × equal side) + third side
= (2 × 30) + 40
= 60 + 40
= 100 cm
Conclusion: Finding the perimeter of all the above figures, it can be concluded that all the figures have the same perimeter.

17. Avneet buys 9 square paving slabs, each with a side of $$\displaystyle \frac{{1}}{2}$$ 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.)
Solution :
(a) What is the perimeter of his arrangement [Fig 10.7(i)]?
Side of the new square = 1.5 m
Perimeter of new square = 4 × side
= 4 × 1.5
= 6.0 m
So the perimeter of the new square will be 6 metres.

(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)]?
Perimeter of the figure made by Shari = Sum of all sides
= 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1
= 10 meters
Hence, the perimeter of the figure made by Sherry is 10 m.

(c) Which has greater perimeter?
Fig. 10.7 (b) has a greater perimeter.

## NCERT Solutions for Class 6 Maths Chapter 10 Mensuration class 6th maths Ex 10.2

1. Find the areas of the following figures by counting square:

Solution :
(a) 9 squre unit
(b) 5 squre unit
(c) 4 squre unit
(d) 8 squre unit
(e) 10 squre unit
(f) 4 squre unit
(g) 6 squre unit
(h) 5 squre unit
(i) 9 squre unit
(j) 4 squre unit
(k) 5 squre unit
(l) 8 squre unit
(m) 14 squre unit
(n) 18 squre unit

## NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Class 6th maths Ex 10.3

1. Find the areas 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

Solution :
(a) 3 cm and 4 cm
Area of Rectangle = Length × Breadth
Area of Rectangle = 3 × 4
= 12 Sq cm

(b) 12 m and 21 m
Area of Rectangle = Length × Breadth
Area of Rectangle = 12 × 21
= 252 Sq meter

(c) 2 km and 3 km
Area of Rectangle = Length × Breadth
Area of Rectangle = 2 × 3
= 6 sq km

(d) 2 m and 70 cm
Area of Rectangle = Length × Breadth
Area of Rectangle = 2 × 0.70 (∵ 70 cm = 0.70 m)
= 1.40 sq meter

2. Find the areas of the squares whose sides are :

 (a) 10 cm (b) 14 cm (c) 5 m

Solution :
(a) 10 cm
Area of Square = (side)2
= (10)2
= 100 sq cm

(b) 14 cm
Area of Square = (side)2
= (14)2
= 196 sq cm

(c) 5 मी
Area of Square = (side)2
= (5)2
= 25 sq meter

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) 9 m and 6 m
Area of Rectangle = Length × Breadth
Area of Rectangle = 9 × 6
= 54 sq meter

(b) 17 m and 3 m
Area of Rectangle = Length × Breadth
Area of Rectangle = 3 × 14
= 42 sq meter

(c) 4 m and 14 m
Area of Rectangle = Length × Breadth
Area of Rectangle = 4 × 14
= 56 sq meter

(c) has the largest area and (b) has the smallest.

4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Solutions :
Rectangular garden length = 50 m
Area of Rectangular garden = 300 sq m
Length × Breadth = 300
50 × Breadth = 300
Breadth = $$\displaystyle \frac{{300}}{50}$$
Breadth = 6 m
The width of the garden is 6 meter.

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 :
Length of Rectangular plot = 500 m
Breadth of Rectangular plot = 200 m
Area of Rectangular plot = Length × Breadth
= 500 × 200
= 100000 sq m

∵ Rate pf per hundred sq m is = ₹ 8
∴ Rate pf per 1 sq m is = $$\displaystyle \frac{{8}}{100}$$
∴ Rate pf per 100000 sq m is = $$\displaystyle \frac{{8}}{100}$$ × 10000
= ₹ 8000

6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solutions :
A table-top measures 2 m by 1 m 50 cm
Area of a table-top = Length × Breadth
= 2 × 1.5
= 3.0 sq m

7. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Solutions :
Length of Room = 4 m
Breadth of Room = 3 m 50 cm
Area of Room = Length × Breadth
= 4 × 3.5
= 14.0 sq meter
So, 14 square metres of carpet is needed to cover the floor of the room?

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 :
Carpet length = 5 m
Carpet breadth = 4 m
Area of carpet = Length × Breadth
= 5 × 4
= 20 sq m

Square carpet side = 3 m
Area of Square carpet = (side)2
= (3)2
= 9 sq meter

The area of the floor that is not carpeted = Area of carpet − Area of Square carpet
= 20 − 9
= 11 sq meter

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 :
Length of rectangular piece of land = 5 m
Breadth of rectangular piece of land = 4 m
Area of rectangular piece of land = Length × Breadth
= 5 × 4
= 20 sq meter

Square flower bed side = 1 m
Area of Square flower bed = (side)2
= (1)2
= 1 sq m
Five square flower bed area  = 1 × 5 = 5 sq m
The area of the remaining part of the land = Area of rectangular piece of land − Five square flower bed area
= 20 − 5
= 15 sq m

10. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Solutions :
Area of figures (a)

First part area = 1 × 1 = 1
Second part area = 4 × 2 = 8
Third part area = 4 × 1 = 4
Fourth part area = 4 × 2 = 8
Fifth part area = 3 × 1 = 3
Total Area of figures (a) = 1 + 8 + 4 + 8 + 3 = 24 sq cm

Area of figures (b)

First part area = 5 × 1 = 5
Second part area = 2 × 1 = 2
Third part area = 2 × 1 = 2
Total Area of figures (b) = 5 + 2 + 2 = 9 sq cm

11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

Solutions :
Area of figures (a)
First part area = 12 × 2 = 24 sq cm
Second part area = 8 × 2 = 16 sq cm
Total Area of figures (a) = 24 + 16 = 40 sq cm

Area of figures (b)
First part area = 7 × 7 = 49 sq cm
Second part area = 21 × 7 = 147 sq cm
Third part area = 7 × 7 = 49 sq cm
Total Area of figures (b) = 49 + 147 + 49 = 245 sq cm

Area of figures (c)
First part area = 5 × 1 = 5 sq cm
Second part area = 4 × 1 = 4 sq cm
Total Area of figures (c) = 5 + 4 = 9 sq cm

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 :
Area of tiles = 5 × 12 = 60 sq cm
(a) 100 cm and 144 cm
Area of rectangular region = Length × Breadth
= 144 × 100
= 14400 sq cm

= $$\displaystyle \frac{{14400}}{60}$$ = 240 tiles

(b) 70 cm and 36 cm
Area of rectangular region = Length × Breadth
= 70 × 36
= 2520 sq cm

= $$\displaystyle \frac{{2520}}{60}$$ = 42 tiles

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