NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

Edited By Ramraj Saini | Updated on Nov 30, 2023 10:07 AM IST

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration are discussed in this article. These NCERT solutions are created by expert team at careers360 considering the latest CBSE syllabus 2023. It is a process of measurement which is needed for different geometric shapes both 2D and 3D shapes. In CBSE NCERT solutions for mensuration class 6, you will get many questions related to calculating the perimeter and area of a rectangle, square and a triangle. You must go through these NCERT Solutions to get a better idea of how to solve a particular problem.

Latest: Important Maths Formulas for Class 6 - Chapterwise

This Story also Contains

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration - Important Formulae
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration - Important Points
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
NCERT Solutions for Class 6 Maths Chapter 10 Mensuration (Intext Questions and Exercise)
NCERT Class 7 Maths Chapter 10 Mensuration Topic 10.2 Perimeter
NCERT Class 7 Maths Chapter 10 Mensuration Topic 10.2.1 Perimeter of a Rectangle
NCERT Class 6 Maths Chapter 10 Mensuration Exercise: 10.1
NCERT Class 6 Maths Chapter 10 Mensuration Exercise 10.2
NCERT Class 6 Maths Chapter 10 Mensuration Topic 10.3.1 Area of A Rectangle
NCERT Class 6 Maths Chapter 10 Mensuration Exercise 10.3
NCERT Solutions for Class 6 Subject Wise
NCERT Class 6 Maths Chapter 10 Mensuration Topics:
NCERT Solutions for Class 6 Mathematics Chapter Wise
Key features of NCERT Solutions for Class 6 Maths chapter 10

$NCERT Solutions for Class 6 Maths Chapter 10 Mensuration$

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

NCERT Class 6 solution for chapter 10 is more focused on the measurement of 2-D shapes. As per the NCERT Syllabus for Class 6 Maths, there are 30 questions given in the 3 exercises of this chapter. In solutions of NCERT for Class 6 Maths chapter 10 mensuration, there are some problems where you have to find areas of figures by counting the number of squares, and sum of their area is the total area of that figure. Refer to the NCERT Books for Class 6 Maths for more clarity of the concepts. All these questions are explained in NCERT Solutions for Class 6 . It will help you to understand the concept of NCERT in a more easy way.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration - Important Formulae

The size of a line segment is its length.
Various units of measurement are connected by the following relations:
- 1 kilometer = 1000 metre (m)
- 1 metre = 100 centimetre (cm)
- 1 decimetre = 10 centimetre (cm)
- 1 metre = 10 decimetre (dm)
- 1 centimetre = 10 millimetre (mm)
- 1 foot= 12 inches
- 1 yard = 3 feet
- 22 yards = 1 chain
A closed plane figure together with its interior is called the region enclosed by it.
Perimeter of the figure = Sum of the lengths of all sides of a plane figure
Area = Measurement of the region enclosed by a closed plane figure
A square centimetre is the area of the region formed by a square of 1 cm.
Standard units of area and their relations are:
- 100 mm² = 1 cm²
- 100 cm² = 1 dm²
- 100 dm² = 1 m²
- 10000 cm² = 1 m²
- 100 m² = 1 are
- 100 ares = 1 hectare
- 100 hectares = 1 sq. km.
Perimeter of a rectangle = 2 (Length + Width) Or, P = 2(l + b)
Perimeter of a square = 4 (Side) Or, P = 4l
Area of a rectangle = Length x Breadth Or, A = lb
Length of a rectangle = Area/Breath Or, l = A/b
Width of a rectangle = Area/Length Or, b= A/l

Area of a square= (Side)²

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration - Important Points

Perimeter: The distance covered along the boundary of a closed figure when you go around the figure once. It represents the total length of the figure's boundary.

(a) Perimeter of a rectangle: It is calculated by adding twice the length and twice the breadth of the rectangle.

(b) Perimeter of a square: It is calculated by multiplying the length of one side by 4.

(c) Perimeter of an equilateral triangle: It is calculated by multiplying the length of one side by 3.

Regular closed figures: These are figures in which all sides and angles are equal. Examples include regular polygons such as equilateral triangles, squares, and regular hexagons.

Area: It refers to the amount of surface enclosed by a closed figure. It is measured in square units and represents the two-dimensional space within the boundaries of the figure.

Conventions for calculating area using squared paper:

(a) Ignore portions of the area that are less than half a square. They are not counted as complete units.

(b) If more than half a square is in a region, count it as one square unit.

(a) Area of a rectangle: It is calculated by multiplying the length and breadth of the rectangle.

(b) Area of a square: It is calculated by multiplying the length of one side by itself.

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NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

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NCERT Solutions for Class 6 Maths Chapter 10 Mensuration (Intext Questions and Exercise)

NCERT Class 7 Maths Chapter 10 Mensuration Topic 10.2 Perimeter

Question:3 Meera went to a park 150 m long and 80 m wide. She took one complete round on its boundary. What is the distance covered by her?

Answer: The park is 150m long and 80 m wide. The parks are of the shape of the rectangle. The distance covered by Meera = perimeter of park = 2 (150+80)= 460 m

Question:4 Find the perimeter of the following figures:

Perimeter = AB + BC + CD + DA
= __+__+__+__
= ______

Perimeter = AB + BC + CD + DA
= __ + __ + __+ __
= ______

Perimeter = AB + BC + CD + DE + EF + FG + GH +HI + IJ + JK + KL + LA
= __ + __ +__ + __ + __ + __ + __ + __ +__+ __ + __ + __
= ______

Perimeter = AB + BC + CD + DE + EF+ FA
= __ + __ + __ + __ + __ + __
= ______

Answer: Find the perimeter of the following figures:

356644

Perimeter = AB + BC + CD + DA
= 40+10+40+10
= 100

487874

Perimeter = AB + BC + CD + DA
=5+5+5+5=20

45454612

Perimeter = AB + BC + CD + DE + EF + FG + GH +HI + IJ + JK + KL + LA
=1+3+3+1+3+3+1+3+3+1+3+3=28 cm

4878454

Perimeter = AB + BC + CD + DE + EF+ FA
=100 + 120 + 90 + 45+ 60 +80=495 cm

NCERT Class 7 Maths Chapter 10 Mensuration Topic 10.2.1 Perimeter of a Rectangle

Question: Find the perimeter of the following rectangles:

Answer:

Length of rectangle	Breadth of rectangle	Perimeter by adding all sides	perimeter by $2\times(length + breadth)$
25 cm	12 cm	=25+12+25+12=74 cm	$\\2\times(25+12)\\2\times37=74\ cm$
0.5 cm	0.25cm	=0.5+0.25+0.5+0.25=1.5cm	$\\2\times(0.25+0.5)\\2\times0.75=1.5\ cm$
18 cm	15 cm	=18+15+18+15=66 cm	$\\2\times(18+15)\\2\times33=66\ cm$
10.5 cm	8.5 cm	=10.5+8.5+10.5+8.5=38 cm	$\\2\times(10.5+8.5)\\2\times19=38\ cm$

NCERT Class 6 Maths Chapter 10 Mensuration Exercise: 10.1

Question: 1 Find the perimeter of each of the following figures :

1643092937153

Answer: The perimeter of the following figures-
Perimeter = sum of all sides

Therefore,
(a) Perimeter = 4 cm + 2cm +1 cm +5cm = 12 cm

(b) Perimeter = 23 cm + 35 cm + 40cm + 35cm = 133 cm

(d) Required perimeter= (4 + 4 + 4 + 4 + 4) cm = 20 cm

(e) Perimeter = 4cm + 0.5 cm +2.5cm +2.5cm + 0.5cm +4cm +1cm = 15cm

(f) Perimeter = Sum of all sides = 4 $\times$ (4cm) + 4 $\times$ (1 cm) + 8 $\times$ (3 cm) + 4 $\times$ (2cm) = (16 + 4 + 24 + 8)cm =52 cm

Question: 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?

Answer: We have,
Length = 40 cm and,
Breadth = 10 cm

So, the length of tape required = perimeter of the rectangular box
= 2 (L + B)
= 2 $\times$ 50 cm
= 100 cm

Hence 100 cm tape is required.

Question: 3 A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?

Answer: We have table-top measures 2 m 25 cm by 1 m 50 cm.
Therefore, the Perimeter of the table-top = sum of all sides
= 2 $\times$ (L+B)
= 2 $\times$ ((2m + 0.25m) + (1m + 0.5m))
= 7.5 m

Question: 4 What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?

Answer: We have,
Length and breadth of a photograph is 32 cm and 21 cm respectively.

So, According to question,
Length of wooden strip required to frame the photograph = 2 [L + B]
= 2 [32 + 21] cm
= 2 $\times$ 53 cm = 106 cm

Hence the required length of the strip is 106 cm or 1.06 m

Question: 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?

Answer: We have,
A rectangular piece of land measures 0.7 km by 0.5 km and side is to be fenced with 4 rows of wires.

As per the question,

Required length of wires
= 4 $\times$ 2 [ 0.7 km + 0.5 km]
= 8 $\times$ 1.2 km
= 9.6 km

Question: 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.

Answer: (a) The perimeter of a triangle = Sum of all the three sides
= 3cm + 4cm + 5cm
= 12 cm

(b) The perimeter of a eqilateral triangle
= Sum of all the sides of the triangle
= (9cm +9cm +9cm)
= 27cm

Question: 7 Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.

Answer: We all know that,
Perimeter of a triangle = sum of all the sides of the triangle
=10cm + 14cm +15cm
= 39 cm

Question: 8 Find the perimeter of a regular hexagon with each side measuring 8 m.

Answer: It is known that,
A regular hexagon has six equal sides.
So, the perimeter of the regular hexagon
= 6 $\times$ side
= 6 $\times$ 8m = 48m

Question: 9 Find the side of the square whose perimeter is 20 m.

Answer: Given that,
The perimeter of the square is 20m
According to question,
The perimeter of square = 4 $\times$ side
$\therefore\ side = \frac{20}{4}=5\ m$

Question: 10 The perimeter of a regular pentagon is 100 cm. How long is its each side?

Answer: Given,
The perimeter of a regular pentagon is 100cm and,
no. of sides in regular pentagon is five (5)
therefore, the length of each side
$=\frac{100}{5}=20cm$

Question: 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?

Answer: Given that,

Length of the string is 30 cm
(a)For square,
Number of sides = 4
Therefore, Length of one side = 30 /4 = 7.5 cm

(b)For equilateral triangle,
Number of sides = 3
Therefore, length of sides = 30/3 = 10cm

Question: 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?

Answer: Given that,
The two sides of the triangle are 12 cm and 14 cm. let the third side be A.
And the perimeter of the triangle is 36cm.

Therefore, the Perimeter of the triangle
= sum of all sides of the triangle
= 12cm +14cm + A = 36cm
= A + 26cm =36cm
= A = 10cm

Question: 13 Find the cost of fencing a square park of side 250 m at the rate of rupees 20 per metre.

Answer: Given that,
Length of the square park = 250 m

Cost of fencing = rs. 20 per meter

Therefore, the Perimeter of the square park =
= 4 $\times$ 250
=1000 m
= 1km

So, total cost = 20 $\times$ 1000 = 20,000 Rs

Question: 14 Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of rupees 12 per metre.

Answer: Given that,
The length of the rectangular park is 15 m and the breadth is 125m.
Cost of fencing rate = Rs 12 per meter

Therefore,
Perimeter of the park
=2 [L + B]
= 2 [175 +125]
=2 $\times$ 300m
=600m

So, the total cost of fencing = 600 $\times$ 12 = Rs. 7,200

Question: 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?

Answer: Given that,
Length of the square park = 75m
Length and breadth of the rectangular park is 60 m and 45m respectively

Therefore,
The perimeter of the square park
= 4 $\times$ 75
=300 m

Also, the Perimeter of the rectangular park
= 2 $\times$ ( 60 + 45)
=2 $\times$ 105m
=210m

Hence Bulbul covers the less distance.

Question: 16 What is the perimeter of each of the following figures? What do you infer from the answers?

1643092972347

Answer: (a) The perimeter of the square
= Sum of all the four side = 4 $\times$ side
= 4 $\times$ 25
= 100 cm

(b)The perimeter of the rectangle
= SUm of all the sides = 2 $\times$ [length + breadth )
= 2 $\times$ (20 + 30)
= 2 $\times$ 50
= 100

(c)The perimeter of the rectangle
= Sum of all the sides = 2 $\times$ [length + breadth ]
= 2 $\times$ (40 + 10)
= 2 $\times$ 50
= 100 cm

(d)The perimeter of the triangle
= Sum of all three sides of the triangle
= 30cm + 30cm + 40cm
=100cm

We conclude all the given figures have the same perimeter.

Question:1 7 Avneet buys 9 square paving slabs, each with a side of $\frac{1}{2}m$ . He lays them in the form of a square.

(a) What is the perimeter of his arrangement ?

(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement?

(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this?

1643093002368

Answer: Given that,
Length of each slab = $1/2 m$
So, the length of the square = $3/2\ m$

(a) The first figure is a square
So, the perimeter of the square = 4 $\times$ side
= $4 \times 3/2 m = 6\ m$

(b) The perimeter of the 2nd figure
= Sum of all the sides
= $(4\times \frac{1}{2} + 8 \times 1)$
= 10 m

(d) Arrange all the slab in horizontally, it forms a rectangle whose dimensions are $\frac{9}{2}\ by\ \frac{1}{2}$

NCERT Class 6 Maths Chapter 10 Mensuration Exercise 10.2

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

1643093035011

Answer: Area of 1 square = 1sq. unit

(a) Number of total squares = 9
Therefore, the area of 9 squares = 9 sq. units

(b) In fig (b) the total number of square = 5
Therefore, are = 5 sq units

(d) In fig (d) number of squares = 8
Therefore, area = 8 sq. units

(e) Total number of square = 10
therefore, area = 10 sq. units

(f) total number of squares = 4
Therefore, area = 4 sq. units

(g) Total number of square = 6
therefore, area = 6 sq. units

(h) Total number of square = 5
therefore, area = 5 sq. units

(i) Total number of square = 9
therefore, area = 9 sq. units

(j) Total number of square = 4
therefore, are = 4 sq. units

(k) Total number of square = 5
therefore, area = 5 sq. units

(l) Total number of full squares = 4
Squares more than half = 3
half squares = 2
Therefore, the total area = 8 sq. units

(m) Count more than half squares as 1
Now, the total number of squares = 14
Therefore, the total area = 14 sq. unit

(n) Count more than half squares as 1
Now, the total number of squares = 18
Therefore, the total area = 18 sq. unit

NCERT Class 6 Maths Chapter 10 Mensuration Topic 10.3.1 Area of A Rectangle

Question:1 Find the area of the floor of your classroom.

Answer: If floor length is 3m and breadth is 2m then

$floor \ area=length\times breadth=3\times2=6m^2$

Question:2 Find the area of any one door in your house.

Answer: Suppose the length of door is 2m and breadth is 0.5 m

then the area of door = $length\times breadth=2\times0.5=1m^2$

NCERT Class 6 Maths Chapter 10 Mensuration Exercise 10.3

Question: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

Answer: We know that area of a rectangle = length $\times$ breadth
Therefore,
(a) Here length is 3cm and the breadth of the rectangle is 4cm
So, the area of the rectangle
= $3 \times 4 = 12\ cm^2$
(b) Here length is 12m and the breadth of the rectangle is 21m
So, the area of the rectangle
= $12 \times 21 = 252\ m^2$
(c) Here length is 3km and the breadth of the rectangle is 3km
So, the area of the rectangle ?
= $2 \times 3 = 6\ km^2$
(d) Here length is 2m and the breadth of the rectangle is 70cm = $70/100 = 0.7 m$
So, the area of the rectangle
= $2 \times 0.7= 1.4\ m^2$

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

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

Answer: It is known that area of a square = $(side)^2$

(a) The length of a side of a square = 10cm
So, the area = $(10)^2 =100cm^2$

(b)The length of a side of a square = 14cm
So, the area = $(14)^2 =196cm^2$

Question: 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?

Answer: We know that area of a rectangle = length $\times$ breadth
Therefore,
(a) Here length is 9m and the breadth of the rectangle is 6m
So, the area of the rectangle
= $9\times 6 = 54m^2$

(b) Here length is 17 m and the breadth of the rectangle is 3m
So, the area of the rectangle
= $17\times 3 = 51m^2$

Hence rectangle C has the greatest area and rectangle B has the least area.

Question: 4 The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.

Answer: Given that,
Area of the rectangular garden = 300 sq. m
and the length = 50 m

Therefore, the Width of the garden = $\frac{Area}{length}$
= $300/50 = 6\ m$

Hence the width of the garden is 6 m

Question: 5 What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of rupees 8 per hundred sq m.?

Answer: Given that,
Length of the rectangular plot = 500m
and the width = 200 m
fencing cost = Rs 8 per hundred sq. meter

Therefore, the area of the plot
= $500\times 200 = 100,000 \ m^2$

Now, Total rate of tiling of 100,1000 m ² = $\frac{8}{100}\times 100,000$
= Rs 8000

Question: 6 A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?

Answer: Given that,
Length of the table-top = 2m
and it's breadth = 1.5 m

Therefore, the area of the tabletop
= length $\times$ breadth
$= 2 \times 1.5$
$=3\ m^2$

Question: 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?

Answer: Given that,
Length of the room = 4 m
And it's breadth = 3.5 m
Area of the room = = length $\times$ breadth
$= 4 \times 3.5$
= $14 m^2$

Hence the area of carpet needed is $14 m^2$

Question: 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.

Answer: Given that,
Length of the floor = 5 m
and its width = 4 m
Therefore, the area of the floor
= length $\times$ breadth
$= 5\times 4 = 20\ m^2$

And, side of the square carpet = 3m
Therefore, the area of the square carpet = $(side)^2$
= 3m $\times$ 3m = 9m ²

Thus, the area of the floor which is not covered with carpet = $20 m^2-9m^2= 11m^2$

Question: 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?

Answer: Given that,
Length and breadth of the land are 5m and 4 m.
Side of the square flower bed = 1m
Therefore the area of the bed = 1 $m^2$

So, Area of 5 square flower bed = 5 $m^2$

Area of the and = length $\times$ breadth
= $5m\times 4m = 20 m^2$

Thus, the area of the remaining part of the land = Area of land - area of the flower bed
= 20 $m^2$ - 5 $m^2$ = 15 $m^2$

Question:1 0 By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

1643093070976

Answer: (a) Split the given figures into small rectangular parts as shown here;
2653
Now, the area of the rectangle (I) = $l\times b$ = 4 $\times$ 3 = 12 $cm^2$
the area of the rectangle (II) = $l\times b$ = 3 $\times$ 2= 6 $cm^2$
the area of the rectangle (III) = $l\times b$ = 4 $\times$ 1 = 4 $cm^2$
the area of the rectangle (IV) = $l\times b$ = 3 $\times$ 2 = 6 $cm^2$

So, the total area of the figure = sum area of all rectangles
= (12 + 6 + 4 + 6) $cm^2$
= 28 $cm^2$

(b) Split the given figure into smalle rectangles as shown here;
3654
the area of the rectangle (I) = $l\times b$ = 3 $\times$ 1 = 3 $cm^2$
the area of the rectangle (II) = $l\times b$ = 3 $\times$ 1= 3 $cm^2$
the area of the rectangle (III) = $l\times b$ = 3 $\times$ 1 = 3 $cm^2$ ?

Therefore, the total area of the figure = sum of the area of all the small rectangles
= $3+3+3$
= $9cm^2$

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

1643093104177

Answer: (a) Split the given figure into a small rectangle, as shown here;
3266465
Now, the area of the rectangle (I) = $l\times b$ = 12 $\times$ 2 = $24cm^2$
the area of the rectangle (II) = $l\times b$ = 8 $\times$ 2= $16cm^2$

Therefore, Total area = $24cm^2 + 16cm^2 = 40cm^2$

(b) Split the given figure into a small rectangle, as shown here;
23654

Now, the area of the rectangle (I) = $l\times b$ = 7 $\times$ 7 = $49 cm^2$
the area of the rectangle (II) = $l\times b$ = 21 $\times$ 7= $147 cm^2$
Now, the area of the rectangle (III) = $l\times b$ = 7 $\times$ 7 = $49 cm^2$

Therefore, Total area = $49cm^2 + 147cm^2 + 49cm^2 = 245 cm^2$

Now, the area of the rectangle (I) = $l\times b$ = 2 $\times$ 1 = $2 cm^2$
the area of the rectangle (II) = $l\times b$ = 5 $\times$ 1= $5 cm^2$
Now, the area of the rectangle (III) = $l\times b$ = 2 $\times$ 1 = $2 cm^2$

Therefore, Total area = $2 cm^2 + 5cm^2 + 2cm^2 = 9cm^2$

Question: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.

Answer: Given that,
the length and breadth of the tiles are 12 cm and 5 cm

Therefore, the area of the tiles
= length $\times$ breadth
= 12 cm $\times$ 5 cm
= 60 cm ²

Now, (a) For 100 cm and 144 cm rectangular region,
Area of region = length $\times$ breadth
= 14,400 cm ²

Therefore, the number of tiles needed = $\frac{14,400}{60}=240$ tiles

(b) For 70 cm and 36 cm rectangular region,
Area of region = length $\times$ breadth
= 70 cm $\times$ 36 cm
= 2520 m ²

Required number of tiles = $\frac{2520}{60}=42$

NCERT Solutions for Class 6 Subject Wise

NCERT Solutions for Class 6 Maths

NCERT Solutions for Class 6 Science

NCERT Class 6 Maths Chapter 10 Mensuration Topics:

Perimeter
The perimeter of a rectangle
The perimeter of regular shapes
Area
Area of a rectangle
Area of a square

NCERT Solutions for Class 6 Mathematics Chapter Wise

Chapters No.	Chapters Name
Chapter - 1	Knowing Our Numbers
Chapter - 2	Whole Numbers
Chapter - 3	Playing with Numbers
Chapter - 4	Basic Geometrical Ideas
Chapter - 5	Understanding Elementary Shapes
Chapter - 6	Integers
Chapter - 7	Fractions
Chapter - 8	Decimals
Chapter - 9	Data Handling
Chapter -10	Mensuration
Chapter -11	Algebra
Chapter -12	Ratio and Proportion
Chapter -13	Symmetry
Chapter -14	Practical Geometry

Key features of NCERT Solutions for Class 6 Maths chapter 10

Conceptual Clarity: The NCERT Maths Class 6 Chapter 10 solutions aim to improve students' conceptual understanding by simplifying complex topics into easier and more understandable explanations.

Comprehensive Coverage: The class 6 chapter 10 maths solutions cover all the important topics and subtopics of Chapter 10, ensuring that students gain a complete grasp of the chapter's content.

Exam-Focused Approach: The NCERT chapter 10 maths class 6 solutions are designed with a focus on exams, equipping students with the necessary tools and techniques to effectively tackle questions and score well in their examinations.

Also Check

NCERT Books and NCERT Syllabus

Happy Reading!!!

Also Read

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NCERT Solutions for Class 6 Science Chapter 2 Components of Food

NCERT Solutions for Class 6 Science Chapter 1 Food Where Does It Come From

NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes

NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry

NCERT Solutions for Class 6 Maths Chapter 13 Symmetry

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

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NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

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