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# NCERT Solutions Class 6 Maths Chapter 10 Mensuration

Last updated date: 15th Sep 2024
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## NCERT Solutions Class 6 Maths Chapter 10 Mensuration - FREE PDF Download

NCERT Solutions for Chapter 10 Mensuration class 6 Maths, deals with the measurement of geometric figures. This chapter focuses on calculating the perimeter and area of basic shapes such as rectangles, squares, and triangles. By understanding these fundamental concepts, students learn how to measure the boundary length and the space enclosed within these shapes. Class 6 Maths NCERT Solutions is essential for solving practical problems in everyday life, such as determining the amount of material needed for construction projects or the fencing required for a garden. The chapter provides a solid foundation for more advanced geometrical concepts, emphasizing practical applications and problem-solving skills.

Table of Content
1. NCERT Solutions Class 6 Maths Chapter 10 Mensuration - FREE PDF Download
2. Glance on Maths Chapter 10 Class 6 - Mensuration
3. Access Exercise wise NCERT Solutions for Chapter 10 Maths Class 6
4. Exercises Under NCERT Solutions for Class 6 Maths Chapter 10 Mensuration
5. Access NCERT Solutions for Class 6 Maths Chapter 10 – Mensuration
6. Class 6 Maths Chapter 10: Exercises Breakdown
7. Other Study Material for CBSE Class 6 Physics Chapter 10
8. Chapter-Specific NCERT Solutions for Class 6 Maths
FAQs

## Glance on Maths Chapter 10 Class 6 - Mensuration

• Chapter 10 of Class 6 Maths introduces the concept of mensuration, focusing on measuring the perimeter and area of various geometric figures.

• This chapter covers how to find the perimeter of simple figures like rectangles, squares, and triangles, helping students grasp the concept of boundary measurement.

• The chapter delves into the area calculation, which measures the space enclosed within a shape. Students learn to calculate the area of rectangles and squares using specific formulas.

• There are three exercises (30 fully solved questions) in class 6th maths chapter 10 Mensuration.

## Access Exercise wise NCERT Solutions for Chapter 10 Maths Class 6

 Current Syllabus Exercises of Class 6 Maths Chapter 10 NCERT Solutions of Class 6 Maths Mensuration Exercise 10.1 NCERT Solutions of Class 6 Maths Mensuration Exercise 10.2 NCERT Solutions of Class 6 Maths Mensuration Exercise 10.3

## Exercises Under NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

• Exercise 10.1: In this exercise, students will learn the perimeter of Simple Shapes. Understand and calculate the perimeter of basic geometric shapes. Calculation of perimeter for rectangles, squares, and triangles. Application of perimeter in real-life scenarios.

• Exercise 10.2: In this exercise, students will learn how to calculate the area of rectangles and squares using specific formulas. Introduction to the area. Solving problems related to finding the area.

• Exercise 10.3: In this exercise, students will learn Mixed Problems on Perimeter and Area. The objective is to solve problems involving both perimeter and area calculations. Combining the concepts of perimeter and area. Solving real-life application problems. Enhancing problem-solving skills with mixed questions.

## Access NCERT Solutions for Class 6 Maths Chapter 10 – Mensuration

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

Ans:  According to this figure:

As we know Perimeter= Sum of all the sides

=$1+2+4+5$

=$3+8$

=$11$

So, the perimeter= $11$ cm

2. Find the perimeter of each of the following figures:

Ans: According to this figure:

As we know Perimeter= Sum of all the sides

=$23+35+35+40$

=$58+75$

=$133$

So, the perimeter= $133$ cm

3. Find the perimeter of each of the following figures:

Ans:  According to this figure:

As we know Perimeter= Sum of all the sides

=$15+15+15+15$

=$30+30$

=$60$

So, the perimeter= $60$ cm

4. Find the perimeter of each of the following figures:

Ans: According to this figure:

As we know Perimeter= Sum of all the sides

=$4+4+4+4+4$

=$8+8+4$

=$20$

So, the perimeter= $20$ cm

5. Find the perimeter of each of the following figures:

Ans: According to this figure:

As we know Perimeter= Sum of all the sides

=$\text{1 + 4 + 0}\text{.5 + 2}\text{.5 + 2}\text{.5 + 0}\text{.5 + 4 }$

=$5+3+3+4$

=$8+7$

So, the perimeter= $15$ cm

6. Find the perimeter of each of the following figures:

Ans: According to this figure:

As we know Perimeter= Sum of all the sides

=$\text{4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 }$

=$5+5+7+4+5+5+5+7+4+5$

=$10+11+10+12+9$

=$21+22+9$

=$52$

So, the perimeter= $52$ cm

7. 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?

Ans:

Here,

Required tape length = Perimeter of rectangle

Given, Length=$40$cm

Breadth=$10$cm

=$2(length+breadth)$

= $2(40+10)$

=$2\times 50$

=$100$

Required tape length = $100$ cm.

8. A table-top measures $2$ m $25$ cm by $1$ m $50$ cm. What is the perimeter of the table-top?

Ans: Here

$25$cm = $0.25$cm

$50$cm = $0.5$cm

So, Length of table top= $2+0.25$= $2.25$m

Breadth of table top = $1+0.5$=$1.5$m

Perimeter of table top = $2(length+breadth)$

=$2(2.25+1.5)$

=$2\times 3.75$

=$7.50$m

So, Perimeter = $7.5$m

9. What is the length of the wooden strip required to frame a photograph of length and breadth $32$ cm and $21$ cm respectively?

Ans: Here

Given, Length=$32$cm

Breadth=$21$cm

Required length of wooden strip  = Perimeter of photograph

=$2(length+breadth)$

=$2(32+21)$

=$2\times 53$cm

=$106$cm

Required Length of wooden strip = $106$cm

10. 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?

Ans: As we know,

Perimeter of field = $2(length+breadth)$

Given, Length=$0.7$km

Breadth=$0.5$km

=$2(length+breadth)$

=$2(0.7+0.5)$

=$2\times 1.2$km

=$2.4$km

A $4$row fence is required on each side =$4\times 2.4$= $9.6$km

So, Total length of the required field =  $9.6$km

11. Find the perimeter of each of the following shapes: A triangle of sides $3$ cm,  $4$cm and $5$ cm.

Ans: We know,

Perimeter of Triangle=$3+4+5$

=$12$cm

So, Perimeter of Triangle=$12$cm

12. Find the perimeter of each of the following shapes:  An equilateral triangle of side $9$ cm

Ans: We know,

Perimeter of Equilateral Triangle=$3\times side$

=$3\times 9$cm

= $27$cm

So, Perimeter of Equilateral  Triangle=$27$cm

13. Find the perimeter of each of the following shapes: An isosceles triangle with equal sides $8$ cm each and third side $6$ cm.

Ans: We know,

Perimeter of isosceles Triangle=$8+8+6$

=$22$cm

So, Perimeter of isosceles Triangle=$22$cm

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

Ans: We know,

Perimeter of Triangle=$10+14+15$

=$39$cm

So, Perimeter of Triangle=$39$cm

15. Find the perimeter of a regular hexagon with each side measuring $8$ cm.

Ans: We know,

Perimeter of Hexagon=$6\times 8$

=$48$m

So, Perimeter of Hexagon=$48$m

16. Find the side of the square whose perimeter is $20$ m.

Ans: We know,

Perimeter of square=$4\times side$

$20=4\times side$

Side=$\frac{20}{4}$

=$5$ m

So, Perimeter of Square=$5$m

17. The perimeter of a regular pentagon is $100$ cm. How long is its each side?

Ans: We know,

Perimeter of regular pentagon=$100$cm

$100=5\times side$

Side=$\frac{100}{5}$

=$20$ cm

So, Perimeter of Square=$20$cm

18. A piece of string is $30$ cm long. What will be the length of each side if the string is used to form: a square

Ans: We know,

Perimeter of square=$4\times side$

$30=4\times side$

Side=$\frac{30}{4}$

=$7.5$ cm

So, Perimeter of Square=$7.5$cm

19. A piece of string is $30$ cm long. What will be the length of each side if the string is used to form: an equilateral triangle

Ans: We know,

Perimeter of Equilateral Triangle=$3\times side$

$30=3\times side$

Side=$\frac{30}{3}$ cm

=$10$ cm

So, Perimeter of Equilateral Triangle=$10$cm

20. A piece of string is $30$ cm long. What will be the length of each side if the string is used to form: a regular hexagon

Ans: We know,

Perimeter of Hexagon=$30$cm

$6\times side=30$cm

Side=$\frac{30}{6}$ cm

=$5$ cm

So, Perimeter of Hexagon=$5$cm

21. Two sides of a triangle are $12$ cm and $14$ cm. The perimeter of the triangle is $36$ cm. What is the third side?

Ans: Assume third side to be $x$ cm

We know,

Perimeter of Triangle=$36$

$12+14+x=36$cm

$26+x=36$cm

$x=36-26$

$x=10$cm

So, third side =$10$cm

22. Find the cost of fencing a square park of side $250$ m at the rate of  $20$ per meter.

Ans: Given, Side of square =$250$m

We know,

Perimeter of square=$4\times side$

$=4\times 250$

=$1000$m

=$7.5$ cm

So, Perimeter of Square=$7.5$cm

Now, Cost of fencing= Rs.$20$per m

Than, for $1000$m cost of fencing = $20\times 1000$

=Rs.$20000$

23. Find the cost of fencing a rectangular park of length $175$ m and breadth $125$ m at the rate of  $12$ per meter.

Ans: Given, Length=$175$m

Breadth=$125$m

We know, Perimeter of park=$2(length+breadth)$

=$2(175+125)$

=$2\times 300$

=$600$m

Cost of fencing = $12\times 600$=Rs.$7200$

Hence, Cost of fencing = Rs.$7200$

24. Sweety runs around a square park of side $75$ m. Bulbul runs around a rectangular park with length of $60$ m and breadth $45$ m. Who covers less distance?

Ans: We know,

Perimeter of square=$4\times side$

=$4\times 75=300$m

Distance covered by Sweety =$300$m

Perimeter of rectangular park=$2(length+breadth)$

=$2(60+45)$

=$2\times 105$

=$210$m

Distance covered by Bulbul =$210$m

So, distance covered by Bulbul is less.

25. What is the perimeter of each of the following figures? What do you infer from the answer?

Ans:

(a) We know,

Perimeter of square=$4\times side$

=$4\times 25$

=$100$cm

(b) We know,

Perimeter of rectangle=$2(length+breadth)$

=$2(40+10)$

=$2\times 50$

=$100$cm

(c) We know,

Perimeter of rectangle=$2(length+breadth)$

=$2(30+20)$

=$2\times 50$

=$100$cm

(d) We know,

Perimeter of Triangle=$30+30+40$

=$100$cm

The perimeter of each fig is the same.

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

(a) What is the perimeter of his arrangement?

Ans: Side of Square= $3\times side$

=$3\times \frac{1}{2}$

=$\frac{3}{2}$m

Perimeter of Square = $4\times \frac{3}{2}$

=$2\times 3$

=$6$m

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

Ans: Perimeter = $\text{ 0}\text{.5 + 1 + 1 + 0}\text{.5 + 1 + 1+ 0}\text{.5 + 1 + 1 + 0}\text{.5 + 1 + }1$

=$10$m

(c) Which has greater perimeter?

Ans: Cross arrangement 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.)

Ans: Yes, the perimeter will be 10 cm if all of the squares are lined up in a straight line.

Exercise 10.2

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

Ans:  Here,

The number of squares is $9$

So, Area = $9$ sq. units

2. Find the areas of the following figures by counting squares:

Ans:  Here,

The number of squares is $5$

So, Area = $5$ sq. units

3. Find the areas of the following figures by counting squares:

Ans:  Here,

The number of complete squares is $2$

The number of half squares is $4$

So,  Total Area = $2+4(0.5)=4$ sq. units

4. Find the areas of the following figures by counting squares:

Ans: Here,

The number of squares is $8$

So, Area = $8$ sq. units

5. Find the areas of the following figures by counting squares:

Ans: Here,

The number of squares is $10$

So, Area = $10$ sq. units

6. Find the areas of the following figures by counting squares:

Ans: Here,

The number of complete squares is $2$

The number of half squares is $4$

So,  Total Area = $2+4(0.5)=4$ sq. units

7. Find the areas of the following figures by counting squares:

Ans: Here,

The number of complete squares is $4$

The number of half squares is $4$

So,  Total Area = $4+4(0.5)=6$ sq. units

8. Find the areas of the following figures by counting squares:

Ans: Here,

The number of squares is $5$

So, Area = $5$ sq. units

9. Find the areas of the following figures by counting squares:

Ans: Here,

The number of squares is $9$

So, Area = $9$ sq. units

10. Find the areas of the following figures by counting squares:

Ans: Here,

The number of complete squares is $2$

The number of half squares is $4$

So,  Total Area = $2+4(0.5)=4$ sq. units

11. Find the areas of the following figures by counting squares:

Ans: Here,

The number of complete squares is $4$

The number of half squares is $2$

So,  Total Area = $4+2(0.5)=5$ sq. units

12. Find the areas of the following figures by counting squares:

Ans: Here,

The number of complete squares is $3$

The number of half squares is $10$

So,  Total Area = $3+10(0.5)=8$ sq. units

13. Find the areas of the following figures by counting squares:

Ans: Here,

The number of complete squares is $7$

The number of half squares is $14$

So, Total Area = $7+14(0.5)=14$ sq. units

14. Find the areas of the following figures by counting squares:

Ans:  Here,

The number of complete squares is $10$

The number of half squares is $16$

So,  Total Area = $10+16(0.5)=18$ sq. units

Exercise 10.3

1. Find the areas of the rectangles whose sides are:$3$cm and $4$cm

Ans: We know, the area of triangle = Length $\times$ Breadth

Here,

Length=$3$cm

Breadth=$4$cm

Area=$3\times 4$= $12$cm2

2. Find the areas of the rectangles whose sides are:$12$m and $21$m

Ans: We know, the area of triangle = Length $\times$ Breadth

Here,

Length=$12$m

Breadth=$21$m

Area=$12\times 21$= $252$m2

3. Find the areas of the rectangles whose sides are:$2$km and $3$km

Ans: We know, the area of triangle = Length $\times$ Breadth

Here,

Length=$2$km

Breadth=$3$km

Area=$2\times 3$= $6$km2

4. Find the areas of the rectangles whose sides are:$2$m and $70$cm

Ans: We know, the area of triangle = Length $\times$ Breadth

Here,

Length=$2$m

Breadth=$70$cm = $0.7$m

Area=$2\times 0.7$= $1.4$m2

5. Find the areas of the squares whose sides are: $10$cm

Ans: We know,

Area of square = Side2

=${{10}^{2}}$

=$100$ cm2

6. Find the areas of the squares whose sides are: $14$cm

Ans: We know,

Area of square = Side2

=${{14}^{2}}$

=$196$ cm2

7. Find the areas of the squares whose sides are: $5$cm

Ans: We know,

Area of square = Side2

=${{5}^{2}}$

=$25$ m2

8. The length and the 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?

Ans:

(a) We know, the area of rectangle = Length $\times$ Breadth

Here,

Length=$9$m

Breadth=$6$m

Area=$9\times 6$= $54$m2

(b) We know, the area of rectangle = Length $\times$ Breadth

Here,

Length=$3$m

Breadth=$17$m

Area=$3\times 17$= $51$m2

(c) We know, the area of rectangle = Length $\times$ Breadth

Here,

Length=$4$m

Breadth=$14$m

Area=$4\times 14$= $56$m2

Rectangle (c) has the largest size, $56$ m2, whereas rectangle (b) has the smallest area, $51$ m2

9. The area of a rectangle garden $50$ m long is $300$ m2, find the width of the garden.

Ans: We know, the area of rectangle = Length $\times$ Breadth

Here,

Area of rectangle =$300$ m2

Length= $50$m

$300=50\times width$

Width=$6$m

10. What is the cost of tilling a rectangular plot of land $500$ m long and $200$ m wide at the rate of ` $8$ per hundred sq. m?

Ans: Area of the land can be calculated as

=$500\times 200$

=$1,00,000$m2

Cost of tiling $1,00,000$m2 of land will be

=$\frac{8\times 100000}{100}$ = Rs.$8000$

11. A table-top measures $2$ m by $1$ m $50$ cm. What is its area in square meters?

Ans: Given,

L=$2$m

B=$1$m$50$cm= $1.5$m

Area = L$\times$B

=$2\times 1.50=3$m2

12. A room us $4$ m long and $3$ m $50$ cm wide. How many square meters of carpet is needed to cover the floor of the room?

Ans:: Given,

L=$4$m

B=$3$m$50$cm= $3.5$m

Area = L$\times$B

=$4\times 3.50=14$m2

13. 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.

Ans:  Here,

Area of floor = L$\times$B=$5$ m $\times$ $4$ m = $20$ m2

Area of square carpet = $3$ $\times$ $3$ = $9$ m2

Area of floor that is not carpeted =  $20-9$=$11$ m2

14. 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?

Ans: Here,

Area of flower square bed =$1\times 1=1$ m2

Area of 5 square beds = $1\times 5=5$ m2

Area of land = $5\times 4=20$m2

Remaining Area = Area of land – Area of 5 flower beds = $20-5=15$ m2

Therefore, remaining part of the land= $15$m2

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

Ans:

Here,

Area of 1st region =$3\times 3=9$ cm2

Area of 2nd region = $1\times 2=2$cm2

Area of 3rd region = $3\times 3=9$ cm2

Area of 4th region = $2\times 4=8$ cm2

Hence, Total area = $9+2+9+8$ = $28$ cm2

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

Ans:

Area of 1st region =$3\times 1=3$ cm2

Area of 2nd region = $3\times 1=3$cm2

Area of 3rd region = $3\times 1=3$ cm2

Hence, Total area = $3+3+3$ = $9$ cm2

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

Ans:

Here,

Area of 1st rectangle= $12\times 2$

Area of 2nd rectangle= $8\times 2$

Total area of the figure = $12\times 2+8\times 2$ = $40$ cm2

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

Ans: Here,

From figure it is clear that there are five square with each side  $7$.

Area of one square=$7\times 7=49$

So, area of five square= $49\times 5=245$cm2

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

Ans:

Area of 1st region = $5\times 1=5$cm2

Area of 2nd region =$2\times 1=2$ cm2

Area of 3rd region = $2\times 1=2$ cm2

Hence, Total area = $2+5+2$ = $9$ cm2

20. 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:$100$ cm and $144$ cm

Ans: Here,

Area of Rectangle= $100\times 144=14400$cm2

Area of one tile = $5\times 12=60$cm2

We know, Number of tiles = Area of rectangle / Area of one tile

=$\frac{14400}{60}$

= $240$cm2

$240$ tiles are required.

21. 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:$70$ cm and $36$ cm

Ans: Here,

Area of Rectangle= $70\times 36=2520$cm2

Area of one tile = $5\times 12=60$cm2

We know, Number of tiles = Area of rectangle / Area of one tile

=$\frac{2520}{60}$

= $42$cm2

$42$ tiles are required.

## Class 6 Maths Chapter 10: Exercises Breakdown

 Exercise Number of Questions Exercise 10.1 17 Questions & Solutions Exercise 10.2 1 Questions & Solutions Exercise 10.3 12 Questions & Solutions

## Conclusion

Class 6 Maths Chapter Mensuration provides students with essential tools to measure and understand the perimeter and area of various geometric shapes like rectangles and squares. Through this exercise, students learn to apply fundamental formulas, enhancing their ability to solve practical problems involving space and boundaries. By mastering these concepts, students gain the ability to solve practical problems and apply their knowledge to real-world scenarios, such as calculating the amount of material needed for different projects. The exercises and problems in this chapter reinforce practical application and critical thinking, laying a strong foundation for future geometrical studies.

## Other Study Material for CBSE Class 6 Physics Chapter 10

 S. No Important Links for Chapter 10 Mensuration 1 Class 6 Mensuration Important Questions 2 Class 6 Mensuration Revision Notes 3 Class 6 Mensuration Important Formulas 4 Class 6 Mensuration NCERT Exemplar Solution 5 Class 6 Mensuration RD Sharma Solutions

## Chapter-Specific NCERT Solutions for Class 6 Maths

Given below are the chapter-wise NCERT Solutions for Class 6 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 6 Maths Chapter 10 Mensuration

1. Calculate the Perimeter of a Square Which is 3 cm as its Length of the Side.

The length of a side = 3 cms

The formula for calculating perimeter of a square = 4* side

= 4*3

= 12 cms

Therefore, Perimeter = 12 cms.

2. Is it easy to Learn Mathematics Using These PDF solutions?

Yes. It is easy to learn mathematics using NCERT PDF books even for the slow learners because every concept has been explained in a detailed manner by a group of well-experienced experts. They provide several solid examples along with the unsolved test papers. It makes the students get a sound knowledge of every concept. If the students have any doubts, the teachers are available on the website to clarify the doubts.

3. What is the tangent of a circle?

A tangent to a circle is a straight line that meets the circle at anyone or only one point, such point is known as tangency, and at this point, the tangent is at 90 degrees with respect to the circle’s radius. Tangent lines to circles are the topic of numerous theorems and are used in a variety of geometrical constructions and proofs.

4. Is NCERT Class 6 Maths chapter 10 easy?

Maths is fun if you know the tricks, but if you don't know the topics and formulae well, things can get complicated and confusing because there are numerous ways to solve a question or equation. This chapter on Mensuration, Chapter 10 Class 6, is a very important one, and it will also form the base for the other chapters that you will learn further. Focus on the fundamentals, and with enough practise, you'll be fine.

5. What do you mean by a minor and a major arc?

A minor arc is the shortest arc that connects two points on a circle. A minor arc has a measure that is less than 180° and equal to the measure of the arc's centre angle. The longer arc connecting two endpoints of a circle is known as a major arc. These two are fundamental parts of a circle and must be learnt well to solve different problems related to circles.

6. How to achieve good marks in Class 6 Maths Chapter 10?

The steps one can take would be to ace well in Class 6 Chapter 10:

• Refresh your memory on the essentials. Try to recall all the basic concepts.

• Go for the basic logic behind any derivation and formula. Learn each formula related to circles judiciously.

• Comprehensively understand the different approaches to solving any equation or any problem.

• Try to answer NCERT questions.

• Determine which topics of Mensuration you are acquainted with and which require additional practice.

• If you are experiencing difficulties comprehending a topic, you may seek the help of NCERT Solutions for Class 6 Maths to remove any doubts.

7. What do you mean by Mensuration?

Mensuration is a branch of mathematics that examines the computation of geometric figures and their characteristics such as area, length, volume, lateral surface area, surface area, and so on. It explains all of the basic equations and characteristics of numerous geometric forms and figures, as well as the foundations of computation. It is an important aspect of Mathematics, and proper attention must be given to this topic along with adequate practice.

8. Why is learning mensuration important in class 6 maths chapter 10 pdf answer?

Learning mensuration in class 6 maths chapter 10 pdf answer is important because it helps in understanding and solving real-life problems involving measurement and space, such as determining the amount of material needed for construction or the area of a garden.

9. What are the key topics covered in this chapter mensuration class 6 solutions?

The key topics covered in this chapter mensuration class 6 solutions include the calculation of perimeter and area for rectangles and squares, understanding different measurement units, and applying these concepts to solve practical problems.

10. What is Mensuration in NCERT class 6 maths chapter 10?

Mensuration in Class 6 math chapter 10 is the branch of mathematics that deals with the measurement of geometric figures, including their perimeter, area, and volume.

11. What are the units used for measuring perimeter and area NCERT for class 6 maths chapter 10 mensuration pdf?

The perimeter is measured in linear units such as meters (m) or centimeters (cm), while the area is measured in square units such as square meters (m²) or square centimeters (cm²).

12. How do NCERT solutions for class 6 maths chapter 10 mensuration pdf help in understanding the practical applications of mensuration?

NCERT solutions for class 6 maths chapter 10 mensuration pdf provides exercises and problems that simulate real-world scenarios, enhancing students' problem-solving skills and helping them understand the practical applications of measuring perimeter and area.