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NCERT Solutions for Class 6 Maths Chapter 10: Mensuration - Exercise 10.1

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MVSAT 2024

NCERT Solutions for Class 6 Maths Chapter 10 (Ex 10.1)

Free PDF download of NCERT Solutions for Class 6 Maths Chapter 10 Exercise 10.1 (Ex 10.1) and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. Class 6 Maths Chapter 10 Mensuration Exercise 10.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.


Class:

NCERT Solutions for Class 6

Subject:

Class 6 Maths

Chapter Name:

Chapter 10 - Mensuration

Exercise:

Exercise - 10.1

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English,Hindi will become easy to study if you have access to NCERT Solution for Class 6 Science , Maths solutions and solutions of other subjects.

Competitive Exams after 12th Science

Access NCERT Solutions for Class 6 Maths Chapter 10 - Mensuration

Exercise 10.1

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

(a)


A Quadrilateral


Ans: Perimeter = Sum of all sides

$ = 4 + 2 + 1 + 5$

$ = 12cm$


(b)


Quadrilateral


Ans: Perimeter = Sum of all sides

$ = 23 + 35 + 40 + 35$

$ = 133cm$


(c)


A Parallelogram


Ans: Perimeter = Sum of all sides

$ = 15 + 15 + 15 + 15$

$ = 60cm$


(d)


A Pentagon


Ans: Perimeter = Sum of all sides

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

$ = 20cm$


(e)


Find perimeter of arrow


Ans: Perimeter = Sum of all sides

$ = 1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4$

$ = 15cm$


(f)


Find Perimeter of given figure


Ans: Perimeter = Sum of all sides

$ = 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3$

$ = 52cm$


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?

Ans: Given:

Length of rectangular box = 40cm

Breadth of rectangular box = 10cm

To find: length of tape required

Length of tape required = Perimeter of rectangle

$ = 2(length + breadth)$

$ = 2(40 + 10)$

$ = 2 \times 50$

$ = 100cm$

Therefore, required length of tape is 100cm.


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

Ans: Given:

Length of table top $ = 2m\,25cm = \,2.25m$

Breadth of table top $ = 1m\,\,50cm = 1.50m$

To find: Perimeter of table top 

Here,

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

$ = 2(2.25 + 1.50)$

$ = 2 \times 3.75 = 7.50m$

Perimeter of table top is $7.50m$


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

Ans: Given: 

Length of frame: 32cm

Breadth of frame: 21cm

To find: length of wooden strip 

Required length of wooden strip required = Perimeter of photograph

$ = 2(length + breadth)$

$ = 2(32 + 21)$

$ = 2 \times 53cm = 106\,cm$

Therefore, 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?

Ans: Given: 

Length of rectangular land = \[0.7km\]

Breadth of rectangular land = \[0.5km\]

To find: 

Length of wire required to be fenced.

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

$ = 2(0.7 + 0.5)$

$ = 2 \times 1.2$

$ = 2.4\,km$

Total length of wire required for each side to be fenced with 4 rows

$ = 4 \times 2.4\,km$

$ = 9.6\,km$

Therefore, total length of the required wire is $9.6\,km$.


6. (a) A triangle of sides 3 cm, 4 cm and 5 cm.

Ans: Given: 

Sides of triangle = 3cm, 4cm , and 5cm.

To find: 

Perimeter of triangle

Perimeter of triangle $ = 10 + 14 + 15$

$ = 12\,cm$


(b) An equilateral triangle of side 9 cm.

Ans: Given: 

Side of equilateral triangle = 9cm

To find: 

Perimeter of equilateral triangle

Perimeter of equilateral triangle$ = 3 \times \,side$

$ = 3 \times 9\,cm$

$ = 27\,cm$


(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Ans: Given: 

Two equal Sides of isosceles triangle = 8cm

Third side of isosceles triangle = 6cm

To find: 

Perimeter of equilateral triangle

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

$ = 22\,cm$


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

Ans: Given: 

Side of triangle = 10cm, 14cm, 15cm

To find: 

Perimeter of triangle

Perimeter of triangle $ = 10 + 14 + 15$

$ = 39\,cm$

Perimeter of triangle is 39 cm.


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

Ans: Given: 

A Regular hexagon has 6 sides

Side of hexagon = 8cm

To find: 

Perimeter of rectangular hexagon

Perimeter of Hexagon $ = 6 \times 8\,cm$

$ = 48\,m$


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

Ans: Given: 

Perimeter of square = 20cm

To find: 

Side of the square.

Perimeter of square$ = 4 \times \,side$

$20 = \,4 \times \,side$

$side\, = \,\dfrac{{20}}{4}\, = 5\,m$


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

Ans: Given: 

Perimeter of pentagon = 100cm

To find: 

Side of pentagon

A Regular pentagon has 5 sides

Perimeter of regular pentagon$ = 100\,cm$

$5 \times \,side\, = \,100\,cm$

$\,side\, = \,\dfrac{{100}}{5}\, = \,20\,cm$


11. A piece of string is 30 cm long. What will be the length of each side if the string is used toform:

(a) a square

Ans: Perimeter of square$ = \,30\,cm$

$4 \times \,side\,\, = \,30\,cm$

Side of square $ = \,\dfrac{{30}}{4}$

$ = \,7.5\,cm$


(b) An equilateral triangle

Ans: Perimeter of equilateral triangle $ = \,30\,cm$

$3\, \times \,side\, = \,30\,cm$

$side\, = \,10\,cm$


(c) A rectangle hexagon

Ans: Perimeter of hexagon $ = \,30\,cm$

$6 \times \,side\, = \,30\,cm$

$side\, = \,5cm$


12. 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: Let the third side be $x$cm.

Perimeter of triangle $ = \,36\,cm$

$12 + 14 + x = 36$

$26 + x = 36$

$x = 36 - 26$

$x = 10\,cm$


13. Find the cost of fencing a square park of side 250 m at the rate of ` 20 permeter.

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

Perimeter of square $ = 4 \times side$

$ = 4 \times 250 = 1000\,m$

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

Hence, cost of fencing for 1000$m$ $ = 20 \times 1000\, = \,Rs.20,000$


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

Ans: Length ,$l = 175\,m$

Breadth ,$b = 125\,m$

Perimeter of park$ = 2\left( {l + b} \right)$

$ = 20\left( {175 + 125} \right)$

$ = 2 \times 300 = 600\,m$

Cost of fencing $ = 12 \times 600\, = \,Rs.7,200$


15. 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: Perimeter of square $ = 4 \times side$

$ = 4 \times 75 = 300\,m$

Hence, distance covered by Sweety is 300 m.

Perimeter of rectangle park$ = 2 \times \left( {l + b} \right)$

$ = 2 \times \left( {60 + 45} \right)$

$ = 2 \times 105\, = \,210\,m$

Hence, distance covered by Bulbul $ = 210\,m$

Bulbul covers lesser distance.


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

(a) 


A Square


Ans: Perimeter of square $ = 4 \times 25 = 100\,cm$


(b)


Rectangle


Ans: Perimeter of rectangle $ = 2(40 + 10)$

$ = 2 \times 50 = 100\,cm$


(c)


A Rectangle


Ans: Perimeter of rectangle $ = 2(30 + 20)$

$ = 2 \times 50 = 100\,cm$


(d)


A Triangle


Ans: Perimeter of triangle$ = 2(30 + 20)$

$ = 2 \times 50 = 100\,cm$

All the figures have same perimeter.


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


Square Paving Slabs


(a) What is the perimeter of his arrangement?

Ans: 6m


(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 $ = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 = 10\,m$


(c) Which has greater perimeter?

Ans: Second


(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, if we put all the squares in a line, then the perimeters will be 10 cm.


NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1

Opting for the NCERT solutions for Ex 10.1 Class 6 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 10.1 Class 6 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.


Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 6 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 6 Maths Chapter 10 Exercise 10.1 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.


Besides these NCERT solutions for Class 6 Maths Chapter 10 Exercise 10.1, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.


Do not delay any more. Download the NCERT solutions for Class 6 Maths Chapter 10 Exercise 10.1 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

FAQs on NCERT Solutions for Class 6 Maths Chapter 10: Mensuration - Exercise 10.1

1. How many questions are present in NCERT Solutions for Class 6 Maths Chapter 10 Mensuration?

The NCERT Class 6 Maths Chapter 10  Mensuration has the following exercise questions-

  • Exercise 10.1- 17 questions

  • Exercise 10.2 - 1 question

  • Exercise 10.3- 12 questions

The solutions to all these questions can be found on Vedantu’s website. The solutions are clearly explained in detail to clarify the concepts so that students can easily clear their doubts and gain confidence in the topic. Students should practice all the questions to understand the topic thoroughly.

2. How many questions have long solutions (more than 180 words) in NCERT Solutions for class 6 maths chapter 10 mensuration?

Chapter 10 of Class 6 Maths NCERT has the following long answer questions-

  • Exercise 10.1- 3 questions(1,15,16)

  • Exercise 10.2 - 1 question 

  • Exercise 10.3- 4 questions (1,10,11,12)

The long answer-type questions have step-by-step detailed explanations provided by Vedantu. The long answer type questions should be practised by students properly because they have numerous steps and each step is important if students want to score full marks in a long answer question.

3. What are the topics covered in Chapter 10 of NCERT Solutions for Class 6 Maths?

Chapter 10 of Class 6 Maths NCERT ‘Mensuration’ introduces students to the concept of comparing and measuring the regions and boundaries of plane figures. The first section of the chapter introduces students to the concept of perimeter and how to calculate it for rectangles and other regular shapes like square and equilateral triangles. The next section covers the concept of measuring the area of a rectangle and square. Students should learn this chapter properly as it is important for the exams.

4. What are the most important formulas that I need to remember in Class 6 Maths Chapter 10?

The most important formulas that students need to remember from the  NCERT Class 6 Maths Chapter 10 are the following-

  • The perimeter of a rectangle

  • The perimeter of a square

  • The perimeter of an equilateral triangle

  • Area of a rectangle

  • Area of a square

These formulas are important to learn as they will be used in higher classes when learning advanced concepts. Students can memorize them well by solving the exercise questions.

5.  What is the use of perimeter in our daily life?

The concept of perimeter means the distance covered along the boundary of a closed figure once. It is an important concept to learn as we see its use in many of our day-to-day situations like when a farmer wants to build a fence around his farm or during sports events when tracks are made. Engineers use it in their work when they build a wall on all sides of a house. In all these situations the perimeter is calculated.