Chapter 11 Algebra Exercise 11.2

NCERT Solutions For Class 6 Maths Chapter 11 Algebra Exercise 11.2 - 2025-26

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In NCERT Solutions Class 6 Maths Chapter 11 Exercise 11 2, you’ll get to play with algebra in a simple way! Here, you’ll learn how to write formulas for the perimeter of shapes, use letters (like l and r) in maths, and see how rules work for numbers. If you ever feel confused about how to start using algebra or solve these new kinds of questions, you’ll find the step-by-step NCERT solutions handy from Vedantu.

Don’t forget, you can also download the exercise solutions as a PDF or check them online if you need quick help for your homework or revision. If you want to see the full syllabus, you can check the Class 6 Maths Syllabus. Practicing these NCERT Solutions will make exam preparation easier and help you score higher in maths!

Access NCERT Solutions for Class 6 Maths Chapter 11 – Algebra

Exercise 11.2

Refer to pages 1-5 for exercise 11.2 in the PDF.

1. The side of an equilateral triangle is shown by \[l\]. Express the perimeter of the equilateral triangle using \[l\].

Ans: Given side of an equilateral triangle is \[l\].

We know that the perimeter of an equilateral triangle is \[P = 3a\], where a is the side of the equilateral triangle.

The perimeter of an equilateral triangle

\[ \Rightarrow P = 3 \times l = 3l\].

2. The side of a regular hexagon is denoted by l. Express the perimeter of the hexagon using l. (Hint: A regular hexagon has all its six sides equal in length.)

Ans: Given side of the hexagon is \[l\].

We know that the perimeter of the hexagon is \[P = 6a\], where a is the side of the hexagon.

Perimeter of hexagon

\[ \Rightarrow P = 6 \times l = 6l\]

3. A cube is a three-dimensional figure. It has six faces and all of them are identical squares. The length of an edge of the cube is given by $l$. Find the formula for the total length of the edges of a cube.

Ans: Given the length of one edge of a cube is \[l\].

We know that the number of edges in a cube is 12.

Hence the total length is given by

\[ \Rightarrow P = 12 \times l = 12l\]

4. The diameter of a circle is a line which joins two points on the circle and also passes through the center of the circle. (In the adjoining figure AB is the diameter of the circle; C is its center.) Express the diameter of the circle (d) in terms of its radius (r).

We know that the length of a diameter is two times the length of the radius. Here they have given the radius as r.

Ans: We know that length of a diameter is two times the length of the radius. Here they have given the radius as r.

\[ \Rightarrow d = 2r\]

5. To find the sum of three numbers 14, 27, and 13, we can have two ways:

(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54 or

(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54. Thus, $(14 + 27) + 13 = 14 + (27 + 13)$

This can be done for any three numbers. This property is known as the associativity of the addition of numbers. Express this property which we have already studied in the chapter on Whole Numbers, in a general way, by using variables a, b and c.

Ans: The given property is stated as following,

For the given 3 variables a, b and c we have

\[\left( {a + b} \right) + c = a + \left( {b + c} \right)\]

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.2

Opting for the NCERT solutions for Ex 11.2 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 11.2 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 Subject 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 11 Exercise 11.2 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 11 Exercise 11.2, 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 11 Exercise 11.2 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.