GSEB Class 6 Maths Solutions Chapter 11 બીજગણિત Exercise 11.2

Get the most accurate GSEB Solutions for Class 6 Mathematics Chapter 11 બીજગણિત here. Updated for the 2026-27 academic session, these solutions are based on the latest GSEB textbooks for Class 6 Mathematics. Our expert-created answers for Class 6 Mathematics are available for free download in PDF format.

Detailed Chapter 11 બીજગણિત GSEB Solutions for Class 6 Mathematics

For Class 6 students, solving GSEB textbook questions is the most effective way to build a strong conceptual foundation. Our Class 6 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 11 બીજગણિત solutions will improve your exam performance.

Class 6 Mathematics Chapter 11 બીજગણિત GSEB Solutions PDF

 

Question 1. Represent the length of the side of an equilateral triangle with \( l \) and use it to show the perimeter of the equilateral triangle.

A B C l l l

Answer: Let \( \triangle ABC \) represent an equilateral triangle. Each side of \( \triangle ABC \) has a length of \( l \). The perimeter of \( \triangle ABC \) is calculated by adding its sides: \( AB + BC + CA = l + l + l = 3l \). So, the equilateral triangle's perimeter equals \( 3l \).
In simple words: For an equilateral triangle, all three sides are the same length. To find its perimeter, you just add the length of its three sides together, which is \( l + l + l \), making it \( 3l \).

Exam Tip: Remember that an equilateral triangle has three equal sides, which simplifies the perimeter calculation to three times the length of one side.

 

Question 2. Represent the sides of the regular hexagon below with \( l \), and using this \( l \), show the perimeter of the regular hexagon. (Hint: All sides of a regular hexagon are equal.)

l l l l l l

Answer: For a regular hexagon, all its sides possess the same length. In this case, the side length of the hexagon is \( l \). The perimeter of the hexagon equals the sum of its six side lengths: \( l + l + l + l + l + l = 6l \). Therefore, the regular hexagon's perimeter comes out to be \( 6l \).
In simple words: A regular hexagon has six equal sides. To find its perimeter, you simply multiply the length of one side by six, because you add \( l \) six times.

Exam Tip: Remember that "regular" means all sides and angles are equal. This makes calculating the perimeter of any regular polygon straightforward: just multiply the number of sides by the length of one side.

 

Question 3. A three-dimensional cube with 6 faces, each of which is square, is shown in the figure below. Represent the length of the edge of this cube with \( l \) and find the formula for the total length of the edges of this cube.

l

Answer: A cube consists of six identical faces. In total, a cube possesses 12 edges. All of these edges maintain an equal length. Here, we denote the edge length of the cube as \( l \). Consequently, the total length of all cube edges becomes \( 12 \times l = 12l \). So, the cube's entire edge length measures \( 12l \).
In simple words: A cube has 12 edges, and they are all the same length. So, to find the total length of all edges, you multiply the length of one edge by 12.

Exam Tip: Visualizing the cube and counting its edges (4 on top, 4 on bottom, 4 vertical connectors) helps confirm the number 12 for edge count.

 

Question 4. The line segment connecting two points on a circle passing through its center is the diameter of the circle. (In the figure, \( \overline{AB} \) is the diameter of the circle. C is its center. Express the diameter in terms of radius r.)

C B A r

Answer: In this case, the provided circle has a radius of \( r \) and a diameter of \( d \). We know that the diameter of any circle is always twice its radius.
\( \implies d = 2 \times r \) or \( d = 2r \). This formula expresses the diameter in terms of the radius.
In simple words: The diameter of a circle is simply two times its radius. If the radius is \( r \), then the diameter \( d \) is \( 2r \).

Exam Tip: Clearly define your variables (d for diameter, r for radius) and remember the fundamental relationship: diameter is always double the radius.

 

Question 5. We have two methods to sum 14, 27, and 13:
(a) First, we add 14 and 27 to get 41, then add 13 to it. The total sum will be 54.
(b) Alternatively, we add 27 and 13 to get 40, and then add 14 to it. This gives \( (14 + 27) + 13 = 14 + (27 + 13) \).
This can be done for any three numbers. This property is known as the Associative Property for addition. This property is shown in the chapter on whole numbers, which we have already studied. Generally, variables \( a, b \), and \( c \) are used.

Answer: As stated in the question, for any three numbers represented by \( a, b \), and \( c \), the associative property in addition can be shown as follows: \( (a + b) + c = (b + c) + a \). This rule allows us to group numbers differently without changing the final sum.
In simple words: The associative property means you can change how you group numbers when adding them, and the total will still be the same. For example, \( (a+b)+c \) is the same as \( a+(b+c) \).

Exam Tip: Clearly write down the expression and show the grouping with parentheses. The associative property applies to addition and multiplication, but not to subtraction or division.

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GSEB Solutions Class 6 Mathematics Chapter 11 બીજગણિત

Students can now access the GSEB Solutions for Chapter 11 બીજગણિત prepared by teachers on our website. These solutions cover all questions in exercise in your Class 6 Mathematics textbook. Each answer is updated based on the current academic session as per the latest GSEB syllabus.

Detailed Explanations for Chapter 11 બીજગણિત

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 6 Mathematics chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 6 students who want to understand both theoretical and practical questions. By studying these GSEB Questions and Answers your basic concepts will improve a lot.

Benefits of using Mathematics Class 6 Solved Papers

Using our Mathematics solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 6 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 11 બીજગણિત to get a complete preparation experience.

FAQs

Where can I find the latest GSEB Class 6 Maths Solutions Chapter 11 બીજગણિત Exercise 11.2 for the 2026-27 session?

The complete and updated GSEB Class 6 Maths Solutions Chapter 11 બીજગણિત Exercise 11.2 is available for free on StudiesToday.com. These solutions for Class 6 Mathematics are as per latest GSEB curriculum.

Are the Mathematics GSEB solutions for Class 6 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the GSEB Class 6 Maths Solutions Chapter 11 બીજગણિત Exercise 11.2 as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Mathematics concepts are applied in case-study and assertion-reasoning questions.

How do these Class 6 GSEB solutions help in scoring 90% plus marks?

Toppers recommend using GSEB language because GSEB marking schemes are strictly based on textbook definitions. Our GSEB Class 6 Maths Solutions Chapter 11 બીજગણિત Exercise 11.2 will help students to get full marks in the theory paper.

Do you offer GSEB Class 6 Maths Solutions Chapter 11 બીજગણિત Exercise 11.2 in multiple languages like Hindi and English?

Yes, we provide bilingual support for Class 6 Mathematics. You can access GSEB Class 6 Maths Solutions Chapter 11 બીજગણિત Exercise 11.2 in both English and Hindi medium.

Is it possible to download the Mathematics GSEB solutions for Class 6 as a PDF?

Yes, you can download the entire GSEB Class 6 Maths Solutions Chapter 11 બીજગણિત Exercise 11.2 in printable PDF format for offline study on any device.