GSEB Class 6 Maths Solutions Chapter 11 Algebra Exercise 11.2

Get the most accurate GSEB Solutions for Class 6 Mathematics Chapter 11 Algebra 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 Algebra 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 Algebra solutions will improve your exam performance.

Class 6 Mathematics Chapter 11 Algebra GSEB Solutions PDF

 

Question 1. The side of an equilateral triangle is shown by \( l \). Express the perimeter of the equilateral triangle using \( l \).
Answer: A B C lThe side of the equilateral triangle ABC is \( l \). Therefore, its perimeter equals AB plus BC plus AC. This gives us \( l + l + l = 3l \). Thus, the perimeter of an equilateral triangle becomes \( 3l \). Note: \( 3 \times l \), 3.1, and \( 3l \) all mean the same thing. They show the multiplication of 3 and \( l \).
In simple words: An equilateral triangle has three equal sides, each measured as \( l \). To find the perimeter, you just add the lengths of all three sides together. So, \( l + l + l \) becomes \( 3l \), which is the total distance around the triangle.

Exam Tip: Remember that an equilateral triangle has three sides of equal length. Its perimeter is always found by multiplying the side length by three.

 

Question 2. The side of a regular hexagon in the figure is denoted by \( l \). Express the perimeter of the hexagon using \( l \).
Answer: lHint: A regular hexagon has all six of its sides the same length. Since all sides of a regular hexagon possess equal lengths, each side of the hexagon is \( l \). Therefore, its perimeter is \( l + l + l + l + l + l = 6l \).
In simple words: A regular hexagon has six sides, all the same length. If each side is \( l \), then its perimeter is found by adding \( l \) together six times, which gives you \( 6l \).

Exam Tip: Always remember that the perimeter of any regular polygon is found by multiplying the number of sides by the length of one side.

 

Question 3. A cube is a three-dimensional figure as shown here. 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.
Answer: lA cube possesses six identical faces and contains 12 edges. Every edge has an equal length. So, the total length of all edges equals \( 12 \times l \) or simply \( 12l \).
In simple words: A cube has 12 edges, and they all have the same length. If one edge is \( l \), then to find the total length of all edges, you multiply \( l \) by 12.

Exam Tip: For problems involving 3D shapes, count the number of edges carefully to avoid errors in your calculations.

 

Question 4. The diameter of a circle is a line that joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure, AB is the diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius (r).
Answer: A B C P rThe radius is represented by \( r \) and the diameter by \( d \). Because the diameter of a circle is twice the radius. Consequently, the diameter equals 2 times the radius. This means \( d = 2 \times r \) or just \( d = 2r \).
In simple words: The diameter of a circle is simply two times its radius. So, if you know the radius \( r \), you just multiply it by 2 to get the diameter \( d \).

Exam Tip: Remember the basic relationships in a circle: diameter is twice the radius, and the radius is half the diameter. These are fundamental for solving circle-related problems.

 

Question 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 of 54 or (b) We may add 27 and 13 to get 40 and then add 14 to get the sum of 54. Thus, (14+27) + 13 = 14 + (27 + 13). 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.
Answer: Let the three numbers be \( a \), \( b \), and \( c \). Then, the property called 'associativity of addition' can be shown as: \( a + (b + c) = (a + b) + c \).
In simple words: This rule means that when you add three numbers together, it doesn't matter which two you add first. You will always get the same total answer.

Exam Tip: The associative property is key for simplifying calculations and understanding number operations. It applies to addition and multiplication, but not subtraction or division.

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GSEB Solutions Class 6 Mathematics Chapter 11 Algebra

Students can now access the GSEB Solutions for Chapter 11 Algebra 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 Algebra

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 Algebra to get a complete preparation experience.

FAQs

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

The complete and updated GSEB Class 6 Maths Solutions Chapter 11 Algebra 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 Algebra 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 Algebra Exercise 11.2 will help students to get full marks in the theory paper.

Do you offer GSEB Class 6 Maths Solutions Chapter 11 Algebra 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 Algebra Exercise 11.2 in both English and Hindi medium.

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