GSEB Class 6 Maths Solutions Chapter 11 બીજગણિત Exercise 11.2

Official GSEB Solutions for Class 6 Mathematics: Chapter 11 બીજગણિત

Explore reliable textbook solutions for Chapter 11 બીજગણિત tailored for Class 6 learners. Utilizing these Mathematics answers ensures thorough preparation and strengthens foundational knowledge before final GSEB evaluations.

Chapter-wise Solutions for Mathematics: Chapter 11 બીજગણિત

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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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Free GSEB Textbook Explanations: Class 6 Mathematics Chapter 11 બીજગણિત

Official GSEB Solutions for Chapter 11 બીજગણિત

Access structured GSEB textbook solutions for Chapter 11 બીજગણિત. Designed in alignment with the latest academic curriculum for Class 6 Mathematics, these answers cover all end-of-chapter exercises to support daily learning and homework completion.

Step-by-Step Explanations for Chapter 11 બીજગણિત

Beyond providing final answers, these guides offer step-by-step breakdowns for complex queries in the Class 6 Mathematics module. This approach helps students balance theoretical depth with practical problem-solving skills required for GSEB exams.

Next Steps in Your Mathematics Revision

Consistent practice with these solution guides cultivates faster problem-solving habits and clearer logical structuring. For a complete preparation experience, pair these textbook answers with our dedicated revision notes and sample papers for Class 6 Mathematics.

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.

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