NCERT Solutions For Class 6 Maths Chapter 11 Algebra Exercise 11.2

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.2 deals with concepts like geometric and arithmetic rules. They provide students with knowledge about the perimeter of squares and rectangles and various properties in arithmetics. The faculty at BYJU’S has explained these NCERT Solutions in an elaborate and interesting manner with the aim of helping students perform well in the Class 6 exam.

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.2

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Access NCERT Solutions for Class 6 Chapter 11: Algebra Exercise 11.2

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

Solutions:

Side of equilateral triangle = l

Perimeter = l + l + l

= 3l

2. The side of the regular hexagon (Fig 11.10) is denoted by l. Express the perimeter of the hexagon using l.

(Hint: A regular hexagon has all its six sides equal in length.)

NCERT Solutions for Class 6 Maths Chapter 11 Exercise 11.2 - 1

Solutions:

Side of a regular hexagon = l

Perimeter = l + l + l + l + l + 1

= 6l

3. A cube is a three-dimensional figure, as shown in Fig 11.11. 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.

NCERT Solutions for Class 6 Maths Chapter 11 Exercise 11.2 - 2

Solutions:

Length of an edge of the cube = l

Number of edges = 12

Total length of the edges = Number of edges × Length of an edge

=12l

4. The diameter of a circle is a line which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure (Fig 11.2), AB is a diameter of a circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius (r).

NCERT Solutions for Class 6 Maths Chapter 11 Exercise 11.2 - 3

Solutions:

Diameter = AB

= AC + CB

= r + r

= 2r

Hence, the diameter of the circle in terms of its radius is 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)

Solutions:

For any three whole numbers a, b and c,

(a + b) + c = a + (b + c)

Also, explore – 

NCERT Solutions for Class 6 Maths Chapter 11

NCERT Solutions for Class 6 Maths

NCERT Solutions for Class 6


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