NCERT Solutions For Class 6 Maths Ex 11.2

NCERT Solutions for Class 6 Maths Chapter 11 Algebra Exercise 11.2 deals with concepts like geometric and arithmetic rules. It provides students with knowledge about perimeter of square and rectangle and various properties in arithmetics. The faculties at BYJU’S have 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 Chapter 11: Algebra Exercise 11.2 Download PDF

 

ncert solutions for class 6 maths chapter 11 ex 2
ncert solutions for class 6 maths chapter 11 ex 2

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 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 sum of three numbers 14, 27 and 13 we can have two ways:

(a) We may first add 14 and 27 to get 41and 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)


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