## NCERT Class 6 Mathematics Eleventh Chapter Algebra Exercise 11.2 Solution

## EXERCISE 11.2

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

Solution: Here, the length of side of the equilateral triangle is l.

Therefore, the perimeter of the triangle = (l + l + l) = 3l.

**(2) The side of a 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.)**

Solution: The length of the side is l.

There are 6 sides of a hexagon.

Therefore, the perimeter of the hexagon = (l + l + l + l + l + l) = 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.**

Solution: The length of the edge of the cube is l.

The number of edge = 12

Hence, the total length of the edge of the cube = 12 × l = 12l

**In case you are missed :- Previous Chapter Exercise Solution**

**(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 11.12) AB is a diameter of the circle; C is its centre.) Express the diameter of the circle (d) in terms of its radius (r).**

Solution: Here, we would write AC = CB = CP (because, both lines is broadest centre to ambits)

There CP = r

We know, AB d) = AC + CB = CP + CP = r + r = 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 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) **

Solution: Therefore, the ways we would write,

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

**In case you are missed :- Next Chapter Exercise Solution**