Haryana State Board HBSE 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Ex 5.1 Textbook Exercise Questions and Answers.

## Haryana Board 6th Class Maths Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.1

Question 1.

What is the disadvantage in comparing line segments by mere observation ?

Solution:

In comparing line segments by mere observation, we cannot always be sure about the usual judgement. For example, look at A the following segments :

The difference in lengths between these two may not be obvious. Actually in this figure \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{PQ}}\) have the same lengths. This is not quite obvious.

Question 2.

Why is it better to use a divider with a ruler, while measuring the length of a linesegment ? ‘

Solution:

It is better to use a divider and a ruler, while measuring the length of a line segment, because by this method we can find the exact length of a line segment.

Question 3.

Draw a line segment, say \(\overline{\mathrm{AB}}\). Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AC = AB + BC ? [Note: If A, B, C are any three points on a line such that AC + CB = AB, then we can be sure that C lies between A and B.]

Solution:

On measuring, we find that

AB = 5 cm

BC = 2 cm

and AC = 3 cm

AB + BC = 5 cm + 2 cm = 7 cm

∴ AC ≠ AB + BC

But AC + CB = 3 cm + 2 cm

= 5 cm = AB.

Thus, AC + CB = AB.

Question 4.

If A, B, C are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two ?

Solution:

AB = 5 cm

BC = 3cm

and AC = 8 cm

AB + BC = 5 cm + 3 cm

= 8 cm = AC

AB + BC = AC

∴ B lies between A and C.

Question 5.

Verify whether D is the mid-point of AG.

Solution:

AD = 4 – 1 = 3 units

DG = 7 – 4 = 3 units

∵ AD = DG

⇒ D is the mid-point of AG.

Question 6.

If B is the mid-point of \(\overline{\mathrm{AC}}\) and C is the mid-point of \(\overline{\mathrm{BD}}\) , where A, B, C, D lie on a straight line, say why AB = CD ?

Solution:

B is the mid-point of \(\overline{\mathrm{AC}}\)

⇒ AB = BC ………… (i)

and C is the mid-point of \(\overline{\mathrm{BD}}\)

⇒ BC = CD …(ii)

Comparing (i) and (ii), we get AB = CD.

Question 7.

Draw five triangles and measure their sides. Check in each case if the sum of two sides is ever less than the third side.

Solution:

(i) AB = BC = AC = 3 cm

AB + BC = 3 cm + 3 cm = 6 cm

∴ AB + BC > AC

BC + AC = 3 cm + 3 cm = 6 cm

∴ BC + AC > AB

AC + AB = 3 cm + 3 cm = 6 cm

∴ AC + AB > BC.

(ii) AB = AC = 3.5 cm

and BC = 3 cm

AB + BC = 3.5 cm + 3 cm

= 6.5 cm

∴ AB + BC > AC

BC + AC = 3 cm + 3.5 cm

= 6.5 cm

∴ BC + AC > AB

AC + AB = 3.5 cm + 3.5 cm

= 7 cm .

∴ AC + AB > BC.

(iii) Scalene Triangle :

AB = 2.5 cm, BC = 4 cm

AC = 3.5 cm

AB + BC = 2.5 cm + 4 cm = 6.5 cm

∴ AB + BC > AC

BC + AC = 4 cm + 3.5 cm = 7.5 cm

∴ BC + AC > AB

AC + AB = 3.5 cm + 2.5 cm = 6 cm

∴ AC + AB > BC.

(iv) AB = 4 cm, BC = 3 cm, AC = 5 cm

AB + BC = 4 cm + 3 cm = 7 cm

AB + BC > AC

BC + AC = 3 cm + 5 cm = 8 cm

BC + AC > AB

AC + AB = 5 cm + 4 cm = 9 cm

AC + AB > BC.

(v) AB = 2.5 cm

BC = 3 cm

AC = 5 cm

AB + BC = 2.5 cm + 3 cm = 5.5 cm AB + BC > AC

BC + AC = 3 cm + 5 cm = 8 cm BC + AC > AB

AC + AB = 5 cm + 2.5 cm = 7.5 cm

AC + AB > BC.