**(01) What is the disadvantage in comparing line segments by mere observation?**

**Sol. **

a. Two line segments with slight difference in length cannot be differentiated by mere observation

b. If you are doing architectural work (designing buildings etc) you need perfect length of lines. Comparing the line segment of wall just by observation can lead to improper design of the structure.

**(02)** **Why is it better to use a divider than a ruler, while measuring the length of a line segment?**

**Sol. **While measuring the length of a line segment, divider is preferred because by using ruler there are more chances of error because of two reason – i) because of thickness of the ruler ii) error due to angular viewing.

So, it is better to use divider for measuring the length of a line segment because it gives accurate result.

**(03) Draw any line segment, say AB. Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB?****[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.]**

Since, C is a point on the mid of the line-segment AB, Therefore, for all the cases in which C lies between AB line segment, AB = AC + CB.

Example –

Here, AB = 10 cm

C is the mid-point of line segment AB

On measuring, AC = 3.5 cm and CB = 6.5 cm

Hence, AB = 10 cm

& AC + CB = 6.5 cm + 3.5 cm = 10 cm = AB

Therefore, AB = AC + CB ** proved**

**(04) 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?**

**. **Given, AB = 5 cm

BC = 3 cm

AC = 8 cm

Clearly, AC = AB + BC

Therefore, point B lies in between A and C

**(05) Verify, whether D is the midpoint of AG**

**Sol. **Here, AG = 6 units

AD = 3 units

DG = 3 units

Clearly, AG = AD + DG = 6 unis

Therefore, D is the midpoint of AB.

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

Given – B is the mid-point line segment AC.

Therefore, AB = BC …………………………eq. 1

Also, C is the mid-point line segment BD.

Therefore, BC = CD …………………………eq. 2

From eq.1 and eq.2

AB = BC = CD

Therefore, AB = CD **proved.**

**(07)** **Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side**

AB= 7.5 cm

BC = 8.8 cm and

AC = 13 cm

AB + BC = 7.5 cm + 8.8 cm

= 16.3 cm

Since, 16.3 > 13

Therefore, AB + BC > AC Hence, the sum of any two sides of a triangle is greater than the third side

AB= 7.5 cm

BC = 7.5 cm and

AC = 12 cm

AB + BC = 7.5 cm + 7.5 cm

= 15 cm

Since, 15 > 12

Therefore, AB + BC > AC Hence, the sum of any two sides of a triangle is greater than the third side

AB= 7.5 cm

BC = 13 cm and

AC = 10 cm

AB + BC = 7.5 cm + 10 cm

= 17.5 cm

Since, 17.5 > 10

Therefore, AB + BC > AC

Hence, the sum of any two sides of a triangle is greater than the third side.

AB= 7.5 cm

BC = 8.8 cm and

AC = 12 cm

AB + BC = 7.5 cm + 8.8 cm

= 16.3 cm

Since, 16.3 > 12

Therefore, AB + BC > AC Hence, the sum of any two sides of a triangle is greater than the third side

PQ = 7 cm

PR = 7 cm and

QR = 7 cm

PQ + QR = 7 cm + 7 cm

= 14 cm

Since, 14 > 7

Therefore, PQ + QR > PR

Hence, the sum of any two sides of a triangle is greater than the third side.