**Understanding Elementary Shapes**

**Question 1.What is the disadvantage in comparing line segments by mere observation?**

**Solution:**

The disadvantage in comparing line segments by mere observation is the inaccuracy in judgement.

**Question 2.Why is it better to use a divider than a ruler, while measuring the length of a line segment?**

**Solution:**

It is better to use a divider with ruler as it given an accurate measurement of the line segment.

**Question 3.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.]

**Solution:**

On measuring the lengths of line segments AB, BC and AC using ruler, we find

AB =6.5 cm,

CB = 4.2 cm,

and AC = 2.3 cm

Now, AC +CB = 2.3 cm +4.2 cm = 6.5 cm = AB

AB = AC+CB.

**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:**

Since, AB + BC = 5 cm +3 cm = 8 cm = AC

∴ A, B, C are collinear and B lies between A and C.

**Question 5**.**Verify, whether D is the mid point of **.**Solution:**

Since AD = AB +BC + CD = 3 units

and, DG = DE +EF + FG = 3 units

∵ AD = DG [∵ Each = 3 units]

Thus, D is the mid-point of .

**Question 6.If B is the mid point of and C is the mid point of , where A, B, C, D lie on a straight line, say why AB = CD?**

**Solution:**

∵ B is the mid point of . Therefore, AB = BC.

∵ C is the mid-point of . Therefore, BC = CD.

Thus, AB =BC = CD i.e., AB = CD.

**Question 7.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.**

**Solution:**

Draw any five triangles, T_{1}, T_{2}, T_{3}, T_{4} and T_{5}. label each one as ∆ ABC. Measure, in each case, the three sides a = BC, b = CA and c = AB.

Let us tabulate the measurements of sides as under:

From the above table, we find that**(i)** each value of b + c – a is positive;**(ii)** each value of c + a – b is positive;**(iii)** each value of a + b – c is positive.

Now, b + c – a is positive ⇒ b + c – a > 0 ⇒ b + c > a

c + a – b is positive ⇒ c + a- b > 0 ⇒ c + a > b

a+ b – c is positive ⇒ a + b – c > 0 ⇒ a + b > c

Thus, it is verified that the sum of any two sides of a triangle is greater than the third side.

∴ The statement ‘the sum of two sides is ever less than the third side’ is never true.