### NCERT Solutions for Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.1

**Ex 5.1 Class 6 Maths Question 1.**

**What is the disadvantage in comparing line segment by metre observation?**

**Solution:**

Comparing the lengths of two line segments simply by ‘observation’ may not be accurate. So we use divider to compare the length of the given line segments.

**Ex 5.1 Class 6 Maths Question 2.**

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

**Solution:**

Measuring the length of a line segment using a ruler, we may have the following errors:

(i) Thickness of the ruler

(ii) Angular viewing

These errors can be eradicated by using the divider. So, it is better to use a divider than a ruler, while measuring the length of a line segment.

**Ex 5.1 Class 6 Maths Question 3.**

**Draw any line segment, say . 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 AC + CB = AB, then we can be sure that C lies between A and B]**

**Solution:**

Let us consider

A, B and C such that C lies between A and B and AB = 7 cm.

AC = 3 cm, CB = 4 cm.

∴ AC + CB = 3 cm + 4 cm = 7 cm.

But, AB = 7 cm.

So, AB = AC + CB.

**Ex 5.1 Class 6 Maths 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:**

We have, AB = 5 cm; BC = 3 cm

∴ AB + BC = 5 + 3 = 8 cm

But, AC = 8 cm

Hence, B lies between A and C.

**Ex 5.1 Class 6 Maths Question 5.**

**Verify, whether D is the mid point of .**

**Solution:**

From the given figure, we have

AG = 7 cm – 1 cm = 6 cm

AD = 4 cm – 1 cm = 3 cm

and DG = 7 cm – 4 cm = 3 cm

∴ AG = AD + DG.

Hence, D is the mid point of \overline { AG }.

**Ex 5.1 Class 6 Maths Question 6.**

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

**Solution:**

We have

B is the mid point of \overline { AC } .

∴ AB = BC …(i)

C is the mid-point of \overline { BD } .

BC = CD

From Eq.(i) and (ii), We have

AB = CD

**Ex 5.1 Class 6 Maths Question 7.**

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

**Solution:**

Case I. In ∆ABC

Let AB = 2.5 cm

BC = 4.8 cm

and AC = 5.2 cm

AB + BC = 2.5 cm + 4.8 cm

= 7.3 cm

Since, 7.3 > 5.2

So, AB + BC > AC

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

Case II. In ∆PQR,

Let PQ = 2 cm

QR = 2.5 cm

and PR = 3.5 cm

PQ + QR = 2 cm + 2.5 cm = 4.5 cm

Since, 4.5 > 3.5

So, PQ + QR > PR

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

Case III. In ∆XYZ,

Let XY = 5 cm

YZ = 3 cm

and ZX = 6.8 cm

XY + YZ = 5 cm + 3 cm

= 8 cm

Since, 8 > 6.8

So, XY + YZ > ZX

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

Case IV. In ∆MNS,

Let MN = 2.7 cm

NS = 4 cm

MS = 4.7 cm

and MN + NS = 2.7 cm + 4 cm = 6.7 cm

Since, 6.7 >4.7

So, MN + NS > MS

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

Case V. In ∆KLM,

Let KL = 3.5 cm

LM = 3.5 cm

KM = 3.5 cm

and KL + LM = 3.5 cm + 3.5 cm = 7 cm

7 cm > 3.5 cm

**Solution:**

(i) For one-fourth revolution, we have

So, KL + LM > KM

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

Hence, we conclude that the sum of any two sides of a triangle is never less than the third side.