## Class 6 Maths NCERT Understanding Elementary Shape Chapter 5 Exercise 5.1

### Understanding Elementary Shapes Class 6 Ex 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 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 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.

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.

Verify, whether D is the mid point of AG¯¯¯¯¯¯¯¯ .

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 AG¯¯¯¯¯¯¯¯.

Ex 5.1 Class 6 Maths Question 6.

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?

Solution:

We have

B is the mid point of AC¯¯¯¯¯¯¯¯ .

∴ AB = BC …(i)

C is the mid-point of 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.