- Chapter 5 Understanding Elementary Shape Ex-5.2
- Chapter 5 Understanding Elementary Shape Ex-5.3
- Chapter 5 Understanding Elementary Shape Ex-5.4
- Chapter 5 Understanding Elementary Shape Ex-5.5
- Chapter 5 Understanding Elementary Shape Ex-5.6
- Chapter 5 Understanding Elementary Shape Ex-5.7
- Chapter 5 Understanding Elementary Shape Ex-5.8
- Chapter 5 Understanding Elementary Shape Ex-5.9

Chapter 5 Understanding Elementary Shape Ex-5.2 |
Chapter 5 Understanding Elementary Shape Ex-5.3 |
Chapter 5 Understanding Elementary Shape Ex-5.4 |
Chapter 5 Understanding Elementary Shape Ex-5.5 |
Chapter 5 Understanding Elementary Shape Ex-5.6 |
Chapter 5 Understanding Elementary Shape Ex-5.7 |
Chapter 5 Understanding Elementary Shape Ex-5.8 |
Chapter 5 Understanding Elementary Shape Ex-5.9 |

**Answer
1** :
By mere observation we can’t compare the line segments with slight difference in their length. We can’t say which line segment is of greater length. Hence, the chances of errors due to improper viewing are more.

**Answer
2** :
While using a ruler, chances of error occur due to thickness of the ruler and angular viewing. Hence, using divider accurate measurement is possible.

**Answer
3** :

Since given that point C lie in between A and B. Hence, all points are lying on same line segment

. Therefore for every situation in which point C is lying in between A and B we may say that

AB = AC + CB

For example:

AB is a line segment of length 7 cm and C is a point between A and B such that AC = 3 cm and CB = 4 cm.

Hence, AC + CB = 7 cm

Since, AB = 7 cm

∴ AB = AC + CB is verified.

**Answer
4** :

Given AB = 5 cm

BC = 3 cm

AC = 8 cm

Now, it is clear that AC = AB + BC

Hence, point B lies between A and C.

**Answer
5** :
Since, it is clear from the figure that AD = DG = 3 units. Hence, D is the midpoint of

**Answer
6** :

Given

B is the midpoint of AC. Hence, AB = BC (1)

C is the midpoint of BD. Hence, BC = CD (2)

From (1) and (2)

AB = CD is verified

**Answer
7** :
Case 1. In triangle ABC

AB= 2.5 cm

BC = 4.8 cm and

AC = 5.2 cm

AB + BC = 2.5 cm + 4.8 cm

= 7.3 cm

As 7.3 > 5.2

∴ AB + BC > AC

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

Case 2. In triangle PQR

PQ = 2 cm

QR = 2.5 cm

PR = 3.5 cm

PQ + QR = 2 cm + 2.5 cm

= 4.5 cm

As 4.5 > 3.5

∴ PQ + QR > PR

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

Case 3. In triangle XYZ

XY = 5 cm

YZ = 3 cm

ZX = 6.8 cm

XY + YZ = 5 cm + 3 cm

= 8 cm

As 8 > 6.8

∴ XY + YZ > ZX

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

Case 4. In triangle MNS

MN = 2.7 cm

NS = 4 cm

MS = 4.7 cm

MN + NS = 2.7 cm + 4 cm

6.7 cm

As 6.7 > 4.7

∴ MN + NS > MS

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

Case 5. In triangle KLM

KL = 3.5 cm

LM = 3.5 cm

KM = 3.5 cm

KL + LM = 3.5 cm + 3.5 cm

= 7 cm

As 7 cm > 3.5 cm

∴ KL + LM > KM

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

Therefore we conclude that the sum of any two sides of a triangle is always greater than the third side.

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