 # Ncert Solutions For Class 6 Maths Ex 5.1

## NCERT Solutions For Class 6 Maths Ex 5.1 PDF Free Download

The shapes which we see around us in our day to day lives are formed using lines and curves. Exercise 5.1 of chapter 5 contains problems which help students understand the steps to be followed for measuring line segments by observation, tracing, ruler or divider. The students gain a better hold on the concepts by using solutions created by faculty at BYJU’S. To understand the concept of measuring line segments, students can access NCERT Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Exercise 5.1 PDF, which are available here.

## NCERT Solutions for Class 6 Chapter 5: Understanding Elementary Shapes Exercise 5.1 Download PDF    ### Access NCERT Solutions for Class 6 Chapter 5: Understanding Elementary Shapes Exercise 5.1

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

Solutions:

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.

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

Solutions:

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

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?

Solutions:

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.

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?

Solutions:

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.

5. Verify, whether D is the mid point of . Solutions:

Since, it is clear from the figure that AD = DG = 3 units. Hence, D is the midpoint of 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?

Solutions: 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

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.

Solutions:

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.

### Access other exercise solutions of Class 6 Maths Chapter 5: Understanding Elementary Shapes

Exercise 5.9 Solutions