NCERT Solutions Class 6th Maths Chapter – 5 Understanding Elementary Shapes Exercise – 5.1

Solution –

One of the main disadvantage in comparing the line segments by mere observation is that there will be high chances of miscalculation.

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

Solution –

We know that rulers are very thick. Hence, when we will measure the length, then we can misread the readings due to the thickness. Therefore, it is better to use a divider than a ruler.

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 –

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

AB^—. 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.

Question 4. If A, B, and 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 –

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.

Question 5. 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¯

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 – 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

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 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.