GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1

   

Gujarat Board GSEB Textbook Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1

Question 1.
What is the disadvantage in comparing line segments by mere observation?
Solution:
Comparing the line segments simply by ‘observation’ may not be accurate. For example, the line segments AB and CD (in the following figure) seem to be equal, but actually they are not.
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-1

GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1

Question 2.
Why is it better 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 may accompany with some errors due to:
(i) Thickness of the ruler
(ii) Angular viewing
These errors can be removed by measuring a line segment with the help of a divider.
So, the use of a divider is better 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:
Measuring the lengths of the segments AB, BC and AC, we have:
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-2
\(\overline{\mathrm{AC}}\) = 6.3 cm
\(\overline{\mathrm{BC}}\) = 2.7 cm
\(\overline{\mathrm{AB}}\) = 9.0 cm
AC + BC = 6.3 cm + 2.7 cm = 9.0 cm and AB = 9.0 cm
AC + BC = AB Hence verified.

GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1

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:
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-3
But AC = 8cm
The point B lies between A and C.

Question 5.
Verify, whether D is the mid-point of \(\overline{\mathrm{AG}}\).
Solution:
Since,
AG = 7 cm – 1 cm = 6 cm
AD = 4cm – 1cm = 3cm
DG = 7cm – 4cm = 3cm
AG = AD + DG
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-4
So, D is the mid-point of AG.

GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1

Question 6.
If B is the mid-point of and C is the mid-point of I5. where A, B, C, D lie on a straight line, say why AB = CD?
Solution:
Since, B is the mid-point of AC
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-5
AB = BC ………….(i)
Similarly, C is the mid-point of BD
BC = CD ………………(ii)
From (i) and (ii),
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-6

Question 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.
Solution:
We draw triangles with different set of lengths:
(i) In ∆ABC, we have
AB = 3.7 cm,
BC = 3.7 cm
and AC = 3.7 cm
= AB + BC
= 3.7 cm + 3.7 cm
= 74cm
7.4 > 3.7
(AB + BC) > AC
Similarly, we can have,
(BC + AC) > AB
(AB + AC) > BC
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-7

(ii) In ∆PQR, we have:
PQ = 4cm,
PR = 2.7 cm
and RQ = 4.7cm
= PQ + RQ
= 4cm + 4.7cm
a = 8.7 cm
PQ + PR = 4 cm + 2.7 cm = 6.7 cm
RQ + PR = 2.7 cm + 4.7 cm = 7.4 cm
(PQ + RQ) > PR
(PQ + PR) > RQ
(RQ + PR) > PQ
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-8

(iii) In ∆BCD, we have
BC = 3 cm,
CD = 4 cm and
BD = 5 cm
Now,
= BC + CD
= 3 cm + 4 cm = 7
= BD + CD = 5 cm +4 cm = 9 cm
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-9
BC + BD = 3 cm + 5 cm = 8 cm
(BC + CD) > BD
(BD + CD) > BC
(BC + BD) > CD

(iv) In ∆XYZ, we have
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-10
XY = 4.5 cm,
YZ = 3.8cm
and XZ = 2cm
XY + YZ = 4.5 cm + 3.8 cm = 8.3 cm
YZ + XZ = 3.8 cm +2 cm = 5.8 cm
XZ + XY = 2 cm + 4.5 cm = 6.5 cm
(XY + YZ) > XZ
(YZ + XZ) > XY
(XZ + XY) > YZ

(v) In ∆KLM, we have:
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1 img-11
KL = 4.5 cm, KM = 3.4 cm and LM = 6 cm
KL + KM = 4.5cm + 3.4cm = 7.9 cm
KM + LM = 3.4 cm + 6 cm = 9.4 cm
KL + LM = 4.5 cm + 6 cm = 10.5 cm
(KL + KM) > LM
(KM + LM) > KL
and (KL + LM) > KM.
In each of the above cases, we have seen that the sum of the lengths of any two sides is always greater than the third side.
Thus, the sum of the lengths of any two sides of a triangle is never less than its third side.
GSEB Solutions Class 6 Maths Chapter 5 Understanding Elementary Shapes Ex 5.1

Leave a Comment

Your email address will not be published. Required fields are marked *