Well-designed AP 6th Class Maths Textbook Solutions Chapter 5 Understanding Elementary Shapes Exercise 5.1 offers step-by-step explanations to help students understand problem-solving strategies.
Understanding Elementary Shapes Class 6 Exercise 5.1 Solutions – 6th Class Maths 5.1 Exercise Solutions
Question 1.
What is the disadvantage in comparing line segments by mere observation?
Solution:
Comparing the lengths of two line segments simply by observation may not be accurate.
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.
The above errors we 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.
Question 3.
Draw any line segment, say \(\overline{\mathrm{AB}})\). 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:
Let A, B and C such that C lies between A and B and AB = 8 cm.
AB = 8 cm, AC = 5 cm, CB = 3 cm
∴ AC + CB = 5 cm + 3 cm = 8 cm
So, AC + CB = AB
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:
Given that, AB = 5 cm,
BC = 3 cm and AC = 8 cm
AB + BC = 5 cm + 3 cm = 8 cm
But AC = 8 cm
AB + BC = AC
Hence, B lies between A and C.
Question 5.
Verify, whether D is the mid point of \(\overline{\mathrm{AG}})\)
Solution:
From the adjacent figure, we have
AG = 6 cm [∵ 7 – 1 = 6]
AD = 3 cm [∵ 4 – 1 = 3]
DG = 3 cm [∵ 7 – 4 = 3]
AD + DG = AG and also AD = DG
Hence, D is the mid point of \(\overline{\mathrm{AG}})\).
Question 6.
If B is the mid point of \(\overline{\mathrm{AC}})\) and C is the midpoint of \(\overline{\mathrm{BD}})\), where A, B, C, D lie on a straight line, say why AB = CD?
Solution:
Given that B is the mid point of \(\overline{\mathrm{AC}})\) and C is the mid point of \(\overline{\mathrm{BD}})\).
B is mid point of \(\overline{\mathrm{AC}})\) ∴ AB = BC ….(1)
Again C is mid point of \(\overline{\mathrm{BD}})\) ∴ BC = CD …..(2)
from (1) and (2) we have AB = CD
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:
1) Case 1 : In △ABC,
AB = 4.7 cm, BC = 5 cm and AC = 4.9 cm
AB + AC = 4.7 cm + 4.9 cm = 9.6 cm
9.6 > 5
∴ AB + AC > BC
Hence, sum of any two sides of a triangle is greater than the third side.
ii) Case 2: In △EFG,
EF = 2.4 cm, FG = 5 cm and EG = 5.6 cm
EF + FG = 2.4 cm + 5 cm = 7.4 cm
7.4 > 5.6
So, EF + FG > EG
Hence, sum of any two sides of a triangle is greater than the third side.
iii) Case 3 : In △XYZ,
XY = 3.4 cm, YZ = 5.8 cm and XZ = 7.7 cm
XY + YZ = 3.4 cm + 5.8 cm = 9.2 cm
9.2 > 7.7
So, XY + YZ = XZ
Hence, sum of any two sides of a triangle is greater than the third side.
iv) Case 4 : In △STU,
SU = 3 cm, ST = 7.7 cm and TU = 6.6 cm
SU + TU = 3 cm + 6.6 cm = 9.6 cm
9.6 = 7.7
So, SU + TU > ST
Hence, sum of any two sides of a triangle is greater than the third side.
v) Case 5 : In ΔKLM,
KL = 5.2 cm, LM = 5 cm and KM = 5 cm
LM + KM = 5 cm + 5 cm = 10 cm
10 > 5.2
So, LM + KM > KL
Hence, sum of any two sides of a triangle is greater than the third side.