GSEB Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2

   

Gujarat Board GSEB Textbook Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2

Question 1.
Draw a line segment of length 7.3 cm using a ruler.
GSEB Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 img 11
Solution:
Steps of construction:
Step I: Mark a point A.
Step II: Place the zero mark of the ruler against point A.
Step III: Mark a point B at a distance of 7.3 cm from A.
Step IV: Join A and B.
Thus, \(\overline{A B}\) is the required line segment of length 7.3 cm.
Note: While marking points A and B, we should look straight down at the measuring device. Otherwise, we will get the incorrect length.

Question 2.
Construct a line segment of length 5.6 cm using a ruler and compasses.
Solution:
Steps of construction:
Step I: Draw a line l and mark a point ‘A’ on it.
GSEB Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 img 12
Step II: Place the steel-end of the compasses on the zero mark of the ruler. Open it such that the pencil tip falls on the 5.6 cm mark.
Step III: Without changing the opening of the compasses, place the steel end on ‘A’ and mark an arc to cut l at ‘B’.
Thus, \(\overline{A B}\) = 5.6 cm is the line segment of the required length.

GSEB Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2

Question 3.
Construct \(\overline{A B}\) of length 7.8 cm. From this, cut-off \(\overline{A C}\) of length 4.7 cm. Measure \(\overline{B C}\)
Solution:
Steps of construction:
Step I: Place the zero mark of the ruler at A.
Step II: Mark a point B at a distance of 7.8 cm from A.
GSEB Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 img 15
Step III: Mark another point C between A and B at a distance of 4.7 cm from A such that \(\overline{A C}\) = 4.7 cm.
Step IV: Measure the line segment \(\overline{B C}\). We find that \(\overline{B C}\) =3.1 cm.

Question 4.
Given \(\overline{A B}\) of length 3.9 cm, construct \(\overline{P Q}\) such that the length of \(\overline{P Q}\) is twice that of \(\overline{A B}\). Verify by measurement.

GSEB Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 img 17

Hint: Construct \(\overline{P X}\) such that length of \(\overline{P X}\) = length of \(\overline{A B}\); then cut-off \(\overline{X Q}\) such that \(\overline{X Q}\) also has the length of \(\overline{A B}\).
Solution:
Steps of construction:
Step I: Draw a line l.
Step II: Draw AB = 3.9 cm.
GSEB Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 img 18
Step III: On line l construct \(\overline{P X}\) = \(\overline{A B}\) (= 3.9 cm).
Step IV: Next construct \(\overline{X Q}\) = \(\overline{A B}\) (=3.9 cm)
Thus, the lengths of \(\overline{P X}\) and \(\overline{X Q}\) are added together to make twice the length of \(\overline{A B}\)
Verification: By measurement, we have:
\(\overline{A B}\) + \(\overline{A B}\) = 3.9 cm + 3.9 cm
2 ( \(\overline{A B}\) ) = 7.8 cm = \(\overline{X Y}\)
Thus, twice of \(\overline{A B}\) is equal to \(\overline{X Y}\)

GSEB Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2

Question 5.
Given length 7.3 cm and \(\overline{C D}\) of length 3.4 cm, construct a line segment \(\overline{X Y}\) such that the length of \(\overline{X Y}\) is equal to the difference between the lengths of \(\overline{A B}\) and \(\overline{C D}\) Verify by measurement.
Solution:
Step I: Draw \(\overline{A B}\) = 7.3 cm and \(\overline{C D}\) =3.4 cm.
Step II: Draw a line l and take a point X on it.
Step III: Construct \(\overline{X R}\) such that length of \(\overline{X R}\) = length \(\overline{A B}\) (= 7.3 cm).
GSEB Solutions Class 6 Maths Chapter 14 Practical Geometry Ex 14.2 img 25
Step IV: Cut-off \(\overline{RY}\) = length of \(\overline{C D}\) (= 3.4 cm) such that the length \(\overline{X Y}\) = length
of \(\overline{A B}\) – length of \(\overline{C D}\).
Verification: By measurement, we have
\(\overline{X Y}\) = 3.9 cm = 7.3 cm – 3.4 cm
= \(\overline{A B}\) – \(\overline{C D}\)
Thus, we get \(\overline{X Y}\) = \(\overline{A B}\) – \(\overline{C D}\)
Now, the opening of compasses is equal to \(\overline{A B}\).
Step IV: Draw a line l and mark a point C on it.
Step V: Without changing the openings of the compasses, place the steel end at C and mark a point D on l.
Now, \(\overline{C D}\) is a copy of \(\overline{A B}\).

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