NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exz 14.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2.
- Practical Geometry Class 6 Ex 14.1
- Practical Geometry Class 6 Ex 14.3
- Practical Geometry Class 6 Ex 14.4
- Practical Geometry Class 6 Ex 14.5
- Practical Geometry Class 6 Ex 14.6
Board | CBSE |
Textbook | NCERT |
Class | Class 6 |
Subject | Maths |
Chapter | Chapter 14 |
Chapter Name | Practical Geometry |
Exercise | Ex 14.2 |
Number of Questions Solved | 5 |
Category | NCERT Solutions |
NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2
Ex 14.2 Class 6 Maths Question 1.
Draw a line segment of length 7.3 cm, using a ruler.
Solution:
Steps of Construction:
-
- Mark a point A on the plane of the paper and place the ruler so that zero mark of the ruler is at A.
- Mark a point A on the plane of the paper and place the ruler so that zero mark of the ruler is at A.
- Mark with pencil a point B against the mark on the ruler which indicates 7.3 cm.
- Join points A and B by moving the tip of the pencil against the straight edge of the ruler.
The line segment AB so obtained is the required line segment.
Ex 14.2 Class 6 Maths Question 2.
Construct a line segment of length 5.6 cm using ruler and compasses.
Solution:
Steps of Construction:
- Mark a point A on the plane of the paper and draw a line, say l, passing through it.
- Place the steel end of the compasses at zero mark on the ruler and open out it such that the pencil end on the mark indicates 5.6 cm.
- Transfer the compasses as it is to the line l so that the steel end is on A.
- With the pencil end make a small stroke on l so as to cut it at B.
- The segment AB so obtained is the required line segment.
Ex 14.2 Class 6 Maths Question 3.
Construct \(\overline { AB } \) of length 7.8 cm. From this, cut off \(\overline { AC } \) of length 4.7 cm. Measure \(\overline { BC } \).
Solution:
Steps of Construction:
- Draw a line segment AB of length 7.8 cm.
- Using compasses find a point C on the line segment AB so that segment AC = 4.7 cm.
- On measuring BC, we find that BC = 3.1 cm.
Ex 14.2 Class 6 Maths Question 4.
Given \(\overline { AB } \) of length 3.9 cm, construct \(\overline { PQ } \) such that the length of \(\overline { PQ } \) is twice that of \(\overline { AB } \). Verify by measurement.
Solution:
Steps of Construction:
- Draw a line l and mark a point P on it.
- Using compasses find a point X so that PX(= AB) = 3.9 cm on the line l.
- Using compasses find a point Q so that XQ = 3.9 cm on the line l.
Thus, PQ = PX + XQ = 3.9 cm + 3.9 cm
= 2(3.9 cm) = 2AB.
Ex 14.2 Class 6 Maths Question 5.
Given \(\overline { AB } \) of length 7.3 cm and \(\overline { CD } \) of length 3.4 cm, construct a line segment \(\overline { XY } \) such that the length of \(\overline { XY } \) is equal to the difference between the lengths of \(\overline { AB } \) and \(\overline { CD } \). Verify by measurement
Solution:
Steps of Construction:
- Draw line segments AB = 7.3 cm and CD = 3.4 cm.
- Draw a line l and mark a point X on it.
- Using compasses find a point P on the line l so that segment XP = segment AB (i.e., 7.3 cm).
- Using compasses find a point Y so that the segment PY = segment CD (i.e., 3.4 cm). The segment XY so obtained is the required segment, because XY = OP -PY – AB -CD.
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