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NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.2

Exercise 14.2

Ex 14.2 Class 6 Maths Question 1.
Draw a line segment of length 7.3 cm using ruler.
Solution:
Step I: Mark at point P.
Step II : Place the O mark of the ruler against the point P.
Step III : Mark a point Q at a distance of 7.3 cm from P.
Step IV : Join P and Q.

Thus (overline { PQ }) is the line segment of length 7.3 cm.

Ex 14.2 Class 6 Maths Question 2.
Construct a line segment of length 5.6 cm using ruler and compass.
Solution:
Step I: Draw any line L of suitable lengths.
Step II : Place the needle of the compass on the zero mark of the ruler and open it upto 5.6 mark.
Step III : Place the needle at any point A at the line and draw an arc to cut l at B.

Thus, (overline { AB }) is the required line segment of length 5.6 cm.

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:
Given that (overline { AB }) = 7.8 cm and (overline { AC }) = 4.7 cm.
Step I : Place zero mark of the ruler at A.
Step II : Mark a point B at a distance of 7.8 cm from A.
Step III : Mark another point C at a distance of 4.7 cm from A such that AC = 4.7 cm.
Step IV : On measuring the length of BC, we find that (overline { 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.

(Hint : Construct (overline { PX }) such that the length of (overline { PX }) = length of (overline { AB }) then cut off (overline { XQ }) such that (overline { XQ }) also has the length of (overline { AB }).
Solution:
Step I: Draw a line l of suitable length.
Step II: Draw (overline { AB }) = 3.9 cm
Step III: From the line, construct (overline { PX }) = (overline { AB }) = 3.9 cm.
Step IV: Again construct (overline { XQ })= (overline { AB }) =3.9 cm
Verification: (overline { PX }) + (overline { XQ }) = (overline { AB }) + (overline { AB })

Thus twice of (overline { AB }) is equal to (overline { PQ })

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 XY is equal to the difference between the length of (overline { AB }) and (overline { CD }) . Verify the measurement.
Solution:
Step I : Construct (overline { AB }) = 7.3 cm and (overline { CD }) = 3.4 cm.

Step II: Take a point P on the given line l.
Step III: Construct (overline { PR }) such that (overline { PR }) = (overline { AB }) = (overline { AB }) = 7.3 cm.
Step IV:Construct (overline { RQ }) = (overline { CD }) = 3.4 cm such that PQ = (overline { AB }) – (overline { CD }) .
Verification : On measuring, we observe that (overline { PQ }) = 3.9 cm = 7.3 cm 3.4 cm.
= (overline { AB }) – (overline { CD })
Thus, (overline { PQ }) = (overline { AB }) – (overline { CD }).

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