Practical Geometry-Solutions 14.2

CBSE Class –VI Mathematics
NCERT Solutions
Chapter 14 Practical Geometry (Ex. 14.2)

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Question 1. Draw a line segment of length 7.3 cm, using a ruler.

Answer: Steps of construction:

(i) Place the zero mark of the ruler at a point A.

(ii) Mark a point B at a distance of 7.3 cm from A.

(iii) Join AB.

AB¯¯¯¯¯¯¯$\overline{\text{AB}}$ is the required line segment of length 7.3 cm.

(or)

(i) Place one of the pointers of the divider anywhere on the ruler say at 2 cm and count 7.3 cm and open the divider till there, in this case, it is 9.3cm. (It need not be started at 0 mark all the time).

(ii) remove the divider from the ruler and place it on the paper without disturbing and apply a small pressure so that the paper gets a prick from the needles of the divider.

(iii) Mark the pricks as points A and B

(iv) Join the points using a ruler.

Question 2. Construct a line segment of length 5.6 cm using ruler and compasses.

Answer: Steps of construction:

(i) Draw a line ′l′.${}^{\prime }{l}^{\prime }.$ Mark a point A on this line.

(ii) Place the compasses pointer on zero mark of the ruler. Open it to place the pencil point up to 5.6 cm mark.

(iii) Without changing the opening of the compasses. Place the pointer on A and cut an arc ′l′${}^{\prime }{l}^{\prime }$ at B.

AB¯¯¯¯¯¯¯$\overline{\text{AB}}$ is the required line segment of length 5.6 cm.

Question 3. Construct AB¯¯¯¯¯¯¯$\overline{\text{AB}}$ of length 7.8 cm. From this cut off AC¯¯¯¯¯¯¯$\overline{\text{AC}}$ of length 4.7 cm. Measure BC¯¯¯¯¯¯¯.$\overline{\text{BC}}.$

Answer: Steps of construction:

(i) Place the zero mark of the ruler at A.

(ii) Mark a point B at a distance 7.8 cm from A.

(iii) Again, mark a point C at a distance 4.7 from A.

By measuring BC¯¯¯¯¯¯¯$\overline{\text{BC}}$, we find that BC = 3.1 cm

Question 4. Given AB¯¯¯¯¯¯¯$\overline{\text{AB}}$ of length 3.9 cm, construct PQ¯¯¯¯¯¯¯$\overline{\text{PQ}}$ such that the length PQ¯¯¯¯¯¯¯$\overline{\text{PQ}}$ is twice that of AB¯¯¯¯¯¯¯$\overline{\text{AB}}$. Verify by measurement.

(Hint: Construct PX¯¯¯¯¯¯¯$\overline{\text{PX}}$ such that length of PX¯¯¯¯¯¯¯$\overline{\text{PX}}$ = length of AB¯¯¯¯¯¯¯$\overline{\text{AB}}$; then cut off XQ¯¯¯¯¯¯¯¯$\overline{\text{XQ}}$ such that XQ¯¯¯¯¯¯¯¯$\overline{\text{XQ}}$ also has the length of AB¯¯¯¯¯¯¯$\overline{\text{AB}}$.

Answer: Steps of construction:

(i) Draw a line ′l′.${}^{\prime }{l}^{\prime }.$

(ii) Construct PX¯¯¯¯¯¯¯$\overline{\text{PX}}$ such that length of PX¯¯¯¯¯¯¯$\overline{\text{PX}}$ = length of AB¯¯¯¯¯¯¯$\overline{\text{AB}}$

(iii) Then cut of XQ¯¯¯¯¯¯¯¯$\overline{\text{XQ}}$ such that XQ¯¯¯¯¯¯¯¯$\overline{\text{XQ}}$ also has the length of AB¯¯¯¯¯¯¯.$\overline{\text{AB}}.$

(iv) Thus the length of PX¯¯¯¯¯¯¯$\overline{\text{PX}}$ and the length of XQ¯¯¯¯¯¯¯¯$\overline{\text{XQ}}$ added together make twice the length of AB¯¯¯¯¯¯¯.$\overline{\text{AB}}.$

Verification:

By measurement we find that PQ = 7.8 cm

= 3.9 cm + 3.9 cm = AB¯¯¯¯¯¯¯$\overline{\text{AB}}$ + AB¯¯¯¯¯¯¯$\overline{\text{AB}}$ = 2 x AB¯¯¯¯¯¯¯$\overline{\text{AB}}$

Question 5. Given AB¯¯¯¯¯¯¯$\overline{\text{AB}}$ of length 7.3 cm and CD¯¯¯¯¯¯¯$\overline{\text{CD}}$ of length 3.4 cm, construct a line segment XY¯¯¯¯¯¯¯¯$\overline{\text{XY}}$ such that the length of XY¯¯¯¯¯¯¯¯$\overline{\text{XY}}$ is equal to the difference between the lengths of AB¯¯¯¯¯¯¯$\overline{\text{AB}}$ and CD¯¯¯¯¯¯¯.$\overline{\text{CD}}.$ Verify by measurement.

Answer: Steps of construction:

(i) Draw a line ′l′${}^{\prime }{l}^{\prime }$ and take a point X on it.

(ii) Construct XZ¯¯¯¯¯¯¯$\overline{\text{XZ}}$ such that length XZ¯¯¯¯¯¯¯$\overline{\text{XZ}}$ = length of AB¯¯¯¯¯¯¯$\overline{\text{AB}}$ = 7.3 cm

(iii) Then cut off ZY¯¯¯¯¯¯¯$\overline{\text{ZY}}$ = length of CD¯¯¯¯¯¯¯$\overline{\text{CD}}$ = 3.4 cm

(iv) Thus the length of XY¯¯¯¯¯¯¯¯$\overline{\text{XY}}$ = length of AB¯¯¯¯¯¯¯$\overline{\text{AB}}$ – length of CD¯¯¯¯¯¯¯$\overline{\text{CD}}$

Verification:

By measurement we find that length of XY¯¯¯¯¯¯¯¯$\overline{\text{XY}}$ = 3.9 cm = AB¯¯¯¯¯¯¯−CD¯¯¯¯¯¯¯$\overline{\text{AB}}-\overline{\text{CD}}$ = 7.3 cm – 3.4 cm