These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry InText Questions

NCERT In-text Question Page No. 286

Question 1.

In Step 2 of the construction using ruler and compasses, what would happen if we take the length of radius to be smaller than half the length of \(\overline{\mathrm{AB}}\) ?

Answer:

In case we take the radius smaller than half of the length of \(\overline{\mathrm{AB}}\), the arcs will not intersect each other at two points P and Q.

NCERT In-text Question Page No. 289

Question 1.

In Step 2 above, what would happen if we take radius to be smaller than half the length BC?

Answer:

If we take radius to be smaller than half of BC, the arcs drawn with centres B and C will not intersect each other.

NCERT In-text Question Page No. 290

Question 1.

How will your construct at 15° angle?

Answer:

Steps of construction:

Step I: Construct an angle ∠ABC of 60°.

Step II: Bisect ∠ABC to get an angle of 30°

i. e. ∠ABD = 30°.

Step III: Bisect ∠ABD, such that \(\overline{\mathrm{BE}}\) is bisector of ∠ABD.

Thus, ∠ABD = \(\frac { 1 }{ 2 }\)(30°) = 15°.

NCERT In-text Question Page No. 291

Question 1.

How will you construct a 150° angle?

Answer:

Steps of construction:

Step I: Draw a line I and mark a point O on it.

Step II: With centre O and a convenient radius, draw an arc intersecting I at A.

Step III: With the same radius and centre at A, draw an arc to cut the first arc at B.

Step IV: Again with the same radius and centre at B, draw another arc to intersect the

first arc at C.

Step V : Once again with the same radius and centre at C, draw an arc to cut the first are at D.

Step VI: Now, bisect ∠COD, such that ∠COE = ∠EOD = 30°.

Step VII: Since, 150° = 120° + 30°, therefore ∠AOC + ∠COE = ∠AOE. Thus, ∠AOE is the required angle whose measure is 150°.

Question 2.

How will you construct a 45° angle?

Answer:

Steps of construction :

Step I: Construct an angle of 90° as shown in

the figure. ∠POQ = 90°.

Step II: Draw OR, the angle bisector of ∠POQ such that \(\frac { 1 }{ 2 }\)– [∠POQ] = \(\frac { 1 }{ 2 }\) (90°) = 45° or ∠POR = 45°