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

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