## NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6

NCERT Solutions For Class 6 Maths Chapter 14 Practical Geometry Ex 14.6

Exercise 14.6

Ex 14.6 Class 6 Maths Question 1.
Draw âˆ POQ of measure 75Â° and find its line of symmetry.
Solution:
Step I : Draw a line segment $$\overline { PQ }$$ .
Step II : With centre Q and suitable radius, draw an arc to cut PQ at R.

Step III : With centre R and radius of the same length, mark S and T on the former arc.
Step IV : With centres S and T and with the same radius, draw two arcs which meet each other at U.
Step V: Join QU such that âˆ PQU = 90Â°.
Step VI : With centres S and W, draw two arcs of the same radius which meet each other at Q.
Step VII: Join Q and O such that âˆ PQO = 75Â°.
Step VIII: Bisect âˆ PQO with QV.
Thus, OV is the line of symmetry of âˆ PQO.

Ex 14.6 Class 6 Maths Question 2.
Draw an angle of measure 147Â° and construct its bisector.
Solution:
Step I : Draw âˆ ABC = 147Â° with the help of protractor.

Step II : With centres B and radius of proper length, draw an arc which meets AB and AC at E and F respectively.
Step III : With centres E and F and the radius more that half of the length of arc EF, draw two arcs which meet each other at D.
Step IV : Join B and D.
Thus, BD is the bisector of âˆ ABC.

Ex 14.6 Class 6 Maths Question 3.
Draw a right angle and construct its bisector.
Solution:
Step I: Draw a line segment AB.
Step II : With centre B and proper radius draw an arc to meet AB at C.

Step III : With centre C and same radius, mark two marks D and E on the former arc.
Step IV : With centres D and E and the same radius, draw two arcs which meet each other at G.
Step V : Join B and G such that âˆ ABG = 90Â°
Step VI : Draw BH as the bisector of âˆ ABG such that âˆ ABH = 45Â°.
Thus âˆ ABG is the right angle and BH is the bisector of âˆ ABG.

Ex 14.6 Class 6 Maths Question 4.
Draw an angle of 153Â° and divide it into four equal parts.
Solution:
Step I : Draw âˆ ABP = 153Â° with the help of protractor.

Step II : Draw BC as the bisector of âˆ ABP which dividers âˆ ABP into two equal parts.
Step III : Draw BD and BE as the bisector of âˆ ABC and âˆ CBP respectively.
Thus, the bisectors BD, BC and BE divide the âˆ ABP into four equal parts.

Ex 14.6 Class 6 Maths Question 5.
Construct with ruler and compasses, angles of the following measures:
(a) 60Â°
(b) 30Â°
(c) 90Â°
(d) 120Â°
(e) 45Â°
(f) 135Â°
Solution:
(a) Angle of 60Â°

Step I: Draw a line segment $$\overline { AB }$$ .
Step II : With centre B and proper radius draw an arc.
Step III : With centre D and radius of the- same length mark a point E on the former arc.
Step IV : Join B to E and product to C. Thus âˆ ABC is the required angle of measure 60Â°.

(b) Step I: Draw âˆ ABC = 60Â° as we have done in section (a).
Step II: Draw BF as the bisector of âˆ ABC.

Thus âˆ ABF = $$\frac { 60 }{ 2 }$$ = 30Â°.

(c) Angle of 90Â°
In the given figure,
âˆ ABC = 90Â°(Refer to solution 3)

(d) Angle of 120Â°.
Step I: Draw $$\overline { AB }$$
Step II : With centre A and radius of proper length, draw an arc.

Step III : With centre D and the same radius, draw two mark E and F on former arc.
Step IV : Join A to F and produce to C. Thus âˆ CAB = 120Â°

(e) Angle of 45s, i.e., $$\frac { 90 }{ 2 }$$ = 45Â°
In the figure âˆ ABD = 45Â° (Refer to solution 3)

(f) An angle of 135Â°
Since 135Â° = 90Â° + 45Â°
= 90Â° + ($$\frac { 90 }{ 2 }$$ )Â°
In this figure âˆ ABC = 135Â°

Ex 14.6 Class 6 Maths Question 6.
Draw an angle of measure 45Â° and bisect it.
Solution:
Step I : Draw a line AB and take any point O on it.
Step II: Construct âˆ AOE = 45Â° at O.

Step III: With centre O and proper radius, draw an arc GF.
Step IV : With centres G and F and proper radius, draw two arcs which intersect each other at D.
Step V : Join O to D.
Thus âˆ AOE = 45Â° and OD is its bisector.

Ex 14.6 Class 6 Maths Question 7.
Draw an angle of measure 135Â° and bisect it.
Solution:
Steps I: Draw a line OA and take any point P on it.

Step II: Construct âˆ APQ = 135Â°.
Step III : Draw PD as the bisector of angle APQ.
Thus âˆ APQ = $$\frac { { 135 }^{ 0 } }{ 2 }$$ = 67 $$\frac { { 1 }^{ 0 } }{ 2 }$$.

Ex 14.6 Class 6 Maths Question 8.
Draw an angle of 70Â°. Make a copy of it using only a straight edge and compasses.
Solution:
Step I : Draw a line AB and take any point 0 on it.
Step II : Draw âˆ COB = 70Â° using protractor.

Step III: Draw a ray $$\overrightarrow { PQ }$$ .
Step IV: With centre O and proper radius, draw an arc which meets $$\overrightarrow { OA }$$ and $$\overrightarrow { OB }$$ at E and F respectively.

Step V : With the same radius and centre at P, draw an arc meeting $$\overrightarrow { PQ }$$ at R.
Step VI: With centre R and keeping and radius equal to EF, draw an arc intersecting the former arc at S.
Step VII : Join P and S and produce it. Thus, QPS is the copy of AOB = 70Â°.

Ex 14.6 Class 6 Maths Question 9.
Draw an angle of 40Â°. Copy its supplementary angle.
Solution:
Step I: Construct âˆ AOB = 40Â° using protractor.
âˆ COF is the supplementary angle of âˆ AOB.

Step II : Draw a ray $$\overrightarrow { PR }$$ and take any point Q on it.
Step III : With centre O and proper radius, draw an arc which intersects $$\overrightarrow { OC }$$ and $$\overrightarrow { OB }$$ at E and F respectively.

Step IV : With centre Q and same radius, draw an arc which intersects $$\overrightarrow { PQ }$$ at L.
Step V: With centre L and radius equal to EF, draw an arc which intersects the former arc at S.
Step VI : Join Q and S and produce.
Thus, âˆ PQS is the copy of the supplementary angle COB.