## 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 ∠POQ of measure 75° 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 ∠PQU = 90°.

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 ∠PQO = 75°.

Step VIII: Bisect ∠PQO with QV.

Thus, OV is the line of symmetry of ∠PQO.

Ex 14.6 Class 6 Maths Question 2.

Draw an angle of measure 147° and construct its bisector.

Solution:

Step I : Draw ∠ABC = 147° 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 ∠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 ∠ABG = 90°

Step VI : Draw BH as the bisector of ∠ABG such that ∠ABH = 45°.

Thus ∠ABG is the right angle and BH is the bisector of ∠ABG.

Ex 14.6 Class 6 Maths Question 4.

Draw an angle of 153° and divide it into four equal parts.

Solution:

Step I : Draw ∠ABP = 153° with the help of protractor.

Step II : Draw BC as the bisector of ∠ABP which dividers ∠ABP into two equal parts.

Step III : Draw BD and BE as the bisector of ∠ABC and ∠CBP respectively.

Thus, the bisectors BD, BC and BE divide the ∠ABP into four equal parts.

Ex 14.6 Class 6 Maths Question 5.

Construct with ruler and compasses, angles of the following measures:

(a) 60°

(b) 30°

(c) 90°

(d) 120°

(e) 45°

(f) 135°

Solution:

(a) Angle of 60°

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 ∠ABC is the required angle of measure 60°.

(b) Step I: Draw ∠ABC = 60° as we have done in section (a).

Step II: Draw BF as the bisector of ∠ABC.

Thus ∠ABF = \(\frac { 60 }{ 2 }\) = 30°.

(c) Angle of 90°

In the given figure,

∠ABC = 90°(Refer to solution 3)

(d) Angle of 120°.

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 ∠CAB = 120°

(e) Angle of 45s, i.e., \(\frac { 90 }{ 2 }\) = 45°

In the figure ∠ABD = 45° (Refer to solution 3)

(f) An angle of 135°

Since 135° = 90° + 45°

= 90° + (\(\frac { 90 }{ 2 }\) )°

In this figure ∠ABC = 135°

Ex 14.6 Class 6 Maths Question 6.

Draw an angle of measure 45° and bisect it.

Solution:

Step I : Draw a line AB and take any point O on it.

Step II: Construct ∠AOE = 45° 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 ∠AOE = 45° and OD is its bisector.

Ex 14.6 Class 6 Maths Question 7.

Draw an angle of measure 135° and bisect it.

Solution:

Steps I: Draw a line OA and take any point P on it.

Step II: Construct ∠APQ = 135°.

Step III : Draw PD as the bisector of angle APQ.

Thus ∠APQ = \(\frac { { 135 }^{ 0 } }{ 2 }\) = 67 \(\frac { { 1 }^{ 0 } }{ 2 }\).

Ex 14.6 Class 6 Maths Question 8.

Draw an angle of 70°. 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 ∠COB = 70° 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 = 70°.

Ex 14.6 Class 6 Maths Question 9.

Draw an angle of 40°. Copy its supplementary angle.

Solution:

Step I: Construct ∠AOB = 40° using protractor.

∠COF is the supplementary angle of ∠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, ∠PQS is the copy of the supplementary angle COB.

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