**NCERT
Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6**

NCERT
Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 are the part of NCERT Solutions for
Class 6 Maths. Here you can find the NCERT Solutions for Class 6 Maths Chapter
14 Practical Geometry Ex 14.6.

**NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.1****NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.2****NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.3****NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.4****NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.5****NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6**

**Ex
14.6 Class 6 Maths Question 1.**

Draw ∠POQ of measure 75° and find its line of symmetry.**Solution:
**Step 1: Draw a line segment PO.

Step 2: With centre O and any suitable radius, draw an arc to cut PO at R.

Step 4: With centres S and T and with the same radius, draw two arcs which intersect each other at U.

Step 5: Join OU such that ∠POU = 90°.

Step 6: With centres S and W, draw two arcs of the same radius which intersect each other at Q.

Step 7: Join O and Q such that ∠POQ = 75°.

Step 8: Bisect ∠POQ with OV.

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

**Ex 14.6 Class 6 Maths Question 2.**

Draw an angle of measure 147° and construct its
bisector.**Solution:
**Step 1: Draw ∠ABC = 147° with the help of a protractor.

Step 3: With centres E and F and the radius more than half of the length of arc EF, draw two arcs which intersect each other at D.

Step 4: 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 1: Draw a line segment AB.

Step 2: With centre B and any suitable radius, draw an arc to intersect AB at C.

Step 4: With centres D and E and the same radius, draw two arcs which intersect each other at G.

Step 5: Join B and G such that ∠ABG = 90°.

Step 6: 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 measure 153° and divide it into
four equal parts.**Solution:
**Step 1: Draw ∠ABP of measure 153° with the help of a protractor.

Step 3: Draw BD and BE as the bisectors 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) Construction of 60°:

Step 2: With centre B and any suitable radius, draw an arc which intersect AB at D.

Step 3: With centre D and radius of the same length, draw an arc E on the previous arc.

Step 4: Join B to E and produce it up to C. Thus, ∠ABC is the required angle of measure 60°.

(b)
Construction of 30°:

Step
1: Draw ∠ABC = 60° as we have drawn in part (a).

Step 2: Draw BF as the bisector of ∠ABC.

(c) Construction of 90°:

Step 1: Draw a line segment AB.

Step 2: With centre B and any suitable radius,
draw an arc to intersect AB at C.

Step 3: With centre C and same radius, mark two arcs
D and E on the previous arc.

Step 4: With centres D and E and the same
radius, draw two arcs which intersect
each other at G.

Step 5: Join B and G such that ∠ABG = 90°.

(d)
Construction of 120°:

Step 1: Draw a line segment AB.

Step 2: With centre A and radius of any suitable length, draw an arc which
intersect AB at D.

Step 4: Join A to F and produce it up to C. Thus, ∠CAB = 120°.

(e)
Construction of 45°:

To
make an angle of 45°, we bisect the angle of 90°.

Step 1: Construct an angle of 90° as in part (c).

Step
2: Bisect the angle of 90° to get an angle of 45°.

In
the figure, ∠ABD = 45°

(f)
Construction of 135°:

Since 135° = 90° + 45° = 90° + (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 1: Draw a line AB and take any point O on it.

Step 2: Construct an angle of 90° on point O. Construct ∠AOE = 45° by bisecting ∠AOC.

Step 4: With centres G and F and a suitable radius, draw two arcs which intersect each other at D.

Step 5: 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 1: Draw a line OA and take any point P on it.

Step 3: Draw PD as the bisector of angle APQ.

Thus, ∠APD = 135/2 = 67½°.

**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 1: Draw a line AB and take any point O on it.

Step 2: Draw ∠COB = 70° using a protractor.

Step 4: With centre O and a suitable radius, draw an arc which meets OB and OC at E and F, respectively.

Step 5: With the same radius and centre at P, draw an arc meeting PQ at R.

Step 6: With centre R and radius equal to EF, draw an arc intersecting the former arc at S.

Step 7: Join P and S and produce it. Thus, QPS is the copy of ∠COB = 70°.

**Ex
14.6 Class 6 Maths Question 9.**

Draw an angle of 40°. Copy its supplementary angle.**Solution:
**Step 1: Construct ∠AOB = 40° using protractor.

∠COB is the supplementary angle of ∠AOB.

Step 2: Draw a ray PR and take a point Q on it.

Step 3: With centre O and a suitable radius, draw an arc which intersects OC and OB at E and F, respectively.

Step 4: With centre Q and same radius, draw an arc which intersects PQ at L.

Step 5: With centre L and radius equal to EF, draw an arc which intersects the previous arc at S.

Step 6: Join Q and S and produce it.

Thus, ∠PQS is the copy of the supplementary angle of 40°.

**You can also like these:**

**NCERT Solutions for Maths Class 7**

**NCERT Solutions for Maths Class 8**

**NCERT Solutions for Maths Class 9**

**NCERT Solutions for Maths Class 10**

**NCERT Solutions for Maths Class 11 **