Chapter 14 Practical Geometry Ex 14.6


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.

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.

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.

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.

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 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 = 602 = 30°.

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

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.

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.

NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q1
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q2
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q3
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q4
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q5
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q6
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q7
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q8
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q9
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q10
NCERT Solutions For Class 6 Maths Practical Geometry Exercise 14.6 Q11

Chapter 14 Practical Geometry Ex 14.6