**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.