These NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Ex 14.6 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 6 Maths Chapter 14 Practical Geometry Exercise 14.6

Question 1.

Draw ∠POQ of measure 75° and find its line of symmetry.

Answer:

Steps of construction:

(a) Draw a line l and mark a point O on it.

(b) Place the pointer of the compasses at O and draw an arc of any radius which intersects the line l at A.

(c) Taking same radius, with centre A, cut the previous arc at B.

(d) Join OB, then ∠BOA = 60 .

(e) Taking same radius, with centre B, cut the previous arc at C.

(f) Draw bisector of ∠BOC. The angle is of 90 . Mark it at D. Thus, ∠DOA = 90°

(g) Draw \(\overline{\mathrm{OP}}\) as bisector of ∠DOB. Thus, ∠POA = 75°

Question 2.

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

Answer:

Steps of construction:

(a) Draw a ray \(\overline{\mathrm{OA}}\) .

(b) With the help of protractor, construct ∠AOB = 147°

(c) Taking centre O and any convenient radius, draw an arc which intersects the arms \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) at P and Q respectively.

(d) Taking P as centre and radius more than half of PQ, draw an arc.

(e) Taking Q as centre and with the same radius, draw another arc which intersects the previous at R.

(f) Join OR and produce it.

(g) Thus, \(\overline{\mathrm{OR}}\) is the required bisector of ∠AOB.

Question 3.

Draw a right angle and construct its bisector.

Answer:

Steps of construction:

(a) Draw a line PQ and take a point O on it.

(b) Taking O as centre and convenient radius, draw an arc which intersects PQ at A and B.

(c) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C.

(d) Join OC. Thus, ∠COQ is the required right angle.

(e) Taking B and E as centre and radius more than half of BE, draw two arcs which intersect each other at the point D.

(f) Join OD. Thus, \(\overline{\mathrm{OD}}\) is the required bisector of ∠COQ.

Question 4.

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

Answer:

Steps of construction:

(a) Draw a ray \(\overline{\mathrm{OA}}\)

(b) At O, with the help of a protractor, construct ∠AOB = 153°.

(c) Draw \(\overline{\mathrm{OC}}\) as the bisector of ∠AOB.

(d) Again, draw \(\overline{\mathrm{OD}}\) as bisector of ∠AOC.

(e) Again, draw \(\overline{\mathrm{OE}}\) as bisector of ∠BOC.

(f) Thus, \(\overline{\mathrm{OC}}\), \(\overline{\mathrm{OD}}\) and \(\overline{\mathrm{OE}}\) divide ∠AOB in four equal arts.

Question 5.

Construct with ruler and compasses, angles of following measures:

(a) 60°

(b) 30°

(c) 90°

(d) 120°

(e) 45°

(f) 135°

Answer:

Steps of construction:

(a) 60°

(i) Draw a ray \(\overline{\mathrm{OA}}\).

(ii) Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{\mathrm{OA}}\) at P.

(iii) Taking P as centre and same radius, cut previous arc at Q.

(iv) Join OQ.

Thus, ∠BOA is required angle of 60°.

(b) 30°

(i) Draw a ray \(\overline{\mathrm{OA}}\) .

(ii) Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{\mathrm{OA}}\) at P.

(iii) Taking P as centre and same radius, cut previous arc at Q.

(iv) Join OQ. Thus, ∠BOA is required angle of 60°.

(v) Put the pointer on P and mark an arc.

(vi) Put the pointer on Q and with same radius, cut the previous arc at C. Thus, ∠COA is required angle of 30°.

(c) 90°

- Draw a ray \(\overline{\mathrm{OA}}\).
- Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{\mathrm{OA}}\) at X.
- Taking X as centre and same radius, cut previous arc at Y.
- Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.
- Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.
- Join OS and produce it to form a ray OB.

Thus, ∠BOA is required angle of 90.

(d) 120°

- Draw a ray \(\overline{\mathrm{OA}}\).
- Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{\mathrm{OA}}\) at P.
- Taking P as centre and same radius, cut previous arc at Q.
- Taking Q as centre and same radius cut the arc at S.
- Join OS.

Thus, ∠AOD is required angle of 120°.

(e) 45°

- Draw a ray \(\overline{\mathrm{OA}}\) .
- Taking O as centre and convenient radius, mark an arc, which intersects \(\overline{\mathrm{OA}}\) at X.
- Taking X as centre and same radius, cut previous arc at Y.
- Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.
- Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.
- Join OS and produce it to form a ray OB. Thus, ∠BOA is required angle of 90°.
- Draw the bisector of ∠BOA.

Thus, ∠MOA is required angle of 45°.

(f) 135°

- Draw a line PQ and take a point O on it.
- Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.
- Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R.
- Join OR. Thus, ∠QOR = ∠POQ = 90° .
- Draw OD the bisector of ∠POR.

Thus, ∠QOD is required angle of 135°.

Question 6.

Draw an angle of measure 45° and bisect it.

Answer:

Steps of construction:

(a) Draw a line PQ and take a point O on it.

(b) Taking O as centre and a convenient radius, draw an arc which intersects PQ at two points A and B.

(c) Taking A and B as centres and radius more than half of AB, draw two arcs which intersect each other at C.

(d) Join OC. Then ∠COQ is an angle of 90.

(e) Draw \(\overline{\mathrm{OE}}\) as the bisector of ∠COE. Thus, ∠QOE = 45°

(f) Again draw \(\overline{\mathrm{OG}}\) as the bisector of ∠QOE.

Thus, ∠QOG = ∠EOG = 22\(\frac { 1 }{ 2 }\)°

Question 7.

Draw an angle of measure 135° and bisect it.

Answer:

Steps of construction:

(a) Draw a line PQ and take a point O on it.

(b) Taking O as centre and convenient radius, mark an arc, which intersects PQ at A and B.

(c) Taking A and B as centres and radius more than half of AB, draw two arcs intersecting each other at R.

(d) Join OR. Thus, ∠QOR = ∠POQ = 90°.

(e) Draw \(\overline{\mathrm{OD}}\) the bisector of ∠POR.

Thus, ∠QOD is required angle of 135°.

(f) Now, draw \(\overline{\mathrm{OE}}\) as the bisector of ∠QOD.

Thus, ∠QOE = ∠DOE =67\(\frac { 1 }{ 2 }\)°

Question 8.

Draw an angle of 70°. Make a copy of it using only a straight edge and compasses.

Answer:

Steps of construction:

Step I: Draw a line 1 and mark a point O on it.

Step II: Using a protractor construct ∠AOB = 70°.

Step III: With centre O and a suitable radius, draw an arc which intersects \(\overline{\mathrm{OA}}\) and \(\overline{\mathrm{OB}}\) at E and F respectively.

Step IV: Draw a ray \(\overline{\mathrm{PQ}}\).

Step V: Keeping the same radius and centre P, draw an arc intersecting \(\overline{\mathrm{PQ}}\) at R.

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

Step VII: Join PS and produce it.

Question 9.

Draw an angle of 40° Copy its supplementary angle.

Answer:

Steps of construction:

Step I: (a) By using protractor draw ∠AOB = 40°

∠COF is the supplementary angle.

Step II: With centre O and a convenient radius, draw an arc which intersects \(\overline{\mathrm{OC}}\) and \(\overline{\mathrm{OB}}\) and E and F respectively.

Step III: Draw a ray \(\overline{\mathrm{OC}}\).

Step IV: With centre Q and same radius, draw an arc intersecting \(\overline{\mathrm{OC}}\) at L.

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

Step VI: Join QS and produce it.

Thus, ZPQS is the copy of the supplementary angle ZCOB.