# Ncert Solutions For Class 6 Maths Ex 14.6

## Ncert Solutions For Class 6 Maths Chapter 14 Ex 14.6

Question 1:

Draw $$\angle 75^{\circ}$$ and find its line of symmetry.

(a) Construct 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 $$\angle BOA=60^{\circ}$$ .

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

(f) Draw bisector of $$\angle BOC$$ . The angle is of $$\angle BOC$$. Mark it at D. Thus, $$\angle DOA=90^{\circ}$$

(g) Draw $$\overline{OP}$$ as bisector of $$\angle DOB$$.

Thus, $$\angle POA=75^{\circ}$$

Question 2:

Draw an angle of measure $$147^{\circ}$$ and construct its bisector.

(a) Draw a line $$\overline{OA}$$ .

(b) Using protractor, construct $$\angle AOB=147^{\circ}$$ .

(c) Taking the centre as O and any suitable radius, draw an arc which cuts the arms $$\overline{OA}$$ and $$\overline{OB}$$ at P and Q respectively.

(d) Taking P as the 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 cut the previous at R.

(f) Join OR and produce it.

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

Question 3:

Construct a right angle and construct its bisector.

(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 centers and radius more than half of AB, draw two arcs which intersect each other at C.

(d) Join OC. Thus, $$\angle COQ$$ 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, $$\overline{OD}$$ is the required bisector of $$\angle COQ$$ .

$$\angle COQ$$ .

Question 4:

Draw an angle of measure $$153^{\circ}$$ and divide it into four equal parts.

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

(b) At O, with the help of a protractor, construct $$\angle AOB=153^{\circ}$$ .

(c) Draw $$\overline{OC}$$ as the bisector of $$\angle AOB$$ .

(d) Again, draw $$\overline{OD}$$ as bisector of $$\angle AOC$$

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

(f) Thus, $$\overline{OC}$$ , $$\overline{OD}$$ and $$\overline{OE}$$ divide $$\angle AOB$$ in four equal parts.

Question 5:

Construct with ruler and compasses, angles of following measures:

(1) 60 (2) 30 (3) 90 (4) 120 (5) 45 (6) 135

Steps of construction:

(1)$$60^{\circ}$$

(i) Draw a ray $$\overline{OL}$$.

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

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

(iv) Join OQ.

Thus, $$\angle$$MOL is required angle of 60 .

(2) $$30^{\circ}$$

(i) Draw a ray $$\overline{OL}$$.

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

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

(iv) Join OQ. Thus, $$\angle$$MOL is required angle of $$60^{\circ}$$.

(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 N.

Thus, $$\angle$$NOL is required angle of $$30^{\circ}$$ .

(3) $$90^{\circ}$$

(i) Draw a ray $$\overline{OL}$$.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects $$\overline{OL}$$ at X.

(iii) Taking X as centre and same radius, cut previous arc at Y.

(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.

(v) Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S. (vi) Join OS and produce it to form a ray OM.

Thus, $$\angle$$MOL is required angle of $$90^{\circ}$$ .

(4) $$120^{\circ}$$

(i) Draw a ray $$\overline{OI}$$.

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

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

(iv) Taking Q as centre and same radius cut the arc at S.

(v) Join OS.

Thus, $$\angle$$ IOL is required angle of $$120^{\circ}$$.

(5) $$45^{\circ}$$

(i) Draw a ray $$\overline{OI}$$.

(ii) Taking O as centre and convenient radius, mark an arc, which intersects $$\overline{OI}$$ at X.

(iii) Taking X as centre and same radius, cut previous arc at Y.

(iv) Taking Y as centre and same radius, draw another arc intersecting the same arc at Z.

(v) Taking Y and Z as centres and same radius, draw two arcs intersecting each other at S.

(vi) Join OS and produce it to form a ray OJ. Thus, $$\angle$$ JOI is required angle of $$90^{\circ}$$ .

(vii) Draw the bisector of $$\angle$$JOI.

Thus, $$\angle$$MOI is required angle of $$45^{\circ}$$ .

(6) $$135^{\circ}$$

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

(ii) Taking O as centre and convenient radius, mark an arc, which intersects PQ at I and J.

(iii) Taking I and J as centres and radius more than half of IJ, draw two arcs intersecting each other at R.

(iv) Join OR. Thus, $$\angle$$QOR = $$\angle$$POQ = $$90^{\circ}$$.

(v) Draw $$\overline{OL}$$ the bisector of $$\angle$$POR.

Thus, $$\angle$$QOL is required angle of $$135^{\circ}$$ .

Question 6:

Draw an angle of measure $$45^{\circ}$$ and bisect it.

Steps of construction:

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

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

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

(iv)Join OC. Then $$\angle$$COQ is an angle of 90

(v) Draw $$\overline{OE}$$ as the bisector of $$\angle$$COE. Thus, $$\angle$$QOE =$$90^{\circ}$$

(vi) Again draw $$\overline{OG}$$ as the bisector of $$\angle$$QOE.

Thus, $$\angle$$QOG = $$\angle$$EOG = $$22\frac{1}{2}^{\circ}$$

Question 7:

Draw an angle of measure $$135^{\circ}$$ and bisect it.

Steps of construction:

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

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

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

(iv) Join OR. Thus, $$\angle$$QOR = $$\angle$$POQ = $$90^{\circ}$$ .

(v) Draw $$\overline{OD}$$ the bisector of $$\angle$$POR. Thus, $$\angle$$QOD is required angle of $$135^{\circ}$$ .

(vi) Now, draw $$\overline{OE}$$ as the bisector of $$\angle$$QOD.

Thus, $$\angle$$QOE = $$\angle$$DOE = $$67\frac{1}{2}^{\circ}$$

Question 8:

Draw an angle of $$70^{\circ}$$ . Make a copy of it using only a straight edge and compasses.

Steps of construction:

(i) Draw an angle $$70^{\circ}$$ with protractor, i.e., $$\angle$$POQ = $$70^{\circ}$$

(ii) Draw a ray $$\overline{AB}$$.

(iii) Place the compasses at O and draw an arc to cut the rays of $$\angle$$POQ at L and M.

(iv) Use the same compasses, setting to draw an arc with A as centre, cutting AB at X.

(v) Set your compasses setting to the length LM with the same radius.

(vi) Place the compasses pointer at X and draw the arc to cut the arc drawn earlier at Y.

(vii) Join AY.

Thus, $$\angle$$YAX = $$70^{\circ}$$

Question 9:

Draw an angle of $$40^{\circ}$$. Copy its supplementary angle.

Steps of construction:

(i) Draw an angle of $$40^{\circ}$$ with the help of protractor, naming $$\angle$$AOB.

(ii) Draw a line PQ.

(iii) Take any point M on PQ.

(iv) Place the compasses at O and draw an arc to cut the rays of $$\angle$$AOB at L and N.

(v) Use the same compasses setting to draw an arc O as centre, cutting MQ at X.

(viii)Join MY. Thus, $$\angle$$QMY = $$40^{\circ}$$ and $$\angle$$PMY is supplementary of it