*Question 1:*

*Draw \(\angle 75^{\circ}\) and find its line of symmetry.*

*Answer:*

Follow the steps of construction:

(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.*

*Answer:*

Follow the steps of construction:

(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 \(\angle AOB\).

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*Question 3:*

*Construct a right angle and construct its bisector.*

*Answer:*

Follow the 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 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.*

*Answer:*

Follow the steps of construction:

(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*

*Answer:*

**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.**

**Answer:**

**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.**

**Answer:**

**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.**

**Answer:**

**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.**

**Answer:**

**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.

(vi) Set your compasses to length LN with the same radius.

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

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