**Answer
1** :

Steps of construction :

(i) Draw an angle BAC.

(ii) Draw a line DF.

(iii) With centre A and D, draw arcs of equal radius which intersect AC at L and AB at M and DF at P.

(iv) Cut off PQ = LM and join DQ and produce it to E, then

∠EDF = ∠BAC.

**Answer
2** :

Steps of construction :

(i) Draw an angle ABC which is an obtuse i.e. more than 90°.

(ii) With centre B and a suitable radius draw an arc meeting BC at E and AB at F.

(iii) With centre E and F, and radius more than half of EF draw arcs intersecting each other at G.

(iv) Join BG and produce it to D.

ThenBD is die bisector of ∠ABC.

On measuring each part, we find each angle = 53°.

**Answer
3** :

Steps of construction :

(i) With the help of protractor draw an angle ABC = 108°.

As 54° = x 108°

∴ We shall bisect it.

(ii) With centre B and a suitable radius, draw an arc meeting BC at E and AB at F.

(iii) With centre E and F, and radius more than half of EF, draw arcs intersecting each other at G

(iv) Join BG and produce it to D.

Then ∠DBC = 54°.

**Answer
4** :

Steps of construction :

(i) Using protractor, draw a right angle ∆ABC.

i. e. ∠ABC = 90°.

(ii) With centre B and a suitable radius, draw an arc which meets BC at E and BA at F.

(iii) With centre E and F, draw arcs intersecting each other at G.

(iv) Join BG and produce it to D.

Then BD is the bisector of ∠ABC.

∴ ∠DBC = x 90° = 45°.

Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.

**Answer
5** :

Steps of construction :

(i) Draw a linear pair ∠DCA and ∠DCB.

(ii) Draw the bisectors of ∠DCA and ∠DCB. Forming ∠ECF on measuring we get ∠ECF = 90°.

Verification : ∵∠DCA + ∠DCB = 180°

⇒ ∠DCA + ∠DCB = 180° x = 90°

∴ ∠ECF = 90°

i.e. EC and FC are perpendicular to each other.

**Answer
6** :

Steps of construction :

(i) Draw two lines AB and CD intersecting each other at O.

(ii) Draw the bisector of ∠AOD and also the bisector of ∠BOC. Which are OP and OQ respectively. We see that OP and OQ are in the same straight line.

**Answer
7** :

Steps of construction :

(i) Draw a line segment BC.

(ii) With centre B and a suitable radius, draw an arc meeting BC at E.

(iii) With centre E and same radius, cut off arcs EF and FG.

(iv) Bisect arc FG at H.

(v) Join BH and produce it to A.

Then ∠ABC = 90°.

**Answer
8** :

Steps of construction :

(i) Draw a line DC and take a point B on it.

(ii) With centre B and a suitable radius draw an arc meeting BC at P.

(iii) With centre P, cut off arcs PQ, QR and RS.

(iv) Bisect as QR at T and join BT and produce it to E.

(v) Now bisect the arc KS at RL.

(vi) Join BL and produce it to A.

Now ∠ABC = 135°.

**Answer
9** :

Steps of construction :

(i) Draw an angle ABC = 12° with the help of protractor.

(ii) With centre B and a suitable radius, draw an arc EF.

(iii) With centre E and F, draw arcs intersecting

each other at G and produce it to D.

Then BD is the bisector of ∠ABC.

∴ ∠DBC = 72° x = 36°.

(iv) Again bisect ∠ABD in the same way then PB is the bisector of ∠ABD.

∴ ∠PBC = 36° + x 36°

= 36° + 18° = 54°

Hence ∠PBC = 54°

Construct the following angles at the initial point of a given ray and justify the construction:

(i) 45°

(ii) 90°

**Answer
10** :

(i) 45°

Steps of construction :

(a) Draw a line segment BC.

(b) With centre B and a suitable radius draw an arc meeting BC at E.

(c) With centre E, cut off equal arcs EF and FG.

(d) Bisect FG at H.

(e) Join BH and produce to X so that ∠XBC = 90°.

(f) Bisect ∠XBC so that ∠ABC = 45°.

(ii) 90°

Steps of construction :

(a) Draw a line segment BC.

(b) With centre B and a suitable radius, draw an arc meeting BC at E and with centre E cut off arcs from E, EF = FG

(c) Now bisect the arc EG at H.

(d) Join BH and produce it to A.

∴ ∠ABC = 90°.

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