Q.1      Construct an angle of 90º at the  initial  point of a given ray and justify the construction.
Sol.         Steps of construction :   

1.  Draw  a ray OA.
2. With its initial point  O as  centre and any radius,  draw an arc CDE, cutting OA at C.
3. With  centre  C and same  radius  (as in step 2), draw  an arc, cutting  the arc CDE at D.
4. With D as centre and the same radius, draw  an arc cutting the arc CDE at E.
5. With D and E as centres, and any convenient radius (morethan12DE), draw two arcs intersecting at P.
6. Join OP. Then  ∠AOP=90o
Justification : –
By  construction , OC = CD = OD
Therefore ΔOCD is an  equilateral triangle. So, ∠COD=60o
Again OD  = DE = EO
Therefore Δ ODE is also an equilateral triangle. So ∠DOE=60o
Since OP bisects ∠DOE,so∠POD=30o.
Now, ∠AOP=∠COD+∠DOP=60∘�+30∘�=90o

Q.2      Construct an angle of 45º at the initial  point of a  given ray and justify the construction.
Sol. 

Steps of Construction :
1.  Draw a ray OA.
2. With O as centre and any suitable  radius  draw an arc cutting OA at B.
3. With B as centre and same radius cut the previous drawn arc at C and then with C as centre  and same radius  cut the arc at D.
4. With C as centre and radius  more  than half CD draw an arc.
5. With D as centre  and same  radius  draw another  arc to cut the  previous arc at E.
6. Join  OE. Then ∠AOE=90o
7. Draw the bisector OF of  ∠AOE.Then∠AOF=45o

Justification :
By construction ∠AOE=90o and OF is the bisector of ∠AOE
Therefore, ∠AOF=12∠AOE=12×90∘�=45o

Q.3       Construct the angles of the following  measurements :
              (i) 30º          (ii) 2212∘           (iii) 15º
Sol.          (i) Steps of Construction :

1. Draw  a ray OA.
2. With its initial  point O as centre and any radius, draw an  arc,cutting  OA at C.
3. With  centre C and Same radius (as in step 2). Draw an arc,cutting the arc of step 2 in D.
4. With C and D as centres, and any convenient radius (morethan12CD),draw two arcs intersecting at B.
5. Join  OB. Then ∠AOB=30o

(ii)  Steps of Construction :

1. Draw an angle AOB  = 90º
2. Draw the bisector OC of  ∠AOB,then∠AOC=45o
3. Bisect ∠AOC, such  that ∠AOD=∠COD=22.5o
Thus ∠AOD=22.5o

(iii) Steps of Construction :

1. Construct an ∠AOB=60o
2. Bisect ∠AOB so that ∠AOC=∠BOC=30o.
3. Bisect ∠AOC, so  that ∠AOD=∠COD=15o
Thus ∠AOD=15o

Q.4       Construct the following  angles and verify by  measuring  them by a protractor :
               (i)  75º           (ii) 105º           (iii) 135º
Sol.          (i)  Steps of Construction :

1.  Draw a ray OA.
2.  Construct ∠AOB=60o
3.  Construct ∠AOP=90o
4.  Bisect ∠BOP so that
∠BOQ=12∠BOP
=12(∠AOP−∠AOB)
=12(90o−60o)=12×30o=15o
So, we obtain
∠AOQ=∠AOB+∠BOQ
= 60º + 15º = 75º

Verification :
On  measuring  ∠AOQ, with  the protractor, we find ∠AOQ=75o

(ii) Steps of Construction :
1. Draw a line  segment  XY.
2. Construct ∠XYT=120o and ∠XYS=90o , so that
∠SYT=∠XYT−∠XYS
= 120º – 90º
= 30º
3.  Bisect angle SYT, by  drawing its bisector YZ.
Then  ∠XYZ is the   required angle  of 105º

(iii) Steps  of Construction :

1. Draw ∠AOE=90∘
Then ∠LOE=90∘
2. Draw   the bisector of  ∠LOE.
Then∠AOF=135o

Q.5       Construct an equilateral triangle, given its side  and justify the construction.
Sol.

Let us draw an equilateral  triangle  of side 4.6 cm (say).
Steps  of Construction :
1. Draw  BC = 4.6 cm
2. With  B and C as centres and radii  equal to BC = 4.6 cm, draw two  arcs on  the same  side of BC,    intersecting each-other at A.
3. Join AB and AC.

Then, ABC is the required equilateral  triangle.
Justification : Since by construction :
AB = BC = CA = 4.6 cm
Therefore  Δ ABC is an equilateral  triangle.

Q.1    Construct a triangle ABC in which  BC = 7 cm, ∠B=75o and  AB + AC = 13 cm
Sol.       Steps of Construction :

1.Draw a ray BX and cut off a line segment BC = 7 cm
2. Construct ∠XBY=75o
3. From BY, cut off BD = 13 cm
4. Join CD.
5. Draw the perpendicular  bisect of CD, intersecting BA at A.
6. Join AC.
The  triangle ABC thus obtained is the required triangle.

Q.2      Construct a triangle ABC in which  BC = 8 cm, ∠B=45o and AB – AC = 3.5 cm
Sol.

Steps of Construction :
1. Draw  a ray BX and cut off a line  segment BC = 8 cm from it.
2. Construct ∠YBC=45o
3. Cut off a line  segment BD = 3.5 cm from BY.
4. Join CD.
5. Draw perpendicular bisector of CD intersecting  BY at a point A.
6. Join AC

Then  ABC is the  required triangle.

Q.3       Construct a triangle PQR in which  QR = 6 cm ∠Q=60o and PR – PQ = 2cm
Sol.

Steps of Construction :
1. Draw a ray QX and cut off a line segment QR = 6 cm  from it.
2. Construct  a ray QY making  an angle  of 60º with QR and produce YQ to form a line YQY’
3. Cut off  a line segment QS = 2cm  from QY’.
4. Join RS.
5. Draw perpendicular bisector  of RS intersecting QY at a point  P.
6. Join  PR.

Then  PQR is the required triangle.

Q.4       Construct a triangle XYZ in which ∠Y=30o,  ∠Z=90o and XY + YZ + ZX = 11 cm.
Sol.

Steps  of Construction :
1. Draw  a line segment PQ = 11 cm

2. At P,  draw a ray PL  such  that ∠LPQ=12×30o=15o
3. At Q, draw ray QM such that ∠MQP=12×90o=45o intersecting  PL at X.
4. Draw perpendicular bisectors of XP and XQ intersecting  PQ in Y and Z respectively.
Then  Δ XYZ is the required  triangle.
Note : –  For  clarity in figure, method of drawing angles of 15º and 45º have not been shown. Students should draw these angles with the help of ruler and  compass only by the method as shown earlier.

Q.5      Construct a right triangle  whose base is 12 cm and sum  of its hypotenuse and other  side is 18 cm.
Sol.          Steps of Construction :

1. Draw a ray BX and cut off a line segment  BC = 12 cm
2. Construct ∠XBY=90o
3. From by  cut off a line segment BD = 18 cm .
4. Join CD.
5. Draw the perpendicular  bisector of CD intersecting BD at A.
6. Join AC
Then  ABC is the  required triangle.