**Q.1Â Â Â Â Â Construct an angle of 90Âº at theÂ initialÂ point of a given ray and justify the construction.**

**Sol.Â Â Â Â Â Â Â Â ****Steps of construction : Â Â ** Â

1.Â DrawÂ a ray OA.

2. With its initial pointÂ O asÂ centre and any radius,Â draw an arc CDE, cutting OA at C.

3. WithÂ centreÂ C and sameÂ radiusÂ (as in step 2), drawÂ an arc, cuttingÂ the arc CDE at D.

4. With D as centre and the same radius, drawÂ an arc cutting the arc CDE at E.

5. With D and E as centres, and any convenient radius , draw two arcs intersecting at P.

6. Join OP. Then Â Â

**Justification : - **

ByÂ construction , OC = CD = OD

Therefore OCD is anÂ equilateral triangle. So,

Again ODÂ = DE = EO

Therefore ODE is also an equilateral triangle. So

Since OP bisects .

Now,

**Q.2Â Â Â Â Â Construct an angle of 45Âº at the initialÂ point of aÂ given ray and justify the construction.**

**Sol.Â **

**Steps of Construction :**

1.Â Draw a ray OA.

2. With O as centre and any suitableÂ radiusÂ 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 centreÂ and same radiusÂ cut the arc at D.

4. With C as centre and radiusÂ moreÂ than half CD draw an arc.

5. With D as centreÂ and sameÂ radiusÂ draw anotherÂ arc to cut theÂ previous arc at E.

6. JoinÂ OE. Then

7. Draw the bisector OF ofÂ

**Justification :**

By construction and OF is the bisector of

Therefore,Â

**Q.3Â Â Â Â Â Â Construct the angles of the followingÂ measurements : **

Â Â Â **Â Â Â Â Â Â Â Â (i) 30ÂºÂ Â Â Â Â Â Â (ii) Â Â Â Â Â Â Â (iii) 15Âº**

**Sol.Â Â Â Â Â Â Â Â Â ****(i) Steps of Construction :**

1. DrawÂ a ray OA.

2. With its initialÂ point O as centre and any radius, draw anÂ arc,cuttingÂ OA at C.

3. WithÂ 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 ,draw two arcs intersecting at B.

5. JoinÂ OB. Then

** (ii)Â Steps of Construction :**

1. Draw an angle AOBÂ = 90Âº

2. Draw the bisector OC ofÂ

3. Bisect , suchÂ that

Thus

** (iii) Steps of Construction :**

1. Construct an

2. Bisect so that .

3. Bisect , soÂ that

Thus

**Q.4Â Â Â Â Â Â Construct the followingÂ angles and verify byÂ measuringÂ them by a protractor : **

Â Â Â Â Â **Â Â Â Â Â (i)Â 75ÂºÂ Â Â Â Â Â Â (ii) 105ÂºÂ Â Â Â Â Â Â (iii) 135Âº**

* Sol.*Â Â Â Â Â Â Â Â Â

1.Â Draw a ray OA.

2.Â Construct

3.Â Construct

4.Â Bisect so thatÂ

So, we obtain

= 60Âº + 15Âº = 75Âº

**Verification :**

OnÂ measuringÂ , withÂ the protractor, we find

**(ii) Steps of Construction :**

1. Draw a lineÂ segmentÂ XY.

2. Construct and , so that

= 120Âº â€“ 90Âº

= 30Âº

3.Â Bisect angle SYT, byÂ drawing its bisector YZ.

ThenÂ is theÂ Â required angleÂ of 105Âº

Â

**(iii) StepsÂ of Construction :**

1. Draw

Then Â

2. DrawÂ Â the bisector ofÂ .

**Q.5Â Â Â Â Â Â Construct an equilateral triangle, given its sideÂ and justify the construction.
**

Let us draw an equilateralÂ triangleÂ of side 4.6 cm (say).

**StepsÂ of Construction :**

1. DrawÂ BC = 4.6 cm

2. WithÂ B and C as centres and radiiÂ equal to BC = 4.6 cm, draw twoÂ arcs onÂ the sameÂ side of BC,Â Â Â intersecting each-other at A.

3. Join AB and AC.

Then, ABC is the required equilateralÂ triangle.

**Justification :** Since by construction :

AB = BC = CA = 4.6 cm

ThereforeÂ ABC is an equilateralÂ triangle.

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