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**Exercise 11.1****1. Construct an angle of 90Â° at the initial point of a given ray and justify the construction. Answer**

Steps of construction:

Step 1: A ray YZ is drawn.

Step 2: With Y as a centre and any radius, an arc ABC is drawn cutting YZ at C.

Step 3: With C as a centre and the same radius, mark a point B on the arc ABC.

Step 4: With B as a centre and the same radius, mark a point A on the arc ABC.

Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.

Step 6: X and Y are joined and a ray XY making an angle 90Â° with YZ is formed.

Justification for construction:

We constructed âˆ BYZ = 60Â° and also âˆ AYB = 60Â°.

Thus, âˆ AYZ = 120Â°.

Also, bisector of âˆ AYB is constructed such that:

âˆ AYB = âˆ XYA + âˆ XYB

â‡’ âˆ XYB = 1/2âˆ AYB

â‡’ âˆ XYB = 1/2Ã—60Â°

â‡’ âˆ XYB = 30Â°

Now,

âˆ XYZ = âˆ BYZ + âˆ XYB = 60Â° + 30Â° = 90Â°**2. Construct an angle of 45Â° at the initial point of a given ray and justify the construction. Answer**

Steps of construction:

Step 1: A ray OY is drawn.

Step 2: With O as a centre and any radius, an arc ABC is drawn cutting OY at A.

Step 3: With A as a centre and the same radius, mark a point B on the arc ABC.

Step 4: With B as a centre and the same radius, mark a point C on the arc ABC.

Step 5: With A and B as centre, draw two arcs intersecting each other with the same radius at X.

Step 6: X and Y are joined and a ray making an angle 90Â° with YZ is formed.

Step 7: With A and E as centres, two arcs are marked intersecting each other at D and the bisector of âˆ XOY is drawn.

Justification for construction:

By construction,

âˆ XOY = 90Â°

We constructed the bisector of âˆ XOY as DOY.

Thus,

âˆ DOY = 1/2 âˆ XOY

âˆ DOY = 1/2Ã—90Â° = 45Â°**3. Construct the angles of the following measurements: (i) 30Â° **

**(ii) 22.5Â° **

**(iii) 15Â° Answer**

(i) 30Â°

Steps of constructions:

Step 1: A ray OY is drawn.

Step 2: With O as a centre and any radius, an arc AB is drawn cutting OY at A.

Step 3: With A and B as centres, two arcs are marked intersecting each other at X and the bisector of is drawn.

Thus, âˆ XOY is the required angle making 30Â° with OY.

(ii) 22.5Â°

Steps of constructions:

Step 1: An angle âˆ XOY = 90Â° is drawn.

Step 2: Bisector of âˆ XOY is drawn such that âˆ BOY = 45Â° is constructed.

Step 3: Again, âˆ BOY is bisected such that âˆ AOY is formed.

Thus, âˆ AOY is the required angle making 22.5Â° with OY.

(iii) 15Â°

Steps of constructions:

Step 1: An angle âˆ AOY = 60Â° is drawn.

Step 2: Bisector of âˆ AOY is drawn such that âˆ BOY = 30Â° is constructed.

Step 3: With C and D as centres, two arcs are marked intersecting each other at X and the bisector of âˆ BOY is drawn.

Thus, âˆ XOY is the required angle making 15Â° with OY.**4. Construct the following angles and verify by measuring them by a protractor: (i) 75Â° (ii) 105Â° (iii) 135Â° Answer**

(i) 75Â°

Steps of constructions:

Step 1: A ray OY is drawn.

Step 2: An arc BAE is drawn with O as a centre.

Step 3: With E as a centre, two arcs are A and C are made on the arc BAE.

Step 4: With A and B as centres, arcs are made to intersect at X and âˆ XOY = 90Â° is made.

Step 5: With A and C as centres, arcs are made to intersect at D

Step 6: OD is joined and and âˆ DOY = 75Â° is constructed.

Thus, âˆ DOY is the required angle making 75Â° with OY.

(ii) 105Â°

Steps of constructions:

Step 1: A ray OY is drawn.

Step 2: An arc ABC is drawn with O as a centre.

Step 3: With A as a centre, two arcs are B and C are made on the arc ABC.

Step 4: With B and C as centres, arcs are made to intersect at E and âˆ EOY = 90Â° is made.

Step 5: With B and C as centres, arcs are made to intersect at X

Step 6: OX is joined and and âˆ XOY = 105Â° is constructed.

Thus, âˆ XOY is the required angle making 105Â° with OY.

(iii) 135Â°

Steps of constructions:Step 1: A ray DY is drawn.

Step 2: An arc ACD is drawn with O as a centre.

Step 3: With A as a centre, two arcs are B and C are made on the arc ACD.

Step 4: With B and C as centres, arcs are made to intersect at E and âˆ EOY = 90Â° is made.

Step 5: With F and D as centres, arcs are made to intersect at X or bisector of âˆ EOD is constructed.

Step 6: OX is joined and and âˆ XOY = 135Â° is constructed.

Thus, âˆ XOY is the required angle making 135Â° with DY.**5. Construct an equilateral triangle, given its side and justify the construction. Answer **

Steps of constructions:

Step 1: A line segment AB=4 cm is drawn.

Step 2: With A and B as centres, two arcs are made.

Step 4: With D and E as centres, arcs are made to cut the previous arc respectively and forming angle of 60Â° each.

Step 5: Lines from A and B are extended to meet each other at C.

Thus, ABC is the required triangle formed.

Justification:

By construction,

AB = 4 cm, âˆ A = 60Â° and âˆ B = 60Â°

We know that,

âˆ A + âˆ B + âˆ C = 180Â° (Sum of the angles of a triangle)

â‡’ 60Â° + 60Â° + âˆ C = 180Â°

â‡’ 120Â° + âˆ C = 180Â°

â‡’ âˆ C = 60Â°

BC = CA = 4 cm (Sides opposite to equal angles are equal)

AB = BC = CA = 4 cm

âˆ A = âˆ B = âˆ C = 60Â°

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