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# NCERT Solutions Chapter 11 - Constructions (I), Class 9, Maths Class 9 Notes | EduRev

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## Class 9 : NCERT Solutions Chapter 11 - Constructions (I), Class 9, Maths Class 9 Notes | EduRev

The document NCERT Solutions Chapter 11 - Constructions (I), Class 9, Maths Class 9 Notes | EduRev is a part of Class 9 category.
All you need of Class 9 at this link: Class 9

Exercise 11.1

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

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.

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°

(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°

(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.

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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