# Class 9 NCERT Solutions- Chapter 11 Constructions – Exercise 11.1

Last Updated : 15 Mar, 2022

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

Solution:

Steps of construction

1. Take a ray with initial point A.
2. Taking care center and same radius draw an Arc of a circle which intersect AB at C.
3. With C as Centre and the same radius, draw an arc intersecting the previous arc at E.
4. With E as Centre and the same radius draw an arc which intersects the arc drawn in step 2 at F.
5. With E as Centre and the same radius, draw another arc, intersecting the previous arc at G.
6. Draw the ray AG.
7. Then âˆ BAG is the required angle 90Â°

Justification:

Join AE, CE, EF and AE, AF

AC = CE = AE      [ by construction]

âˆ´ ACE is an equilateral Triangle

â‡’ âˆ CAE = 60Â°    —————–1

Similarly, AE = EF = AF

âˆ´Triangle AEF is an equilateral Triangle

â‡’ âˆ EAF = 60Â°

Because AG bisects â‡’ âˆ EAF

âˆ´âˆ GAE = 1\2 = 30Â° = 30Â°————2

1+2

âˆ´âˆ CAE + âˆ GAE = 60Â°+30Â°

âˆ GAB=30Â°

### Question 2. Construct an angle of 45Â° at the initial point of a given ray and justify the construction.

Solution:

Step of Construction:

1. Take a ray AB with initial point A
2. Draw âˆ BAF=90Â°
3. Taking C as Centre and radius more than   draw an arc.
4. Taking G as Centre and the same radius as before, draw another arc.
5. Taking G as Centre and the same radius as before, draw another arc. Intersecting previous arc at H.
6. Draw the ray AH.
7. Then âˆ BAH is the required angle of 45Â°

Justification:

Join GH and HC    (construct)

In âˆ† AHG and âˆ† AHC

AH=AHâ€¦â€¦â€¦â€¦â€¦â€¦[common]

AHGâ‰…AHC             [S.S.S]

âˆ HAG=âˆ HAC                [C.P.C.T]

But âˆ HAG+âˆ HAC=90

âˆ HAG=âˆ HAC=90\2=45

âˆ´âˆ BAH=45

### Question 3. Draw the angles of the following measurement

i) 30Â°

Solution:

Step of construction

1. Draw a ray AB with initial point A.
2. With A as centre, draw an arc intersecting AB at c.
3. With c as centre and the same radius, draw another arc, intersecting the previously drawn arc at D.
4. Taking C and D as centre  and with the radius more than 1\2 DC draws arcs to intersect each other at E.
5. Draw ray AE. âˆ EAB is the required angle of 30.

ii) 22 Â½Â°

Solution:

Steps of construction

1. Take a ray AB
2. Draw an angle âˆ AB=90Â° on point A.
3. Bisect âˆ CAB and draw âˆ DAB=45Â°
4. Bisect âˆ DAB and draw âˆ EAB
5. âˆ EAB is required angle of 22 Â½Â°

iii) 15Â°

Solution:

Steps of construction

1. Take a ray AB.
2. Draw an arc on AB, by taking A a center, which intersect AB at c.
3. From C with the same radius draw another re which intersect the previous  arc at D.
4. Join DA.
5. âˆ DAB =60Â°
6. Bisect  âˆ DAB and draw angle  EAB=30Â°
7. Bisect âˆ EAB and draw âˆ FAB
8. âˆ FAB is the required angle.

### (i) 75Â°

Solution:

Steps of construction

1. Draw a ray AB with initial point  A.
2. At point A draw an angle âˆ CAB=90Â°
3. At point A draw âˆ DAB=60Â°
5. âˆ EAB=75Â°        {âˆ EAB=âˆ EAD+âˆ DAB=15Â°+60Â°=75Â°}

### (ii) 105Â°

Solution:

Steps of construction

1. Draw a ray AB with initial point  A.
2. At point A draw an angle âˆ CAB=90Â°
3. At point A draw âˆ DAB=120Â°
5. âˆ EAB=75Â°        {âˆ EAB=âˆ EAC+âˆ CAB=15Â°+90Â°=105Â°}

### (iii) 135Â°

Solution:

Steps of construction

1. Draw a ray AB with initial point A.
2. At point A draw an angle âˆ CAB=120Â°
3. At point A draw âˆ DAB=150Â°
4. Bisect âˆ CAD, now âˆ EAC=15Â°
5. âˆ EAB=135Â°        {âˆ EAB=âˆ EAC+âˆ CAB=15Â°+120Â°=135Â°}

### Question 5. Construct an equilateral triangle, given its side and justify the construction.

Solution:

Steps of construction

1. Draw a line segment of AB of a given length.
2. With A and B as centre and radius equal to AB draw arcs to intersect each other at c.
3. Join AC and BC.

Then ABC is the required equilateral triangle.

Justification:

AB=AC    â€¦â€¦â€¦â€¦â€¦. [by construction]

AB=BC    â€¦â€¦â€¦â€¦â€¦..[by construction]

AB=AC=BC

Hence, âˆ†ABC is required equilateral triangle.

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