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

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

**Solution:**

Steps of construction

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

Justification:Join AE, CE, EF and AE, AF

AC = CE = AE [ by construction]

∴ ACE is an equilateral Triangle

⇒ ∠CAE = 60° —————–1

Similarly, AE = EF = AF

∴Triangle AEF is an equilateral Triangle

⇒ ∠EAF = 60°

Because AG bisects ⇒ ∠EAF

∴∠GAE = 1\2 = 30° = 30°————2

1+2

∴∠CAE + ∠GAE = 60°+30°

∠GAB=30°

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

**Solution:**

Step of Construction:

- Take a ray AB with initial point A
- Draw ∠BAF=90°
- Taking C as Centre and radius more than draw an arc.
- Taking G as Centre and the same radius as before, draw another arc.
- Taking G as Centre and the same radius as before, draw another arc. Intersecting previous arc at H.
- Draw the ray AH.
- Then ∠BAH is the required angle of 45°

Justification:Join GH and HC (construct)

In ∆ AHG and ∆ AHC

HG=HC…………….[arc of equal radii]

AG=AC……………..[radii of same arc]

AH=AH………………[common]

AHG≅AHC [S.S.S]

∠HAG=∠HAC [C.P.C.T]

But ∠HAG+∠HAC=90

∠HAG=∠HAC=90\2=45

∴∠BAH=45

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

**i) 30°**

**Solution:**

Step of construction

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

**ii) 22 ½°**

**Solution:**

Steps of construction

- Take a ray AB
- Draw an angle ∠AB=90° on point A.
- Bisect ∠CAB and draw ∠DAB=45°
- Bisect ∠DAB and draw ∠EAB
- ∠EAB is required angle of 22 ½°

**iii) 15°**

**Solution:**

Steps of construction

- Take a ray AB.
- Draw an arc on AB, by taking A a center, which intersect AB at c.
- From C with the same radius draw another re which intersect the previous arc at D.
- Join DA.
- ∠DAB =60°
- Bisect ∠DAB and draw angle EAB=30°
- Bisect ∠EAB and draw ∠FAB
- ∠FAB is the required angle.

### Question 4. Construct the following angles and verify by measuring them by a protractor

**(i) 75°**

**Solution:**

Steps of construction

- Draw a ray AB with initial point A.
- At point A draw an angle ∠CAB=90°
- At point A draw ∠DAB=60°
- Bisect ∠CAD, now ∠EAD=15°
- ∠EAB=75° {∠EAB=∠EAD+∠DAB=15°+60°=75°}

### (ii) 105°

**Solution:**

Steps of construction

- Draw a ray AB with initial point A.
- At point A draw an angle ∠CAB=90°
- At point A draw ∠DAB=120°
- Bisect ∠CAD, now ∠EAD=15°
- ∠EAB=75° {∠EAB=∠EAC+∠CAB=15°+90°=105°}

### (iii) 135°

**Solution:**

Steps of construction

- Draw a ray AB with initial point A.
- At point A draw an angle ∠CAB=120°
- At point A draw ∠DAB=150°
- Bisect ∠CAD, now ∠EAC=15°
- ∠EAB=135° {∠EAB=∠EAC+∠CAB=15°+120°=135°}

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

**Solution:**

Steps of construction

- Draw a line segment of AB of a given length.
- With A and B as centre and radius equal to AB draw arcs to intersect each other at c.
- Join AC and BC.
Then ABC is the required equilateral triangle.

Justification:AB=AC ……………. [by construction]

AB=BC ……………..[by construction]

AB=AC=BC

Hence, ∆ABC is required equilateral triangle.