**Chapter 11 Constructions**

**Ex 11.1**

**Ex 11.1 Class 9 Maths Question 1.**

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

Solution:

Steprf of Construction:

Step I : Draw

Step II : Taking O as centre and having a suitable radius, draw a semicircle,
which cuts

Step III : Keeping the radius same, divide the semicircle into three equal
parts such that

Step IV: Draw

Step V: Draw**∠****COD.**

Thus,

Thus,

**∠**

**AOF = 90°**

Justification:

Justification:

**∵**

**O is the centre of the semicircle and it is divided into 3 equal parts.**

**∴**

**⇒**

**∠**

**BOC =**

**∠**

**COD =**

**∠**

**DOE [Equal chords subtend equal angles at the centre]**

And,

And,

**∠**

**BOC +**

**∠**

**COD +**

**∠**

**DOE = 180°**

**⇒**

**∠**

**BOC +**

**∠**

**BOC +**

**∠**

**BOC = 180°**

**⇒**

**3**

**∠**

**BOC = 180°**

**⇒**

**∠**

**BOC = 60°**

Similarly,

Similarly,

**∠**

**COD = 60° and**

**∠**

**DOE = 60°**

**∵**

is the bisector of

**∠**

**COD**

**∴**

**∠**

**COF =**

**∠**

**COD =**
(60°) = 30°

Now,

Now,

**∠**

**BOC +**

**∠**

**COF = 60° + 30°**

**⇒**

**∠**

**BOF = 90° or**

**∠**

**AOF = 90°**

**Ex 11.1 Class 9 Maths Question 2.**

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

Solution:

Steps of Construction:

Step I : Draw

Step II: Taking O as centre and with a suitable radius, draw a semicircle such
that it intersects

Step III : Taking B as centre and keeping the same radius, cut the semicircle
at C. Now, taking C as centre and keeping the same radius, cut the semicircle at
D and similarly, cut at E, such that

Step IV : Draw

Step V: Draw**∠****BOC.
Step VI: Draw
**

**∠**

**FOC.**

Thus,

Thus,

**∠**

**BOG = 45° or**

**∠**

**AOG = 45°**

Justification:

Justification:

**∵**

**∴**

**∠**

**BOC =**

**∠**

**COD =**

**∠**

**DOE [Equal chords subtend equal angles at the centre]**

Since,

Since,

**∠**

**BOC +**

**∠**

**COD +**

**∠**

**DOE = 180°**

**⇒**

**∠**

**BOC = 60°**

**∵**

is the bisector of

**∠**

**BOC.**

**∴**

**∠**

**COF =**

**∠**

**BOC =**
(60°) = 30° …(1)

Also,
is the bisector of

Also,

**∠**

**COF.**

**∠**

**FOG =**

**∠**

**COF =**
(30°) = 15° …(2)

Adding (1) and (2), we get

Adding (1) and (2), we get

**∠**

**COF +**

**∠**

**FOG = 30° + 15° = 45°**

**⇒**

**∠**

**BOF +**

**∠**

**FOG = 45° [**

**∵**

**∠**

**COF =**

**∠**

**BOF]**

**⇒**

**∠**

**BOG = 45°**

**Ex 11.1 Class 9 Maths Question 3.**

**Construct the angles of the following
measurements
(i) 30°
(ii) 22 **

(iii) 15°

(iii) 15°

Solution:

(i) Angle of 30°

Steps of Construction:

Step I : Draw
.

Step II : With O as centre and having a suitable radius, draw an arc cutting
at B.

Step III : With centre at B and keeping the same radius as above, draw an arc to cut the previous arc at C.

Step IV : Join
which gives

Solution:

(i) Angle of 30°

Steps of Construction:

Step I : Draw

Step II : With O as centre and having a suitable radius, draw an arc cutting

Step III : With centre at B and keeping the same radius as above, draw an arc to cut the previous arc at C.

Step IV : Join

**∠**

**BOC = 60°.**

Step V: Draw
, bisector of

Step V: Draw

**∠**

**BOC, such that**

**∠**

**BOD =**

**∠**

**BOC =**
(60°) = 30°

Thus,

Thus,

**∠**

**BOD = 30° or**

**∠**

**AOD = 30°**

**(ii) Angle of 22
**

**∠**

**AOB = 90°**

Step III: Draw
, the bisector of

Step III: Draw

**∠**

**AOB, such that**

**∠**

**AOC =**

**∠**

**AOB =**
(90°) = 45°

Step IV: Now, draw OD, the bisector of

Step IV: Now, draw OD, the bisector of

**∠**

**AOC, such that**

**∠**

**AOD =**

**∠**

**AOC =**
(45°) = 22

Thus,

Thus,

**∠**

**AOD = 22**

**(iii) Angle of 15°
Steps of Construction:
Step I: Draw**

**∠**

**AOB = 60°.**

Step III: Draw OC, the bisector of

Step III: Draw OC, the bisector of

**∠**

**AOB, such that**

**∠**

**AOC =**

**∠**

**AOB =**
(60°) = 30°

i.e.,

i.e.,

**∠**

**AOC = 30°**

Step IV: Draw OD, the bisector of

Step IV: Draw OD, the bisector of

**∠**

**AOC such that**

**∠**

**AOD =**

**∠**

**AOC =**
(30°) = 15°

Thus,

Thus,

**∠**

**AOD = 15°**

**Ex 11.1 Class 9 Maths Question 4.**

**Construct the following angles and verify
by measuring them by a protractor
(i) 75°
(ii) 105°
(iii) 135°**

Solution:

Step I : Draw
.

Step II: With O as centre and having a suitable radius, draw an arc which cuts
at B.

Step III : With centre B and keeping the same radius, mark a point C on the previous arc.

Step IV: With centre C and having the same radius, mark another point D on the arc of step II.

Step V: Join
and
, which gives

Solution:

Step I : Draw

Step II: With O as centre and having a suitable radius, draw an arc which cuts

Step III : With centre B and keeping the same radius, mark a point C on the previous arc.

Step IV: With centre C and having the same radius, mark another point D on the arc of step II.

Step V: Join

**∠**

**COD = 60° =**

**∠**

**BOC.**

Step VI: Draw
, the bisector of

Step VI: Draw

**∠**

**COD, such that**

**∠**

**COP =**

**∠**

**COD =**
(60°) = 30°.

Step VII: Draw
, the bisector of

Step VII: Draw

**∠**

**COP, such that**

**∠**

**COQ =**

**∠**

**COP =**
(30°) = 15°.

Thus,

Thus,

**∠**

**BOQ = 60° + 15° = 75°**

**∠**

**AOQ = 75°**

**(ii) Steps of Construction:
Step I : Draw
**

**∠**

**BOP = 90°.**

Step VI: Draw
, the bisector of
such that

Step VI: Draw

**∠**

**POQ = 15°**

Thus,

Thus,

**∠**

**AOQ = 90° + 15° = 105°**

**(iii) Steps of Construction:
Step I : Draw
**

**∠**

**POQ = 135°**

**Ex 11.1 Class 9 Maths Question 5.**

**Construct an equilateral triangle, given
its side and justify the construction.**

Solution:

pt us construct an equilateral triangle, each of whose side = 3 cm(say).

Steps of Construction:

Step I : Draw

Step II : Taking O as centre and radius equal to 3 cm, draw an arc to cut

Step III : Taking B as centre and radius equal to OB, draw an arc to intersect
the previous arc at C.

Step IV : Join OC and BC.

Thus, ∆OBC is the required equilateral triangle.

**Justification:
**

**∵**

**The arcs**
and
are drawn with the same radius.

**∴**

=

**⇒**

**OC = BC [Chords corresponding to equal arcs are equal]**

**∵**

**OC = OB = BC**

**∴**

**OBC is an equilateral triangle.**

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