**NCERT Solutions for Class 9 Maths Chapter 11 Constructions
Ex 11.1**

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for 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:****
Steps of Construction:
**Step I : Draw a ray OB.

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

Step III : Keeping the radius same and centre as B, draw an arc at C and again keeping the same radius and centre as C, draw an arc at D.

Step IV : Draw the rays OC and OD.

Step V : Draw OF, the bisector of ∠COD.

**Justification:**

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

∴ BC = CD = DE

⇒ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre]

And, ∠BOC + ∠COD + ∠DOE = 180°

⇒ ∠BOC + ∠BOC + ∠BOC = 180°

⇒ 3∠BOC = 180°

⇒ ∠BOC = 60°

Similarly, ∠COD = 60° and ∠DOE = 60°

∵ OF is the
bisector of ∠COD.

∴ ∠COF = ½ ∠COD = ½ (60°) = 30°

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 a ray OA.

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

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 BC = CD = DE

Step IV : Draw OC and OD.

Step V : Draw OF, the angle bisector of ∠BOC.

Step
VI : Draw OG, the angle
bisector of ∠FOC.

**Justification:**

∵ BC = CD = DE

∴ ∠BOC = ∠COD = ∠DOE [Equal chords subtend equal angles at the centre]

Since, ∠BOC + ∠COD + ∠DOE = 180°

⇒ ∠BOC = 60°

∵ OF is the bisector of ∠BOC.

∴ ∠BOF = ∠COF = ½ ∠BOC = ½ (60°) = 30° …..…(1)

Also, OG is the bisector of ∠COF.

∠FOG = ½ ∠COF = ½ (30°) = 15° …..…(2)

Adding (1) and (2), we get

∠BOF + ∠FOG = 30° + 15° = 45°

⇒ ∠BOG = 45°

**Ex 11.1 Class 9 Maths Question 3.****
**Construct the angles of the
following measurements.

(i) 30°

(ii) 22½ ∘

(iii) 15°

**Solution:**

**(i)** Angle of 30°

**Steps of Construction:
**Step I : Draw a ray OA.

Step II : With O as centre and having a suitable radius, draw an arc cutting OA 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 OC which gives ∠BOC = 60°.

Step V : Draw OD, bisector of ∠BOC, such that ∠BOD = ½ ∠BOC = ½ (60°) = 30°

**(ii)**
Angle of 22½ ∘

**Steps of Construction:
**Step I : Draw a ray OA.

Step II : Construct ∠AOB = 90°

Step III : Draw OC, the bisector of ∠AOB, such that ∠AOC = ½ ∠AOB = ½ (90°) = 45°

Step IV : Now, draw OD, the bisector of ∠AOC, such that ∠AOD = ½ ∠AOC = ½ (45°) = 22½ ∘

**(iii)**
Angle of 15°

**Steps of Construction:
**Step I : Draw a ray OA.

Step II : Construct ∠AOB = 60°.

Step III : Draw OC, the bisector of ∠AOB, such that ∠AOC = ½ ∠AOB = ½ (60°) = 30°. Step IV : Draw OD, the bisector of ∠AOC such that ∠AOD = ½ ∠AOC = ½ (30°) = 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:**

(i) Steps of Construction:

Step
I : Draw a ray OA.

Step II : With O as centre and having a suitable
radius, draw an arc which cuts OA 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 OC and OD, which gives ∠COD = 60° = ∠BOC.

Step VI : Draw OP, the bisector of ∠COD, such that ∠COP = ½ ∠COD = ½ (60°) = 30°.

Step VII: Draw OQ, the bisector of ∠COP, such that ∠COQ = ½ ∠COP = ½ (30°) = 15°.

Thus,
∠BOQ = 60° + 15° = 75°or ∠AOQ = 75°

**(ii) Steps of Construction:****
**Step I : Draw a ray OA.

Step II : With centre O and having a suitable radius, draw an arc which cuts OA 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 drawn in step II.

Step V : Draw OP, the bisector of CD which cuts CD at E such that ∠BOP = 90°.

Step VI : Draw OQ, the bisector of ∠EOD such that ∠POQ = 15°

**(iii)
Steps of Construction:**

Step I : Draw a ray OP.

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

Step III : Keeping the same radius and starting from A, mark points Q, R and S
on the arc of step II such that AQ = QR = RS.

Step IV : Draw OL, the bisector
of RS which cuts the arc RS at T.

Step V : Draw OM, the bisector
of RT.

Thus, ∠POM = 135°

**Ex
11.1 Class 9 Maths Question 5.
**Construct an equilateral triangle, given its side
and justify the construction.

**Solution:
**Let us construct an equilateral triangle, each of
whose side = 3 cm (say).

**Steps of Construction:**

Step I : Draw a ray OA.

Step II : Taking O as centre and radius equal to 3 cm, draw an arc to cut OA at B such that OB = 3 cm.

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.

**Justification:****
**∵ The arcs OC and BC 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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