NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.1.

NCERT Solutions for Class 9 Maths 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:

Step of Construction:
Step I : Draw AB¯¯¯¯¯¯¯¯.
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, divide the semicircle into three equal parts such that BC˘=CD˘=DE˘.
Step IV : Draw OC¯¯¯¯¯¯¯¯ and OD¯¯¯¯¯¯¯¯.
Step V : Draw OF¯¯¯¯¯¯¯¯, the bisector of ∠COD.

Thus, ∠AOF = 90°
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 = 12 ∠COD = 12 (60°) = 30°
Now, ∠BOC + ∠COF = 60° + 30° ⇒ ∠BOF = 90° or ∠AOF = 90°

Also See: NCERT Solutions for Class 9 Maths Chapter 14 Statistics

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:
Stept I : Draw 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.

Thus, ∠BOG = 45° or ∠AOG = 45°
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.
∴ ∠COF = 12 ∠BOC = 12(60°) = 30° …(1)
Also, OG¯¯¯¯¯¯¯¯ is the bisector of ∠COF.
∠FOG = 12∠COF = 12(30°) = 15° …(2)
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 12∘
(iii) 15°

Solution:
(i) Angle of 30°
Steps of Construction:
Step I : Draw 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 = 12∠BOC = 12(60°) = 30°

Thus, ∠BOD = 30° or ∠AOD = 30°

(ii) Angle of 22 12∘
Steps of Construction:
Step I : Draw OA¯¯¯¯¯¯¯¯.
Step II : Construct ∠AOB = 90°
Step III : Draw OC¯¯¯¯¯¯¯¯, the bisector of ∠AOB, such that
∠AOC = 12∠AOB = 12(90°) = 45°
Step IV : Now, draw OD, the bisector of ∠AOC, such that
∠AOD = 12∠AOC = 12(45°) = 22 12∘

Thus, ∠AOD = 22 12∘

(iii) Angle of 15°
Steps of Construction:
Step I : Draw OA¯¯¯¯¯¯¯¯.
Step II : Construct ∠AOB = 60°.
Step III : Draw OC, the bisector of ∠AOB, such that
∠AOC = 12∠AOB = 12(60°) = 30°
i.e., ∠AOC = 30°
Step IV : Draw OD, the bisector of ∠AOC such that
∠AOD = 12∠AOC = 12(30°) = 15°

Thus, ∠AOD = 15°

Also See: NCERT Solutions for Class 9 Maths Chapter 15 Probability

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 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 = 12∠COD = 12(60°) = 30°.

Step VII: Draw OQ¯¯¯¯¯¯¯¯, the bisector of ∠COP, such that
∠COQ = 12∠COP = 12(30°) = 15°.

Thus, ∠BOQ = 60° + 15° = 75°∠AOQ = 75° (ii) Steps of Construction:
Step I : Draw 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 BC˘ such that ∠POQ = 15°
Thus, ∠AOQ = 90° + 15° = 105°

(iii) Steps of Construction:
Step I : Draw 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˘ .
StepIV :Draw OL¯¯¯¯¯¯¯, the bisector of RS˘ which cuts the arc RS˘ at T.
Step V : Draw OM¯¯¯¯¯¯¯¯¯, the bisector of RT˘.

Thus, ∠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 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.

Thus, ∆OBC is the required equilateral triangle.

Justification:
∵ The arcs OC˘ and BC˘ are drawn with the same radius.
∴ OC˘ = BC˘
⇒ OC = BC [Chords corresponding to equal arcs are equal]
∵ OC = OB = BC
∴ OBC is an equilateral triangle.

Step IV : Join OC and BC.

Thus, ∆OBC is the required equilateral triangle.

Justification:
∵ The arcs OC˘ and BC˘ are drawn with the same radius.
∴ OC˘ = BC˘
⇒ OC = BC [Chords corresponding to equal arcs are equal]
∵ OC = OB = BC
∴ OBC is an equilateral triangle.

Step VI : Join AC.

Thus, ∆ABC is the required triangle..