# NCERT Solutions for Class 9 Maths Chapter 11 Constructions￼

The National Council of Educational Research and Training (NCERT) is an autonomous body of the Indian government that formulates the curricula for schools in India that are governed by the Central Board of Secondary Education (CBSE) and certain state boards. Therefore, students who will be taking the Class 10 tests administered by various boards should consult this NCERT Syllabus in order to prepare for those examinations, which in turn will assist those students get a passing score.

When working through the exercises in the NCERT textbook, if you run into any type of difficulty or uncertainty, you may use the swc NCERT Solutions for class 9 as a point of reference. While you are reading the theory form textbook, it is imperative that you always have notes prepared. You should make an effort to understand things from the very beginning so that you may create a solid foundation in the topic. Use the NCERT as your parent book to ensure that you have a strong foundation. After you have finished reading the theoretical section of the textbook, you should go to additional reference books.

## 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:
Steprf 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°

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 ajngle 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°

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¯¯¯¯¯¯¯, thebisector 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.

## Conclusions for NCERT Solutions for Class 9 Maths Chapter 11 Constructions

SWC academic staff has developed NCERT answers for this chapter of the ninth grade mathematics curriculum. We have solutions prepared for all of the exercises of this chapter. The answers, broken down into steps, to all of the questions included in the NCERT textbook’s chapter are provided here. Read this chapter on theory. Be certain that you have read the theory section of this chapter of the NCERT textbook and that you have learnt the formulas for the chapter that you are studying.     