# NCERT Solutions for Class 9 Mathematics Chapter 11 – Constructions

Mathematics is a subject that is better demonstrated and explained through solved illustrations and concise study notes. While the diagrams aid to prove a theorem, the geometrical construction requires preparing geometrical figures with precision. Basic constructions refer to the different types of geometrical shapes that can be made with the help of a ruler and a pair of compasses. The basic constructions covered in Class 9 Mathematics Chapter 11 – Constructions include construction of angles, construction of triangles,  to bisect a given angle,  to draw  the perpendicular bisector of a given line segment, to construct a triangle given its perimeter and its two base angles , and so on.

Extramarks has developed online study materials that are beneficial for students to study from and also help them revise the chapter and prepare them for their exams. Students can  access  our NCERT Solutions for Class 9 Mathematics Chapter 11 to learn more about the chapter-Constructions.  .

Students can properly study for CBSE examinations through the best study materials available on the Extramarks official website. Moreover, the website provides them with past years’ question papers and revision notes on the NCERT-based questions related to Constructions.

Students can regularly visit the Extramarks website for the latest updates and notifications about the CBSE examinations. They can also access notes and answer solutions for other classes, including NCERT Solutions Class 5, NCERT Solutions Class 6, NCERT Solutions Class 8, NCERT Solutions Class 12, etc. in case they want to refer to solutions of prior classes.

## Key Topics Discussed Within The NCERT Solution for Class 9 Mathematics Chapter 11 – Constructions

Extramarks NCERT Solutions for Class 9 Mathematics Chapter 11 – Constructions analyses every idea and concept as per the NCERT course syllabus. The vital concepts and fundamental definitions that will be explained in this section are related to construction of angles, to construct a triangle, Alternative methods,  construction of a triangle given its perimeter and base angles, and  to draw a perpendicular bisector of a given line segment, etc. A significant portion of the topics covered in this section is also an important part of the syllabus assigned for major entrance examinations such as WB-JEE, NEET, etc. Students will definitely get more practice and superior knowledge of the subject to solve any tricky question and get an edge over their peers.

The list of key topics covered in the Class 9 Mathematics NCERT Solutions Chapter 11 – Constructions include –

 Exercise Topic 11.1 Introduction 11.2 Basic Construction 11.3 Some Constructions of  Triangles Others Alternative methods, Examples and Exercise-11.1 & 11.2

11.1 Introduction

Scope  and Meaning of the  Linear Pair Axiom –

By the Linear Pair Axiom, we must infer that if two adjacent angles are termed as being a linear pair to each other, then the uncommon rays subtending these two adjacent angles must form a straight line.

In this figure, you can understand the meaning of supplementary angles, whose sum is always 180 degrees. The second portion of the illustration explains the application of the Linear Pair Axiom. The same has been elaborated and explained further in our NCERT Solutions for Class 9 Mathematics Chapter 11.

Scope and  Meaning of  Angle Bisector –

An angle bisector can be understood as a straight line which divides a given angle into two halves. It can be constructed with the help of a compass and a ruler after measuring the value at  the initial angle.

In this figure, if we assume that the straight line dividing the acute angle BON is an angle bisector, then the value of x based upon the given  example will be ‘4’. The two angles must be equal in their numerical values to prove that the given straight line is an angle bisector.

Students will find many solved examples and illustrations to improve their skills in our study notes for NCERT Solutions for Class 9 Mathematics Chapter 11. Students are recommended to  register on our website to excel in their CBSE examinations.

9.2 Basic Constructions

Construction of an  Angle Bisector

In Mathematics, the angle bisector hypothesis is concerned with the lengths of the two line segments that form a triangle’s side when divided into two halves by an angle bisector. You can find this illustration in Figure C.

The  line segment BF is the straight line which bisects the angle ABC into two halves at the point F.

Construction of a  Perpendicular Bisector

The line segment CD is the perpendicular bisector to the angle AMB, as it cuts through at 90 degrees to the line segment AB.

9.3 How to  Construct a  Triangle

Below we have discussed a few important examples of construction, such as the construction of angles, triangles, etc. We highly recommend students also register on reliable online learning platforms such as Extramarks which strictly follows NCERT books and provides solved exercises and practice questions to step up their learning experience and eliminate “mathematics phobia” among students.

How to  Construct an  Angle –

In this figure, one can understand the construction of an acute angle COB and the construction of an obtuse angle COA at the point O.   Students can use a compass and a ruler. They need to follow  steps of construction  to construct certain kinds  of angles. .

In this figure, the lengths of one of the sides of the triangle ABC has been used to construct the given mathematical shape. You can use a ruler and compass to construct the two acute angles intersecting at point C of the triangle.

More details about the chapter are covered in our NCERT Solutions for Class 9 Mathematics Chapter 11. These NCERT Solutions study notes give complete information about the ideas and concepts connected with the geometrical construction of angles, to construct a triangle, to bisect an angle, alternative methods  , etc. It  encourages the students to learn with a step-by-step approach and to master the basics of  construction, to build a strong foundation helping them to complete   their school assignments, prepare for term tests  as well as entrance examinations.

### NCERT Solutions for Class 9 Mathematics Chapter 11- Constructions: Exercise &  Solutions

Mathematics is a graded subject and may not be easy for every student to crack the assignments or class tests.  Therefore, students must refer to  high-quality study notes prepared by subject matter experts and be a smart learner. At Extramarks, all our study materials are prepared by experienced faculty  with decades of experience. Students can increase their problem-solving skills and conceptual clarity by accessing our study materials and question banks for each subject. Extramarks, one of the leading e-learning platforms, is known as the most reliable and trusted  learning website  in India.

The NCERT Solutions for Class 9 Mathematics Chapter 11 – Constructions provides a comprehensive guide and updated course outline to help students prepare for Class 9 Mathematics. The different topics elaborated in our study notes are basic constructions, linear pair axioms, construction of angles, construction of triangles with steps of construction, alternative methods and so on. .

The exercise and solutions covered in our NCERT Solutions for Class 9 Mathematics Chapter 11 can be accessed by clicking on the link given below –

• Chapter 11 –  Exercise and Answer Solutions

There is a repository of  study materials for all primary and secondary classes that can be accessed on the Extramarks official website are:

NCERT Solutions Class 1,

• NCERT Solutions Class 2,
• NCERT Solutions Class 3,
• NCERT Solutions Class 4,
• NCERT Solutions Class 5,
• NCERT Solutions Class 6,
• NCERT Solutions Class 7,
• NCERT Solutions Class 8,
• NCERT Solutions Class 9,
• NCERT Solutions Class 10,
• NCERT Solutions Class 11
• NCERT Solutions Class 12

#### NCERT Exemplar for Class 9 Mathematics

Mathematics as a subject is better taught through a set of detailed notes and solved examples. Students should be clear about the various techniques for constructing angles, triangles, etc., to get good grades in their examinations. .

The NCERT Exemplar booklets comprise a detailed set of questions and answers to assist the students in their revisions. While solving questions from NCERT Exemplar, students  will get a thorough understanding of the Mathematical  concepts and gain deeper insights into various  interconnected  topics and how to solve them step wise within the timeframe.  .

Exemplar books have proved not only beneficial for students in CBSE exams but also for other competitive exams such as JEE, NEET, etc. It covers advanced level questions that prepare students to face the upcoming examinations. Students appearing for Class 9 are advised to refer to the NCERT Exemplar Class 9 Mathematics and use  it in their core study materials.

After referring to the NCERT Solutions and NCERT Exemplar, the students think logically about a problem. As a result, students get better equipped to solve more advanced and higher-level conceptual questions.

#### Key Features of NCERT Solutions for Class 9 Mathematics Chapter 11 – Constructions

The NCERT Solutions for Class 9 Mathematics Chapter 11 – Constructions helps the students to better comprehend the central ideas and fundamental concepts in the chapter on geometrical constructions. Unlike theorems which require proper reasoning, the geometrical construction requires preparing geometrical figures using only two instruments- a graded ruler and a compass.   A compact and outlined set of notes assists them in getting good grades in their school assessments. The key features of Extramarks NCERT Solutions are given below –

• Our experienced faculty and subject experts diligently follow the latest CBSE standards. Needless to say, they  completely understand what is legitimate as per the board’s standards.
• They understand the key topics which students might find challenging while studying, therefore the solutions are prepared in such a way that they can get to the point answers without wasting much time on a single subject.
• These solutions help students understand the concepts and not just blindly memorise them.

Students swear by Extramarks Solutions because of the absolute trust and confidence it has built over the years

• These solutions are written  in the most straightforward and easy language possible so  that students can study from it independently without any assistance from  teacher or parents.
•  Resolve any doubt through f Extramarks NCERT Solutions and enjoy the process of learning conveniently and effortlessly at your  own pace
• Extramarks leaves no stone unturned when it comes to providing the best learning material with unmatchable speed and accuracy for students irrespective of the class and subject.  We have all the answers to your queries. This encourages the students to master the topic and increases their confidence in achieving a high grade

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

Ans.

Given:

A ray OA with initial point O is given.

To construct:

An angle of 90°

Steps of Construction:

(i) Draw a ray OA with initial point O.

(ii) Mark an arc of any radius with centre O.

(iii) Mark two arcs with same radius and centre at the points M1 and M2 respectively.

(iv) Now mark an arc with centre M2 and another arc with centre M3 with any radius. These arcs intersect at N.

(v) Join O and N by a ray OB.

(vi) ∠BOA is required angle of 90°.

Justification:

(i) Since, ∠M2OA = ∠M3OM2 = 60°

(ii) Ray OB is bisector of ∠M3OM2 i.e.

∠BOM2 = (1/2) ∠M3OM2

= (1/2) 60°

= 30°

(iii) ∠BOA = ∠BOM2 + ∠M2OA

= 30° + 60°

= 90°

Hence, it is justified that ∠BOA = 90°.

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

Ans.

Given:

A ray with initial point O is given.

To construct:

An angle of 45°

Steps of Construction:

(i) Draw a ray OA with initial point O.

(ii) Mark an arc of any radius with centre O.

(iii) Mark two arcs with same radius and centre at the points M1 and M2 respectively.

(iv) Now mark an arc with centre M2 and another arc with centre M3 with any radius. These arcs intersect at N.

(v) Join O and N by a ray OB.

(vi) ∠BOA is an angle of 90°.

(vii) Now, mark arcs of any radius with centre M1 and Q respectively, which intersect at P.

(viii) Draw ray OC through P, we get ∠COA which is equal to 45°.

(ix) Thus, ∠COA is required angle.

Justification:

∠COA = (1/2) ∠BOA
= (1/2) 90°
= 45°

Q.3 Construct the angles of the following measurements: (i) 30° (ii) 22.5° (iii) 15°

Ans.

(i)

Given:

A ray with initial point O is given.

To construct:

An angle of 30°

Steps of Construction:

(i) Draw a ray OA with initial point O.
(ii) Mark an arc of any radius with centre O, which intersects ray OA at the point M1.
(iii) Mark another arc with centre M1 and same radius, which intersects earlier arc at M2.
(iv) Now, we draw arcs with any radius and centres M1 and M2 respectively. These arcs intersect at N.

(v) Join O and N with the help of ray OB.

(vi) ∠BOA is required angle of 30°.

(ii)

Given:

A ray with initial point O

To construct:

An angle of 22.5°

Steps of Construction:

(i) Draw a ray OA with initial point O.

(ii) Mark an arc of any radius with centre O.

(iii) Mark two arcs with same radius and centre at the points M1 and M2 respectively.

(iv) Now mark an arc with centre M2 and another arc with centre M3 with any radius. These arcs intersect at N.

(v) Join O and N by a ray OB.

(vi) ∠BOA is an angle of 90°.

(vii) Now, mark arcs of any radius with centre M1 and X respectively, which intersect at P.

(viii) Draw ray OC through P, we get ∠COA which is equal to 45°.

(ix) Now we draw bisector of ∠COA and we get ∠DOA, which is equal to 22.5°.

(x) Thus, ∠DOA is required angle.

(iii)

Given:

A ray with initial point O is given.

To construct:

An angle of 15°.

Steps of Construction:

(i) Draw a ray OA with initial point O.

(ii) Mark an arc of any radius with centre O, which intersects ray OA at the point M1.

(iii) Mark another arc with centre M1 and same radius, which intersects earlier arc at M2.

(iv) Now, we draw arcs with any radius and centres M1 and M2 respectively. These arcs intersect at N.

(v) Join O and N with the help of ray OB.

(vi) ∠BOA is angle of 30°.

(vii) Now, draw bisector OC of ∠BOA.

(viii) ∠COA is required angle of 15°.

Q.4 Construct the following angles and verify by measuring them by a protractor:
(i) 75° (ii) 105° (iii) 135°

Ans.

(i)

Given:

A ray with initial point O is given.

To construct:

An angle of 75°

Steps of Construction:

(i) Draw ray OP and construct angle of 90°.

(ii) Mark arcs with centre P and R respectively with any radius. These arcs intersect at Q.

(iii) Draw ray OC throw point Q.

(iv) ∠COA is required angle.

(ii)

Given:

A ray with initial point O is given.

To construct:

An angle of 105°

Steps of Construction:

(v) Draw ray OP and construct angle of 90°.

(vi) Mark arcs with centre P and Q respectively with any radius. These arcs intersect at R.

(vii) Draw ray OC throw point R.

(viii) ∠COA is required angle.

(iii)
Given:

A ray with initial point O is given.

To construct:

An angle of 135°

Steps of Construction:

(i) Draw ray OP and construct angle of 90°.

(ii) Mark arcs with centre P and Q respectively with any radius. These arcs intersect at R.

(iii) Draw ray OC throw point R.

(iv) ∠COA is required angle.

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

Ans.

Given:

A side of triangle is given.

To construct:

An equilateral triangle ABC

Steps of Construction:

(i) Draw a side of given length.

(ii) Draw two arcs of radius AB with centres A and B respectively. These arcs intersect at C.

(iii) Join AC and BC.

(iv) ΔABC is required triangle.

Q.6 Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.

Ans.

Given:

In triangle ABC, BC = 7 cm, ∠B = 75° and
AB + AC = 13 cm

To construct:

A triangle ABC

Steps of Construction:

(i) Cut a line segment BC = 7cm from a ray BX.

(ii) Draw ray BY at 75° and cut BD = 13 cm from ray BY.

(iii) Join DC and draw perpendicular bisector PQ of DC, which intersect BD at A.

(iv) Join AC.

(v) DABC is required triangle.

Q.7 Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB – AC = 3.5 cm.

Ans.

Given:

In triangle ABC, BC = 8 cm, ∠B = 45° and
AB – AC = 3.5 cm

To construct:

A triangle ABC

Steps of Construction:

(i) Cut a line segment BC = 8 cm from a ray BX.

(ii) Draw ray BY at 45° and cut BD = 3.5 cm from ray BD.

(iii) Join DC and draw perpendicular bisector of DC, which intersect BD at A.

(iv) Join AC.

(v) ΔABC is required triangle.

Q.8 Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.

Ans.

Given: In triangle ABC, BC = 6 cm, ∠B = 60° and PR – PQ = 2 cm

To construct:

A triangle PQR

Steps of Construction:

(i) Cut a line segment QR = 8 cm from a ray QX.

(ii) Draw line l at 45° and cut QS = 2 cm from line l in opposite side of QR.

(iii) Join SR and draw perpendicular bisector AB of SR, which intersect line l at P.

(iv) Join PR.

(v) ΔPQR is required triangle.

Q.9 Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Ans.

Given: In triangle XYZ, ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
To construct:

A triangle XYZ

Steps of Construction:

(i) Draw a line segment AB = 8 cm.

(ii) Draw ray AP at 30° and ray BC at 45°.

(iii) Draw angle bisectors of 30° and 45°, which intersect at X.

(iv) Draw perpendicular bisectors of AX and BX intersect AB at Y and Z points respectively.

(v) Join XY and XZ.

(vi) ΔXYZ is required triangle.

Q.10 Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Ans.

Given:

In triangle ABC, BC = 12 cm, ∠B = 90° and
AB + AC = 18 cm

To construct:

A triangle ABC

Steps of Construction:

(i) Cut a line segment BC = 12 cm from a ray BX.

(ii) Draw ray BY at 90° and cut BD = 18 cm from ray BY.

(iii) Join DC and draw perpendicular bisector of DC, which intersect BD at A.

(iv) Join AC.

(v) ΔABC is required triangle.

## FAQs (Frequently Asked Questions)

### 1. How many questions and illustrations are given in the NCERT Solutions for Class 9 Mathematics Chapter 11?

The NCERT Solutions for Class 9 Mathematics Chapter 11 – Constructions covers 15 solved questions and illustrations under the exercises 11.1, and  11.2.. The different exercises in this chapter are as follows – short answer type, long answer type, and optional type one.

### 2. What do you mean by basic construction?

Basic constructions refer to the different types of geometrical shapes that can be made with the help of a ruler and a pair of compasses. The basic constructions covered in the Class 9 Mathematics include construction of angles, construction of triangles, construction of angle bisector, construction of perpendicular bisector, , and so on.

### 3. How can I become good at Mathematics?

First and foremost step is to clear your fundamental concepts learnt in prior classes before you begin with the new syllabus. Second, follow a strict schedule and study regularly.  Regular practice of the various sets of numerical can help you become better  in Mathematics. You  may  practice and do exercises as well as extra questions in our NCERT Solutions for Class 9 Mathematics Chapter 11 available on the Extramarks website, however your first priority should be NCERT books.