NCERT Solutions for Class 9 Maths Chapter 11 - Constructions

NCERT Solutions for Class 9 Maths Chapter 11, Construction, has two exercises that cover the following topics:

Exercise 11.1:

This exercise consists of five questions based on constructing different geometrical figures, such as perpendicular bisectors, angles, triangles, and parallelograms, using a straight edge (ruler) and a compass. Some of the topics covered in this exercise are:

• Construction of the bisector of a given line segment

• Construction of an angle bisector

• Construction of a perpendicular bisector of a line segment

• Construction of a perpendicular to a given line at a given point

• Construction of an angle of 60 degrees and 120 degrees

• Construction of a triangle given its three sides

• Construction of a parallelogram given its adjacent sides and an angle

• Verification of the properties of a parallelogram

In this exercise, students will learn how to use a compass and a straight edge to construct different geometrical figures.

Exercise 11.2:

This exercise consists of five questions that ask students to apply their knowledge of geometrical constructions to solve real-life problems. Some of the topics covered in this exercise are:

• Constructing a quadrilateral with given measurements of four sides and a diagonal

• Constructing a triangle when the perimeter and base are given

• Constructing a triangle when two sides and the angle between them are given

• Constructing a regular hexagon inscribed in a circle

• Constructing a triangle given the sum of its two sides and a median

• Constructing a triangle given the difference between its two sides and the median to the third side

In this exercise, students will learn how to use geometrical constructions to solve real-life problems that involve different shapes and measurements.

These exercises in NCERT Solutions for Class 9 Maths Chapter 11, Construction, are designed to help students understand the principles of geometrical constructions and how they can be applied to solve various problems. These exercises also help in developing their analytical and problem-solving skills.

NCERT Solutions for Class 9 Maths Chapter 11 Construction - PDF Download

1.Construct an angle of $\mathbf{90}{}^\circ $ at the initial point of a given ray and justify the construction.

Ans: The steps given below will be followed to construct an angle of\[90{}^\circ \].

(i) Take the given ray $PQ$. Draw an arc of some radius by taking point $P$ as its centre, which intersects $PQ$ at $R$.

(ii) Take $R$ as centre & with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.

(iii) Now take $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$(see figure).

(iv) Take $S$ and $T$ as centre, draw an arc of same radius to intersect each other at $U$.

(v) Join $PU$, which is the required ray making \[90{}^\circ \] with the given ray $PQ$.

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Justification

We can justify the construction, if we can prove \[\angle UPQ=90{}^\circ .\] For this, join \[~PS\] and \[PT.\]

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We have, $\angle SPQ=\angle TPS=60{}^\circ .$In (iii) and (iv) steps of this construction, $PU$was drawn as the bisector of $\angle TPS.$

$\angle UPS=\frac{1}{2}\angle TPS $

 $ =\frac{1}{2}\times {{60}^{o}} $

 $ ={{30}^{o}}  $

Also, 

$ \angle UPQ=\angle SPQ+\angle UPS $

 $ ={{60}^{o}}+{{30}^{o}} $

 $ ={{90}^{o}}$  

2. Construct an angle of $\mathbf{45}{}^\circ $at the initial point of a given ray and justify the construction.

Ans: The steps given below will be followed to construct an angle of $45{}^\circ $.

1) Take the given ray $PQ$. Draw an arc of some radius taking point $P$as its centre, which intersects $PQ$ at $R$.

2) Take $R$ as centre & with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.

3) Take $S$ as centre & with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).

4) Take $S$ and $T$ as centre, draw an arc of the same radius to intersect each other at $U$.

5) Join $PU$. Let it intersect the arc at point $V$.

6) From $R$ and $V$, draw arcs with radius more than $\frac{1}{2}RV$to intersect each other at $W$. Join $PW$. $PW$ is the required ray making $45{}^\circ $with $PQ.$

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Justification of Construction:

We can justify the construction if we can prove ∠WPQ = 45°. For this, join PS and PT.

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We have,$\angle SPQ=\angle TPS=60{}^\circ $. In (iii) and (iv) steps of this construction, $PU$was drawn as the bisector of \[\angle TPS.\]

$ \angle UPS=\frac{1}{2}\angle TPS $

$ =\frac{1}{2}\times {{60}^{o}} $

$ ={{30}^{o}}  $

Also,

$\angle UPQ=\angle SPQ+\angle UPS $

$ ={{60}^{o}}+{{30}^{o}} $

$ ={{90}^{o}} $

\[\]In step (vi) of this construction, PW was constructed as the bisector of ∠UPQ.

$ \angle WPQ=\frac{1}{2}\angle UPQ $

$=\frac{1}{2}\times {{90}^{o}}$

$={{45}^{o}}  $

3. Construct the angles of the following measurements:

(i) $30{}^\circ $  

Ans: The steps given below will be followed to construct an angle of $30{}^\circ $.

(1) Draw the given ray $PQ$. Taking $P$ as centre and with some radius, draw an arc of a circle that intersects $PQ$ at $R$.

(2) Take $R$ as centre and with the same radius as before & draw an arc intersecting the previously drawn arc at point $S$.

(3) Take $R$ and $S$ as centre and with radius more than $\frac{1}{2}RS$, draw arcs to intersect each other at $T$. Join $PT$which is the required ray making $30{}^\circ $ with the given ray $PQ$.

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(ii) ${{22}^{o}}\frac{1}{2}$

Ans: The steps given below will be followed to construct an angle of ${{22}^{o}}\frac{1}{2}$.

Take the given ray $PQ$. Draw an arc of some radius, taking point $P$ as its centre, which intersects $PQ$ at $R$.

Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.

Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).

Taking $S$ and $T$ as centre, draw an arc of same radius to intersect each other at $U$.

Join$PU$. Let it intersect the arc at point $V$.

From $R$ and $V$, draw arcs with radius more than $\frac{1}{2}RV$ to intersect each other at $W$. Join $PW$.

Let it intersect the arc at $X$. Taking $X$ and $R$as centre and radius more than half $RX$, draw arcs to intersect each other at $Y$.

Join $PY$ which is the required ray making ${{22}^{o}}\frac{1}{2}$ with the given ray $PQ$.

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(iii) $15{}^\circ $

Ans: The below given steps will be followed to construct an angle of $15{}^\circ $.

(1) Draw the given ray $PQ$. Taking $P$ as centre and with some radius, draw an arc of a circle which intersects $PQ$ at $R$.

(2) Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point $S$.

(3) Taking $R$ and $S$ as centre and with radius more than $\frac{1}{2}RS$, draw arcs to intersect each other at $T$. Join $PT$.

(4) Let it intersect the arc at $U$. Taking $U$ and $R$ as centre and with radius more than $\frac{1}{2}RU$, draw an arc to intersect each other at $V$. Join $PV$ which is the required ray making $15{}^\circ $ with the given ray $PQ$.

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4. Construct the following angles and verify by measuring them by a protractor: 

(i) $75{}^\circ $

Ans: The below given steps will be followed to construct an angle of $75{}^\circ $.

1) Take the given ray $PQ$. Draw an arc of some radius taking point $P$ as its centre, which intersects $PQ$ at $R$.

2) Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.

3) Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).

4) Taking $S$ and $T$ as centre, draw an arc of same radius to intersect each other at $U$.

5) Join $PU$. Let it intersect the arc at $V$. Taking $S$ and $V$ as centre, draw arcs with radius more than $\frac{1}{2}SV$. Let those intersect each other at $W$. Join $PW$ which is the required ray making $75{}^\circ $ with the given ray $PQ$ .

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The angle so formed can be measured with the help of a protractor. It comes to be $75{}^\circ $.

(ii) $105{}^\circ $

Ans: The below given steps will be followed to construct an angle of $105{}^\circ $

1) Take the given ray $PQ$. Draw an arc of some radius taking point $P$ as its centre, which intersects $PQ$ at $R$.

2) Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S$.

3) Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).

4) Taking $S$ and $T$as centre, draw an arc of same radius to intersect each other at U.

5) Join $PU$. Let it intersect the arc at $V$. Taking $T$ and $V$ as centre, draw arcs with radius more than $\frac{1}{2}TV$. Let these arcs intersect each other at W. Join $PW$ which is the required ray making $105{}^\circ $ with the given ray $PQ$.

(Image will be uploaded soon)

The angle so formed can be measured with the help of a protractor. It comes to be $105{}^\circ $

 (iii) $135{}^\circ $

Ans: The steps given below will be followed to construct an angle of $135{}^\circ $.

1) Take the given ray $PQ$. Extend $PQ$ on the opposite side of $Q$. Draw a semi-circle of some radius taking point P as its centre, which intersects $PQ$ at $R$and$W$.

2) Taking $R$ as centre and with the same radius as before, draw an arc intersecting the previously drawn arc at $S$. 

3) Taking $S$ as centre and with the same radius as before, draw an arc intersecting the arc at $T$ (see figure).

4) Taking $S$ and $T$ as centre, draw an arc of same radius to intersect each other at U. 

5) Join $PU$. Let it intersect the arc at $V$. Taking $V$and $W$as centre and with radius more than $\frac{1}{2}TV$, draw arcs to intersect each other at $X$. Join $PX,$which is the required ray making $135{}^\circ $ with the given line $PQ$.

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The angle so formed can be measured with the help of a protractor. It comes to be $135{}^\circ $.

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

Ans: Let us draw an equilateral triangle of side $5cm$. We know that all sides of an equilateral triangle are equal. Therefore, all sides of the equilateral triangle will be $5cm$. We also know that each angle of an equilateral triangle is$60{}^\text{o}$. The steps given below will be followed to draw an equilateral triangle of $5cm$ side. 

Step I: Draw a line segment $AB$of $5cm$ length. Draw an arc of some radius, while taking $A$ as its centre. Let it intersect $AB$ at $P$. 

Step II: Take $P$ as centre & draw an arc intersecting the previous arc at $E$. Join $AE$. 

Step III: Take $A$ as centre & draw an arc of $5cm$ radius, which intersects extended line segment $AE$ at $C$. Join $AC$ and $BC$. $\Delta ABC$is the required equilateral triangle of side $5cm$. 

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Justification of Construction: 

We can justify the construction by showing $ABC$as an equilateral triangle i.e., 

$AB\text{ }=\text{ }BC\text{ }=\text{ }AC\text{ }=\text{ }5cm$

and $\angle A\text{ }=\angle B\text{ }=\angle C\text{ }=\text{ }60{}^\circ .$

In $\Delta ABC$, we have $AB\text{ }=\text{ }BC\text{ }=\text{ }AC\text{ }=\text{ }5cm$and $\angle A\text{ }=\text{ }60{}^\circ $. 

Since $AC\text{ }=\text{ }AB,\angle B\text{ }=\angle C$(Angles opposite to equal sides of a triangle) 

In $\Delta ABC$,

 $\angle A\text{ }+\angle B\text{ }+\angle C\text{ }=\text{ }180{}^\circ $ (Angle sum property of a triangle)

$~\Rightarrow 60{}^\circ \text{ }+\angle C\text{ }+\angle C\text{ }=\text{ }180{}^\circ $

$\Rightarrow 60{}^\circ \text{ }+\text{ }2\angle C\text{ }=\text{ }180{}^\circ $

$\Rightarrow 2\angle C\text{ }=\text{ }180{}^\circ \text{ }-\text{ }60{}^\circ \text{ }=\text{ }120{}^\circ $

$\Rightarrow \angle C\text{ }=\text{ }60{}^\circ $

$\Rightarrow \angle B\text{ }=\angle C\text{ }=\text{ }60{}^\circ $

We have,$\angle A\text{ }=\angle B\text{ }=\angle C\text{ }=\text{ }60{}^\circ $... (1) 

$\angle A\text{ }=\angle B$and $\angle A\text{ }=\angle C$

$BC\text{ }=\text{ }AC$and $BC\text{ }=\text{ }AB$(Sides opposite to equal angles of a triangle) 

$AB\text{ }=\text{ }BC\text{ }=\text{ }AC\text{ }=\text{ }5\text{ }cm$... (2) 

From Equations (1) and (2), $\Delta ABC$is an equilateral triangle.

Exercise (11.2)

1. Construct a triangle $\mathbf{ABC}$ in which $\mathbf{BC}\text{ }=\text{ }\mathbf{7}\text{ }\mathbf{cm},\angle \mathbf{B}\text{ }=\text{ }\mathbf{75}{}^\circ $and $\mathbf{AB}\text{ }+\text{ }\mathbf{AC}\text{ }=\text{ }\mathbf{13}\text{ }\mathbf{cm}.$

Ans: The steps given below will be followed to construct the required triangle. 

Step I: Draw a line segment $BC$of$7\text{ }cm$. At point $B$, draw an angle of$75{}^\circ $, say$\angle XBC$.

Step II: Cut a line segment $BD\text{ }=\text{ }13\text{ }cm$(that is equal to$AB\text{ }+\text{ }AC$) from the ray$BX$.

Step III: Join $DC$and make an angle $DCY$equal to $\angle BDC$.

Step IV: Let \[CY\]intersect \[BX\]at \[A\]. $\Delta ABC$is the required triangle.

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2. Construct a triangle \[\mathbf{ABC}\] in which $\mathbf{BC}\text{ }=\text{ }\mathbf{8}\text{ }\mathbf{cm},\angle \mathbf{B}\text{ }=\text{ }\mathbf{45}{}^\circ $ and \[\mathbf{AB}\text{ }-\text{ }\mathbf{AC}\text{ }=\text{ }\mathbf{3}.\mathbf{5}\text{ }\mathbf{cm}\]. 

Ans: The steps given below will be followed to draw the required triangle. 

Step I: Draw the line segment $BC\text{ }=\text{ }8\text{ }cm$and at point $B$, make an angle of$45{}^\circ $, say$\angle XBC$. 

Step II: Cut the line segment $BD\text{ }=\text{ }3.5\text{ }cm$(equal to$AB\text{ }-\text{ }AC$) on ray$BX$. 

Step III: Join $DC$and draw the perpendicular bisector $PQ$of$DC$. 

Step IV: Let it intersect $BX$ at point $A$. Join $AC$. $\Delta ABC$is the required triangle.

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3. Construct a triangle $\mathbf{PQR}$ in which $\mathbf{QR}\text{ }=\text{ }\mathbf{6}\text{ }\mathbf{cm},\angle \mathbf{Q}\text{ }=\text{ }\mathbf{60}{}^\circ $and $\mathbf{PR}\text{ }-\text{ }\mathbf{PQ}\text{ }=\text{ }\mathbf{2}\text{ }\mathbf{cm}$

Ans: The steps given below will be followed to construct the required triangle. 

Step I: Draw line segment $QR$ of $6\text{ }cm$. At point $Q$, draw an angle of $60{}^\circ $, say $\angle XQR$. 

Step II: Cut a line segment $QS$ of $2\text{ }cm$from the line segment $QT$extended in the opposite side of line segment$XQ$. (As $PR\text{ }>\text{ }PQ$and$PR\text{ }-\text{ }PQ\text{ }=\text{ }2\text{ }cm$). Join $SR$. 

Step III: Draw perpendicular bisector $AB$of line segment $SR$. Let it intersect $QX$at point $P$. Join$PQ,\text{ }PR$. $\Delta PQR$is the required triangle.

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4. Construct a triangle $\mathbf{XYZ}$ in which $\angle \mathbf{Y}\text{ }=\text{ }\mathbf{30}{}^\circ ,\angle \mathbf{Z}\text{ }=\text{ }\mathbf{90}{}^\circ $and $\mathbf{XY}\text{ }+\text{ }\mathbf{YZ}\text{ }+\text{ }\mathbf{ZX}\text{ }=\text{ }\mathbf{11}\text{ }\mathbf{cm}.$

Ans: The steps given below will be followed to construct the required triangle. 

Step I: Draw a line segment $AB$ of $11cm$. (As $XY\text{ }+\text{ }YZ\text{ }+\text{ }ZX\text{ }=\text{ }11\text{ }cm$) 

Step II: Construct an angle, $\angle PAB$, of $30{}^\circ $ at point A and an angle, $\angle QBA$, of $90{}^\circ $ at point $B$. 

Step III: Bisect $\angle PAB$ and $\angle QBA$. Let these bisectors intersect each other at point $\angle QBA$. 

Step IV: Draw perpendicular bisector $ST$ of $AX$ and $UV$ of $BX$.

Step V: Let $ST$ intersect $AB$ at $Y$ and $UV$ intersect $AB$ at $Z$. Join $XY,\text{ }XZ$. $\Delta XYZ$is the required triangle.

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5. Construct a right triangle whose base is $\mathbf{12}\text{ }\mathbf{cm}$and sum of its hypotenuse and other side is $\mathbf{18}\text{ }\mathbf{cm}.$

Ans: The steps given below will be followed to construct the required triangle. 

Step I: Draw line segment $AB$ of $12cm$. Draw a ray $AX$ making $90{}^\circ $with $AB$

Step II: Cut a line segment $AD$ of $18cm$ (as the sum of the other two sides is $18$) from ray $AB$. 

Step III: Join $DB$ and make an angle $DBY$equal to $ADB$. 

Step IV: Let $BY$ intersect $AX$ at $C$. Join$AC,\text{ }BC$. \[\Delta ABC\]is the required triangle.

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NCERT Solutions for Class 9 Maths Chapter 11 Construction - PDF Download

If you are looking for online study resources for CBSE Class 9 Mathematics, Vedantu can be the best e-platform for you. Solutions of all the lessons in the book have been structured in a concise, easy to understand manner by experts who have years of experience in their field of study. The questions asked in the CBSE Maths Class 9 Textbook are crucial from an exam point of view, and a student should understand the way of solution writing to achieve good grades. The answers to problems are presented in a step by step manner for students to grasp them quickly. 

Class 9 Maths syllabus includes chapters like Number Systems, Algebra, Coordinate Geometry, Geometry, Mensuration, Statistics, and Probability. To answer the textbook's questions, and in the exam, one needs to understand various complex concepts. The explanations to these complex questions are provided in the Vedantu platform in a simple, lucid manner for the students to understand and produce in their own words in the examination. One can download exclusive and best in class study material including Maths Class 9 Chapter 11, Constructions in easy to read pdf format by merely visiting the Vedantu official website, vedantu.com.

You can opt for Chapter 11 - Constructions NCERT Solutions for Class 9 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclid's Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Construction Class 9 NCERT Solutions

Class 9 Mathematics, NCERT book, Chapter 11 is 'Constructions.' In this chapter, students would learn about the methodology of precisely constructing angles and various 2D shapes of the required measurements. Students in the process would learn about the use of Compass, Protractor, Dividers, and Set-Squares. Although the chapter is regarded as reasonably comfortable in the students' point of view, a student should still adequately understand the methodologies and the logic involved in construction to score full marks in the examination. Proper understanding and practice of all the NCERT questions in a stepwise and thorough manner are essential. 

Our subject experts at Vedantu have taken utmost care in preparing the solutions by going through years of answer keys from CBSE. A student who is well versed with these NCERT questions and our solutions can easily score full marks in the CBSE Board exams. Additionally, students would get themselves acquainted with the type and pattern of questions asked in the exams. The NCERT book itself has numerous questions from varying degrees of difficulty that access a student's understanding. Mastery over the NCERT book will help you breeze over your exams.

Things Learnt from NCERT Solutions for Class 9 Maths Chapter 11 Constructions 

• The Construction chapter is mainly framed for students to make them familiar with compass and rulers. 

• Students can learn how to bisect an angle.

• To draw a perpendicular bisector for bisecting a line segment.

• To make angles with certain degrees like 30°, 60°, 90°, etc.

• To design a triangle by providing its perimeter and its two base angles. 

• To construct a triangle with the help of a given base angle and calculate the remaining two angles.

Importance of Solving the NCERT Solutions

The following are some of the important benefits of solving the NCERT Solutions for Class 9 Maths Chapter 11.

• Whether you're studying for a board test or a competitive exam, the NCERT solutions will help students strengthen their fundamentals.

• It aids in the preparation for competitive exams

• The answers are thoroughly discussed, using relevant and easily understandable examples thus helping in developing a precise understanding of the concept. .

• It aids in the conceptualization of all topics by offering adequate instances for practice.

• Because the CBSE uses the NCERT textbooks, it is critical for students to study the textbooks and complete the exercises and examples provided.

• It aids in the development of a thorough comprehension of the concepts.

• It assists students in becoming familiar with patterns and trends that may be asked in exams.

• Students can improve their efficiency in solving problems by developing the habit of performing NCERT exercises.

• It allows pupils to evaluate their grasp of ideas.

• It aids in the growth of critical thinking skills.

This was a complete discussion of the key features, topics, and advantages of NCERT solutions. To summarise the post, students should practise the questions presented here in order to ace upcoming exams.

We Cover All Exercises in The Chapter Given Below

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