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NCERT Solutions for Class 9 Maths Chapter 9 - Circles Exercise 9.3

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NCERT Solutions for Maths Class 9 Chapter 9 Circles Exercise 9.3 - FREE PDF Download

NCERT Class 9 Maths Chapter 9 Exercise 9.3 Solutions covers the fundamentals of circles and their properties. Experts at Vedantu provide you with Maths Exercise 9.3 Solutions to help you get a comprehensive understanding of the chapter and its concepts. The solutions are designed by our Mathematics experts and are developed following the latest CBSE guidelines and syllabus.

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Table of Content
1. NCERT Solutions for Maths Class 9 Chapter 9 Circles Exercise 9.3 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 9 Exercise 9.3 Class 9 | Vedantu
3. Access NCERT Solutions for Maths Class 9 Chapter 9 - Circles
4. Conclusion
5. Class 9 Maths Chapter 9: Exercises Breakdown
6. CBSE Class 9 Maths Chapter 9 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 9 Maths
8. Important Study Materials for CBSE Class 9 Maths
FAQs


The NCERT Solutions for Maths Class 9 allows you to revise the concepts whenever needed and solve the questions without any confusion. You can also download the CBSE Class 9 Maths Syllabus to help you revise the complete syllabus and score more marks in your examinations.


Glance on NCERT Solutions Maths Chapter 9 Exercise 9.3 Class 9 | Vedantu

  • Exercise 9.3 explains that the angle is subtended by an arc of a circle, which is the angle formed at the center or any point on the circle by lines drawn from the arc's endpoints.

  • If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal.

  • Congruent arcs (or equal arcs) of a circle subtend equal angles at the centre.

  • The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

  • Angles in the same segment of a circle are equal

  • If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle they are concyclic.

  • The sum of either pair of opposite angles of a cyclic quadrilateral

  • If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.

  • There are 12 fully solved questions in Chapter 9 Exercise 9.3 Circles.

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NCERT Solutions for Class 9 Maths Chapter 9 - Circles Exercise 9.3
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Access NCERT Solutions for Maths Class 9 Chapter 9 - Circles

Exercise 9.3

1. In the given figure, $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ are three points on a circle with center $\mathbf{O}$ such that $\angle \mathbf{BOC}=\mathbf{3}{{\mathbf{0}}^{\circ }}$ and \[\angle \mathbf{AOB}=\mathbf{6}{{\mathbf{0}}^{\circ }}\]. If $\mathbf{D}$ is a point on the circle other than the arc $\mathbf{ABC}$, find $\angle \mathbf{ADC}$.

circle with triangle


Ans:

We can notice that,

$ \angle AOC=\angle AOB+\angle BOC $ 

$ ={{60}^{\circ }}+{{30}^{\circ }} $ 

$ ={{90}^{\circ }} $ 

Now, recall that a subtended angle at its centre is double the angle it has at any point on the remaining section of the circle.

Thus, 

$ \angle ADC=\dfrac{1}{2}\angle AOC $ 

$ =\dfrac{1}{2}\times {{90}^{\circ }} $ 

$ ={{45}^{\circ }}. $ 

Hence, $\angle ADC={{45}^{\circ }}$.


2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Ans:

A chord of a circle


By the given information, in $\Delta \,OAB$,

$AB=OA=OB$, since each is a radius of the circle.

Therefore, the triangle $\Delta \,OAB$ is equilateral.

So, the value of each of the interior angles of $\Delta \,AOB$ is ${{60}^{\circ }}$.

That is, $\angle AOB={{60}^{\circ }}$.

$ \Rightarrow \angle ACB=\dfrac{1}{2}\angle AOB $ 

$ =\dfrac{1}{2}\times {{60}^{\circ }} $ 

$ ={{30}^{\circ }}. $ 

Now, in the cyclic quadrilateral $ACBD$,

$\angle ACB+\angle ADB={{180}^{\circ }}$, since the sum of the opposite angles in cyclic quadrilateral is ${{180}^{\circ }}$.

$\Rightarrow \angle ADB={{180}^{\circ }}-{{30}^{\circ }}={{150}^{\circ }}$.

Hence, the chord's angle at the points on the minor arc and on the major arc, respectively, are  ${{30}^{o}}$ and ${{150}^{o}}$.


3. In the given figure, $\angle \mathbf{PQR}=\mathbf{10}{{\mathbf{0}}^{o}}$, where $\mathbf{P}$, $\mathbf{Q}$ and $\mathbf{R}$ are points on a circle with center $\mathbf{O}$. Find $\angle \mathbf{OPR}$.

a chord of the circle centered

Ans:

a chord of the circle centered

First, take $PR$as a chord of the circle centered at $O$.

Then, consider any point $S$ on the major arc of the circle.

Therefore, we get the equilateral quadrilateral $PQRS$.

So, $\angle PQR+\angle PSR={{180}^{o}}$, since the sum of the opposite angles of a cyclic quadrilateral is ${{180}^{o}}$.

$\Rightarrow \angle PSR={{180}^{o}}-{{100}^{o}}={{80}^{o}}$.

Now, it is generally known that the angle subtended by an arc at the center is twice the angle subtended by it at any point on the remaining part of the circle.

Thus, $\angle POR=2\angle PSR=2\times {{80}^{o}}={{160}^{o}}$.

Then, in the triangle $\Delta POR$,

$OP=OR$, since each of them are radius of the circle.

$\angle OPR=\angle ORP$, angles opposite to the equal sides of the triangle.

Therefore,

$\angle OPR+\angle ORP+\angle POR={{180}^{o}}$, by the angle sum property of a triangle.

$\Rightarrow 2\angle OPR+{{160}^{o}}={{180}^{o}}$

$\Rightarrow 2\angle OPR={{180}^{o}}-{{160}^{o}}={{20}^{o}}$.

Hence, $\angle OPR={{10}^{o}}$.


4. In fig. $\mathbf{10}.\mathbf{38}$, $\angle \mathbf{ABC}=\mathbf{6}{{\mathbf{9}}^{o}}$, $\angle \mathbf{ACB}=\mathbf{3}{{\mathbf{1}}^{o}}$, find $\angle \mathbf{BDC}$?

the triangle delta abc

Ans:

In the triangle $\Delta \,ABC$,

$\angle BAC+\angle ABC+\angle ACB={{180}^{o}}$, by the angle sum property of a circle.

$\Rightarrow \angle BAC+{{69}^{o}}+{{31}^{o}}={{180}^{o}}$

$ \Rightarrow \angle BAC={{180}^{o}}-{{100}^{o}} $ 

$ \Rightarrow \angle BAC={{80}^{o}} $ 

Now, it is known that angles in the same segment of the circle are equal. So here, $\angle \text{BAC}=\angle \text{BDC}$.

Hence, $\angle BDC={{80}^{o}}$.


5. In the given figure, $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and $\mathbf{D}$ are four points on a circle. $\mathbf{AC}$ and $\mathbf{BD}$ intersect at a point $\mathbf{E}$ such that $\angle \mathbf{BEC}=\mathbf{13}{{\mathbf{0}}^{o}}$ and $\angle \mathbf{ECD}=\mathbf{2}{{\mathbf{0}}^{o}}$. Find$\angle \mathbf{BAC}$.

he exterior angle in circle

Ans:

It is known that, the exterior angle equals the sum of the opposite interior angles.

So, in the triangle $\Delta \,CDE$,

$\angle CDE+\angle DCE=\angle CEB$

Therefore,

$ \angle CDE+{{20}^{o}}={{130}^{o}} $ 

$ \Rightarrow \angle CDE={{110}^{o}}. $ 

Again, since the angles $\angle BAC$ and $\angle CDE$ are in the same segment of the circle, so $\angle BAC=\angle CDE$.

Hence, $\angle BAC={{110}^{o}}$.


6. $\mathbf{ABCD}$ is a cyclic quadrilateral whose diagonals intersect at a point $\mathbf{E}$. If $\angle \mathbf{DBC}=\mathbf{7}{{\mathbf{0}}^{o}}$, $\angle \mathbf{BAC}=\mathbf{3}{{\mathbf{0}}^{o}}$, find $\angle \mathbf{BCD}$. Further, if $\mathbf{AB}=\mathbf{BC}$, find $\angle \mathbf{ECD}$.

Ans:

the same segment of the circle.


Note that, here the angles $\angle CBD$ and $\angle CAD$ are in the same segment of the circle.

So, $\angle CBD=\angle CAD={{70}^{o}}$.

Now, $\angle BAD=\angle BAC+\angle CAD={{30}^{o}}+{{70}^{o}}={{100}^{o}}$.

Also, since sum of the opposite angles in a cyclic quadrilateral is ${{180}^{o}}$,so

$\angle BCD+\angle BAD={{180}^{o}}$

$\Rightarrow \angle BCD+{{100}^{o}}={{180}^{o}}$

$\Rightarrow \angle BCD={{80}^{o}}$.

Again, in the tringle $\Delta \,ABC$,

$AB=BC$,

$\angle BCA=\angle CAB$, since angles opposite to equal sides of a triangle are equal.

Therefore, $\angle BCA={{30}^{o}}$.

Now, $\angle BCD={{80}^{o}}$

$\Rightarrow \angle BCA+\angle ACD={{80}^{o}}$

$\Rightarrow {{30}^{o}}+\angle ACD={{80}^{o}}$

$\Rightarrow \angle ACD={{50}^{o}}$

Hence, $\angle ECD={{50}^{o}}$.


7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Ans:

a cyclic quadrilateral are diameters of the circle

Suppose that $ABCD$ is a cyclic quadrilateral whose diagonals $BD$ and $AC$ intersecting each other at point the point $O$.

Then, $\angle BAD=\dfrac{1}{2}\angle BOD=\dfrac{{{180}^{o}}}{2}={{90}^{o}}$ (The angle at the circumference in a semicircle).

Also, $\angle BCD$ and $\angle BAD$ are opposite angles in the cyclic quadrilateral.

So, $\angle BCD+\angle BAD={{180}^{o}}$

$\Rightarrow \angle BCD={{180}^{o}}-{{90}^{o}}={{90}^{o}}$.

Therefore,

$\angle ADC=\dfrac{1}{2}\angle AOC=\dfrac{1}{2}\times {{180}^{o}}={{90}^{o}}$. (The angle at the circumference in a semicircle).

Also, $\angle ADC+\angle ABC={{180}^{o}}$, since $\angle ADC$, $\angle ABC$ are the opposite angles in the cyclic quadrilateral.

$\Rightarrow {{90}^{o}}+\angle ABC={{180}^{o}}$

$\Rightarrow \angle ABC={{90}^{o}}$.

Hence, all the interior angles of a cyclic quadrilateral is of ${{90}^{o}}$, and so it is a rectangle.


8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Ans:

he non-parallel sides of a trapezium

Let $ABCD$ is a trapezium such that $AB\parallel CD$ and $BC=AD$.

Now, draw perpendicular lines $AM$ and $BN$ on the line $CD$.

Then, in the triangles $\Delta \,AMD$ and $\Delta \,BNC$,

$AD=BC$ (provided)

$\angle AMD=BNC$, by the formation, each angle is ${{90}^{o}}$.

$AM=BN$, since $AB\parallel CD$.

Thus, $\Delta \,AMD\cong \Delta \,BNC$ (By S-A-S congruence rule)

Therefore,

$\angle ADC=\angle BCD$ (Corresponding angle of congruence triangle) ……(i)

Also, $\angle BAD+\angle ADC={{180}^{o}}$ (The angles are on the same side)

$\Rightarrow \angle BAD+\angle BCD={{180}^{o}}$, by the equation (i).

Thus, the sum of the opposite angles is ${{180}^{o}}$.

Hence, $ABCD$ is a cyclic quadrilateral.


9. Two circles intersect at two points $\mathbf{B}$ and $\mathbf{C}$. Through $\mathbf{B}$, two-line segments $\mathbf{ABD}$ and $\mathbf{PBQ}$ are drawn to intersect the circles at $\mathbf{A}$, $\mathbf{D}$ and $\mathbf{P}$, $\mathbf{Q}$ respectively (see the given figure). Prove that $\angle \mathbf{ACP}=\angle \mathbf{QCD}$.


Two circles intersect at two points

Ans:


The angles are on the chord AP


First, join the chords $AP$, $DQ$.

Then, $\angle PBA=\angle ACP$, (The angles are on the chord $AP$) …… (i)

and $\angle DBQ=\angle QCD$ (The angles are on the chord $DQ$)   …… (ii)

Now, the line segments $ABD$ and $PBQ$ intersects at the point $B$.

So, $\angle PBA=\angle DBQ$                                                            …… (iii)

(Since, $\angle PBA$, $\angle DBQ$ are vertically opposite angles)

Then the equations (i), (ii), and (iii) yields

$\angle ACP=\angle QCD$.


10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Ans:

circles are drawn taking two sides of a triangle as diameters

Let $\Delta \,ABC$ be a triangle.

By the given information, two circles intersect each other in such a way that $AB$ and $AC$ are diameters. 

If possible, let the circles intersect at the point $D$, that does not lie on the line $BC$.

Now, join the line $AD$.

Then, $\angle ADB={{90}^{o}}$, since it is an angle subtended by semicircle.

Similarly, $\angle ADC={{90}^{o}}$.

Therefore, $\angle BDC=\angle ADB+\angle ADC={{90}^{o}}+{{90}^{o}}={{180}^{o}}$.

Thus, $BDC$ is a straight line and so, it is a contradiction.

Hence, the point of intersection of the circles lies on the third side $BC$ of the triangle $\Delta \,ABC$.

the point of intersection of the circles lies on the third side


11. $\mathbf{ABC}$ and $\mathbf{ADC}$ are two right triangles with common hypotenuse $\mathbf{AC}$. Prove that $\angle \mathbf{CAD=}\angle \mathbf{CBD}$.

Ans:

two right triangles with common hypotenuse

In triangle $\Delta \,ABC$, we have

$\angle ABC+\angle BCA+\angle CAB={{180}^{o}}$, by the angle sum property.

$\Rightarrow {{90}^{o}}+\angle BCA+\angle CAB={{180}^{o}}$

$\Rightarrow \angle BCA+\angle CAB={{90}^{o}}$                                                     …… (i)

Again, in the triangle $\Delta \,ADC$, we obtain

$\angle CDA+\angle ACD+\angle DAC={{180}^{o}}$, by the angle sum property.

$\Rightarrow {{90}^{o}}+\angle ACD+\angle DAC={{180}^{o}}$

$\Rightarrow \angle ACD+\angle DAC={{90}^{o}}$                                                …… (ii)

Now, adding (i) and (ii), yields, 

$\angle BCA+\angle CAB+\angle ACD+\angle DAC={{180}^{o}}$

$\Rightarrow \left( \angle BCA+\angle ACD \right)+\left( \angle CAB+\angle DAC \right)={{180}^{o}}$

$\Rightarrow \angle BCD+\angle DAB={{180}^{o}}$                                               …… (iii)

Also, we are provided that,

$\angle B+\angle D={{90}^{o}}+{{90}^{o}}={{180}^{o}}$                                            …… (iv)

By observing the equations (iii) and (iv), it can be concluded that since the sum of the measures of opposite angles of quadrilateral $ABCD$ is ${{180}^{o}}$, so it must be a cyclic quadrilateral.

Thus, $\angle CAD=\angle CBD$, since both the angles are on the chord $CD$.

the angles are on the chord

Hence, the required result is proved.


12. Prove that a cyclic parallelogram is a rectangle.


Ans:

a cyclic parallelogram is a rectangle

Consider that cyclic parallelogram $ABCD$.

Then, $\angle A+\angle C={{180}^{o}}$, (Opposite angles of the cyclic quadrilateral) …… (i)

Again, in a parallelogram, opposite angles are the same.

Therefore, $\angle A=\angle C$ and $\angle B=\angle D$.

So, the equation (i) can be written as

$\angle A+\angle C={{180}^{o}}$

$\Rightarrow \angle A+\angle A={{180}^{o}}$

$\Rightarrow 2\angle A={{180}^{o}}$

$\Rightarrow \angle A={{90}^{o}}$.

Thus, one interior angle of the parallelogram $ABCD$ is of ${{90}^{o}}$. That means, it is a rectangle.

Hence, it is proved that a cyclic parallelogram is a rectangle.


Conclusion

NCERT Class 9 Maths Chapter 9 Exercise 9.3 Solutions focuses on the concepts of angles subtended by an arc and properties of cyclic quadrilaterals. It is important to understand how angles are formed and related within circles and cyclic quadrilaterals. Pay attention to the key formulas and theorems used to solve these problems. Practicing these exercises will help strengthen your understanding of these geometric properties and improve your problem-solving skills. Vedantu provides detailed solutions to help you grasp these concepts effectively.


Class 9 Maths Chapter 9: Exercises Breakdown

Exercise

Number of Questions

Exercise 9.1

2 Questions & Solutions (2 Short Answers)

Exercise 9.2

6 Questions & Solutions (6 Long Answers)



CBSE Class 9 Maths Chapter 9 Other Study Materials


Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Study Materials for CBSE Class 9 Maths

FAQs on NCERT Solutions for Class 9 Maths Chapter 9 - Circles Exercise 9.3

1. What are the theorems about angles in a circle in NCERT Class 9 Maths Chapter 9 Exercise 9.3 Solutions?

  • Theorem 1: The angle subtended by an arc at the centre is twice the angle subtended at any other remaining part of the circle. 

  • Theorem 2: Angles lying in the same segment of a circle are equal.

2. What is a cyclic quadrilateral in NCERT Class 9 Maths Chapter 9 Exercise 9.3 Solutions?

A cyclic quadrilateral is defined as a quadrilateral all of whose vertices lie on the same circle.

3. Give the theorems about cyclic quadrilaterals in Class 9 Maths Chapter 9 Circles Exercise 9.3.

  • Theorem 1: The sum of the opposite angles in a quadrilateral is 180°.

  • Theorem 2 (Converse): A quadrilateral is said to be cyclic if the sum of opposite angles in a quadrilateral is 180°.

4. Where can I get easy solutions to the sums in Class 9 Maths Chapter 9 Circles Exercise 9.3?

Refer to the NCERT Solutions of Class 9 Maths Chapter 9 Circles Exercise 9.3 for easy solutions to the sums covered in the exercise. Download the solutions PDF for free from the Vedantu mobile app or website. Here, you can get the easiest step-wise explanation for each sum given in Exercise 9.3. The subject matter experts at Vedantu have prepared these solutions PDF to make your learning experience easy. 

5. How many sums are there in Class 9 Maths Chapter 9 Circles Exercise 9.3?

There are 12 sums in Exercise 9.3 of Class 9 Chapter 9. A few sums have sub-parts and the sums vary from one another. Every sum needs you to devise the theorems and concepts of circles. The concepts of cyclic quadrilaterals and theorems are covered in this exercise of the chapter on Circles. The sums are application-based and one sum may require more than one theorem or concept to be applied.

6. Are the NCERT Solutions for Class 9th Maths Chapter 9 Exercise 9.3 available for free?

You can download the NCERT Solutions of Class 9th Maths Chapter 9 Exercise 9.3 from the Vedantu website as well as a mobile app for absolutely free of cost. You can refer to these study resources online as well as offline by signing up on Vedantu. There is no doubt that Vedantu is the best learning platform for students. Apart from providing quality study material, it is helping students to gain knowledge for free.

7. Is Class 9th Maths Chapter 9 Exercise 9.3 important for exams?

Each exercise is important from the viewpoint of the examination. Try to practice each exercise thoroughly. All the sums given in the examples and exercises are to be solved and practised regularly to develop a good grasp of the concepts. So, Exercise 9.3 is important for the Class 9 Maths examination. Every question from this exercise is of utmost importance. You can refer to the NCERT Solutions of Class 9th Maths Chapter 9 Exercise 9.3 on the Vedantu website in case of any doubt.

8. What are the important topics in Class 9th Maths Chapter 9 Exercise 9.3?

To practice the exercise, it is important to analyze the important topics covered in it. A few of the important topics covered in Class 9th Maths NCERT Solutions Chapter 9 Exercise 9.3 are given below.      

  • Introduction to Circles

  • Angles that are subtended by an arc of a circle

  • Angles that are subtended by a chord of a circle

  • Cyclic quadrilaterals

  • Theorems used in the chapter.

9. What topics are covered in Class 9th Maths NCERT Solutions Chapter 9 Exercise 9.3?

Class 9th Maths NCERT Solutions Chapter 9 Exercise 9.3 covers the angle subtended by an arc of a circle and the properties of cyclic quadrilaterals. These topics include understanding how angles are formed within circles and the relationship between angles in cyclic quadrilaterals.

10. Why is Class 9th Maths Chapter 9 Circles Exercise 9.3 important for students?

Class 9th Maths Chapter 9 Circles Exercise 9.3 is important because it helps students grasp essential geometric concepts related to circles. Understanding these properties is crucial for solving various geometry problems and is frequently tested in exams.

11. How do the NCERT solutions help in solving Class 9th Maths Chapter 9 Circles Exercise 9.3?

The NCERT solutions provide step-by-step explanations for each problem in Exercise 9.3. These detailed solutions help students understand the methods and concepts, making it easier to solve similar problems on their own.

12. Where can I find the NCERT Solutions for Class 9th Maths Chapter 9 Circles Exercise 9.3?

The NCERT Solutions for Exercise 9.3 can be downloaded from Vedantu's website. They are available in PDF format, making it easy for students to access and use them for their studies and exam preparations.

13. What is the circle of theorem 9.3 Class 9?

The circle in Theorem 9.3 refers to the shape where all points on the edge are the same distance from a central point. The theorem explains how a perpendicular line from the centre to a chord will divide the chord into two equal parts.