NCERT Solutions Class 9 Maths Chapter 9 Circles Exercise 9.3

Students investigate geometrical construction related to circles and angles in NCERT Solutions Class 9 Maths Chapter 9 Circles Exercise 9.3. This exercise is a continuation of their previous knowledge and understanding of theorems to develop their ability to solve problems involving angles by reasoning and drawing diagrams. Building good skills here will assist in developing analytical skills for subsequent geometry chapters.

1.0Download Class 9 Maths Chapter 9 Ex 9.3 NCERT Solutions PDF

You can download a high-quality and precise set of NCERT Solutions for Class 9 Maths Chapter 9 Circles Exercise 9.3 in PDF format. Each of these solutions is so the student can complete the textbook problems with confidence and study for the examinations with preparation too.

2.0Key Concepts of Exercise 9.3

• Understanding angles inscribed in a circle
• Angles subtended by a chord in a circle and at other locations 
• Angles subtended by an arc or chord, at the center of the circle and other locations  
• Using properties and theorems such as:
• • Angles subtended by the same arc are equal 
  • The angle in a semicircle is a right angle. 
  • Angles of a cyclic quadrilateral 
  • Understanding Angles subtended by Chords

3.0CBSE Class 9 Chapter 9 Linear Equations Exercise 9.3 Comprises

1. Angles Subtended by Chords. 

• Learn how a chord on a circle subtends an angle at the center and at other locations on the circle. 
• Understand and learn that the angle at the center of the circle is twice that of the circumference. 

2. Understanding Equal Angles and Arcs

• Recognise how equal chords subtend equal angles at the center. 
• Recognise and relate equal arcs and equal angles in relation to the center. 

3. Angle in a Semicircle

• Identify and prove why the angle in a semicircle is a right angle. 
• Use this reasoning to logically solve other angle problems. 

4. Cyclic Quadrilaterals 

• Identify quadrilaterals that have all of their vertices on a circle. 
• Use the property that the opposite angles of a cyclic quadrilateral sum to 180°. 

5. Using Diagrams and Theorems 

• Create a diagram based on the given conditions. 
• Use mathematical justifications to solve angles.

4.0NCERT Solutions Class 9 Maths Chapter 9: All Exercises

Here is a quick overview and access to solutions for all exercises from Chapter 9, Circles.

5.0Detailed CBSE Class 9 Chapter 9 Exercise 9.3 Solutions

• In Fig. A, B and C are three points on a circle with centre 0 such that $\angle \mathrm{BOC}=30^{\circ }$ and $\angle \mathrm{AOB}=60^{\circ }$. If D is a point on the circle other than the arc ABC , find $\angle \mathrm{ADC}$.
  
  Sol. $\angle \mathrm{ADC}=\frac{1}{2}\angle \mathrm{AOC}=\frac{1}{2}\left(60^{\circ }+30^{\circ }\right)=45^{\circ }$ [The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle]
• 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. Sol.
  
  $\because \mathrm{OA}=\mathrm{OB}=\mathrm{AB}$ [Given] $\therefore △\mathrm{OAB}$ is equilateral $\therefore \angle \mathrm{AOB}=60^{\circ }$ and $\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}$ $=\frac{1}{2}\times 60=30^{\circ }$ $\because \mathrm{ADBC}$ is a cyclic quadrilateral. $\therefore \angle \mathrm{ADB}+\angle \mathrm{ACB}=180^{\circ }$ [The sum of either pair of opposite angles of a cyclic quadrilateral is $180^{\circ }$ ] $\Rightarrow \angle \mathrm{ADB}+30^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{ADB}=180^{\circ }-30^{\circ }$ $\Rightarrow \angle \mathrm{ADB}=150^{\circ }$
• In figure, $\angle \mathrm{PQR}=100^{\circ }$, where $\mathrm{P},\mathrm{Q}$ and R are points on a circle with centre 0 . Find $\angle \mathrm{OPR}$.
  
  Sol.
  
  Take a point $S$ in the major arc. Join PS and RS. $\because \mathrm{PQRS}$ is a cyclic quadrilateral. $\therefore \angle \mathrm{PQR}+\angle \mathrm{PSR}=180^{\circ }$ [The sum of either pair of opposite angles of a cyclic quadrilateral is $180^{\circ }$ ] $\Rightarrow 100^{\circ }+\angle \mathrm{PSR}=180^{\circ }$ $\Rightarrow \angle \mathrm{PSR}=180^{\circ }-100^{\circ }$ $\Rightarrow \angle \mathrm{PSR}=80^{\circ }$ Now $\angle \mathrm{POR}=2\angle \mathrm{PSR}$ [The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle] $\angle \mathrm{POR}=2\times 80^{\circ }=160^{\circ }$ [Using (1)] In $△\mathrm{OPR}$, $\because \mathrm{OP}=\mathrm{OR}$ [radii of a circle] $\therefore \angle \mathrm{OPR}=\angle \mathrm{ORP}$ [Angles opposite to equal sides of a triangle are same] In $△\mathrm{OPR}$, $\angle \mathrm{OPR}+\angle \mathrm{ORP}+\angle \mathrm{POR}=180^{\circ }$ [Sum of all the angles of a triangle is $180^{\circ }$ ] $\Rightarrow \angle \mathrm{OPR}+\angle \mathrm{OPR}+160^{\circ }=180^{\circ }$ [Using (2) and (3)] $\Rightarrow 2\angle \mathrm{OPR}+160^{\circ }=180^{\circ }$ $\Rightarrow 2\angle \mathrm{OPR}=180^{\circ }-160^{\circ }=20^{\circ }$ $\Rightarrow \angle \mathrm{OPR}=10^{\circ }$
• In fig. $\angle \mathrm{ABC}=69^{\circ },\angle \mathrm{ACB}=31^{\circ }$, find $\angle \mathrm{BDC}$.
  
  Sol. $\angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{BAC}=180^{\circ }$ (By angle sum property) $\Rightarrow 69^{\circ }+31^{\circ }+\angle \mathrm{BAC}=180^{\circ }$ $\Rightarrow \angle \mathrm{BAC}=180^{\circ }-100^{\circ }=80^{\circ }$ Since, angles in the same segment are equal $\begin{array}{rl}\angle \mathrm{BDC}& =\angle \mathrm{BAC}\\ \Rightarrow \angle \mathrm{BDC}& =80^{\circ }.\end{array}$
• In figure, $\mathrm{A},\mathrm{B},\mathrm{C}$ and D are four points on a circle. AC and BD intersect at a point E such that $\angle \mathrm{BEC}=130^{\circ }$ and $\angle \mathrm{ECD}=20^{\circ }$. Find $\angle \mathrm{BAC}$.
  
  Sol. $\angle \mathrm{CED}+\angle \mathrm{BEC}=180^{\circ }$ [Linear Pair] $\begin{array}{rl}\Rightarrow & \angle \mathrm{CED}+130^{\circ }=180^{\circ }\\ \Rightarrow & \angle \mathrm{CED}=180^{\circ }-130^{\circ }=50^{\circ }\\ & \angle \mathrm{ECD}=20^{\circ }\end{array}$ In $△\mathrm{CED}$, $\angle \mathrm{CED}+\angle \mathrm{ECD}+\angle \mathrm{CDE}=180^{\circ }$ [Sum of all the angles of a triangle is $180^{\circ }$ ] $\Rightarrow 50^{\circ }+20^{\circ }+\angle \mathrm{CDE}=180^{\circ }$ [Using (i) and (ii)] $\Rightarrow 70^{\circ }+\angle \mathrm{CDE}=180^{\circ }$ $\Rightarrow \angle \mathrm{CDE}=180^{\circ }-70^{\circ }$ $\Rightarrow \angle \mathrm{CDE}=110^{\circ }$ Now $\angle \mathrm{BAC}=\angle \mathrm{CDE}=110^{\circ }$ [Angle in the same segment of a circle are equal]
• ABCD is a cyclic quadrilateral whose diagonals intersect at a point E . If $\angle \mathrm{DBC}=$ $70^{\circ },\angle \mathrm{BAC}=30^{\circ }$, find $\angle \mathrm{BCD}$. Further, if AB $=\mathrm{BC}$, find $\angle \mathrm{ECD}$. Sol.
  
  Since angles in the same segment of a circle are equal $\therefore \angle \mathrm{BAC}=\angle \mathrm{BDC}$ $\Rightarrow \angle \mathrm{BDC}=30^{\circ }$ Also $\angle \mathrm{DBC}=70^{\circ }$ (Given) $\therefore$ In $△\mathrm{BCD}$, we have $\Rightarrow \angle \mathrm{BCD}+\angle \mathrm{DBC}+\angle \mathrm{CDB}=180^{\circ }$ [sum of angles of a triangle is $180^{\circ }$ ] $\Rightarrow \angle \mathrm{BCD}+70^{\circ }+30^{\circ }=180^{\circ }$ $\Rightarrow \angle \mathrm{BCD}=80^{\circ }$ Now, in $△\mathrm{ABC},\mathrm{AB}=\mathrm{BC}$ (Given) $\therefore \angle \mathrm{BCA}=\angle \mathrm{BAC}$ (angles opp. to equal sides of a triangle are equal) $\Rightarrow \angle \mathrm{BCA}=30^{\circ }$ $\left[\angle \mathrm{BAC}=30^{\circ }\right]$ Now, $\angle \mathrm{BCA}+\angle \mathrm{ECD}=\angle \mathrm{BCD}$ $\Rightarrow 30^{\circ }+\angle \mathrm{ECD}=80^{\circ }$ $\Rightarrow \angle \mathrm{ECD}=50^{\circ }$
• If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle. Sol.
  
  Since, AC and BD are diameters. $\Rightarrow \mathrm{AC}=\mathrm{BD}$ [all diameters of a circle are equal] Also, $\angle \mathrm{BAD}=90^{\circ }$ [angle formed in a semicircle is $90^{\circ }$ ] Similarly, $\angle \mathrm{ABC}=90^{\circ },\angle \mathrm{BCD}=90^{\circ }$ and $\angle \mathrm{CDA}=90^{\circ }$. Now, in right $△\mathrm{ABC}$ and $△\mathrm{BAD}$, we have $\mathrm{AC}=\mathrm{BD}$ (from (1)) $\mathrm{AB}=\mathrm{BA}$ (common) $\angle \mathrm{ABC}=\angle \mathrm{BAD}$ (each equal to $90^{\circ }$ ) $\therefore △\mathrm{ABC}\cong △\mathrm{BAD}$ (By RHS congruence) $\Rightarrow \mathrm{BC}=\mathrm{AD}$ (CPCT) Similarly, $\mathrm{AB}=\mathrm{DC}$ Thus, ABCD is a rectangle.
• If the non-parallel sides of a trapezium are equal, prove that it is a cyclic. Sol. Given : ABCD is a trapezium whose two non-parallel sides AD and BC are equal. To Prove : Trapezium ABCD is a cyclic. Construction : Draw BE||AD Proof: $\because \mathrm{AB}\Vert \mathrm{DE}$ [Given] $\mathrm{AD}\Vert$ BE [By construction] $\therefore$ Quadrilateral ABED is a parallelogram.
  
  $\therefore \angle \mathrm{BAD}=\angle \mathrm{BED}$ [Opp. $\angle$ s of a||gm] and $\mathrm{AD}=\mathrm{BE}$ [Opp. sides of a || gm] But $\mathrm{AD}=\mathrm{BC}$ [Given] From (ii) and (iii) $\mathrm{BE}=\mathrm{BC}$ $\therefore \angle \mathrm{BEC}=\angle \mathrm{BCE}$ [Angle opposite to equal sides] $\angle \mathrm{BEC}+\angle \mathrm{BED}=180^{\circ }$ [Linear pair] $\Rightarrow \angle \mathrm{BCE}+\angle \mathrm{BAD}=180^{\circ }$ [From (iv) and (i)] $\Rightarrow$ Trapezium ABCD is cyclic. $[\because$ If a pair of opposite angles of a quadrilateral is $180^{\circ }$, then the quadrilateral is cyclic]
• Two circles intersect at two points B and C. Through B, two line segment ABD and PBQ are drawn to intersect the circles at $\mathrm{A},\mathrm{D}$ and $\mathrm{P},\mathrm{Q}$ respectively. Prove that $\angle \mathrm{ACP}=\angle \mathrm{QCD}$.
  
  Sol. Given : Two circles intersect at two points B and C . Through B , two line segments ABD and PBQ are drawn to intersect the circles at $\mathrm{A},\mathrm{D}$ and $\mathrm{P},\mathrm{Q}$ respectively. To Prove : $\angle \mathrm{ACP}=\angle \mathrm{QCD}$ Proof: $\angle \mathrm{ACP}=\angle \mathrm{ABP}$ [Angles in the same segment of a circle are equal] $\angle \mathrm{QCD}=\angle \mathrm{QBD}$ [Angles in the same segment of a circle are equal] $\angle \mathrm{ABP}=\angle \mathrm{QBD}$ [Vertically Opposite Angles] From (i), (ii) and (iii), $\angle \mathrm{ACP}=\angle \mathrm{QCD}$
• 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. Sol. We have $△\mathrm{ABC}$, and two circles described with diameter as AB and AC respectively. They intersect at a point D , other than A . Let us join A and D.
  
  AB is a diameter $\therefore \angle \mathrm{ADB}$ is an angle formed in a semicircle. $\Rightarrow \angle \mathrm{ADB}=90^{\circ }$ Similarly, $\angle \mathrm{ADC}=90^{\circ }$ adding (1) and (2) $\angle \mathrm{ADB}+\angle \mathrm{ADC}=180^{\circ }$ i.e., $\mathrm{B},\mathrm{D}$ and C are collinear points BC is a straight line. Thus, D lies on BC .
• ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle \mathrm{CAD}=\angle \mathrm{CBD}$. Sol. AC is a hypotenuse $\angle \mathrm{ADC}=90^{\circ }=\angle \mathrm{ABC}$
  
  $\therefore$ Both the triangles are in the same semicircle. $\Rightarrow \mathrm{A},\mathrm{B},\mathrm{C}$ and D are concyclic. Join BD DC is chord $\therefore \angle \mathrm{CAD}$ and $\angle \mathrm{CBD}$ are formed on the same segment $\therefore \angle \mathrm{CAD}=\angle \mathrm{CBD}$
• Prove that a cyclic parallelogram is a rectangle. Sol.
  
  We have a cyclic parallelogram ABCD. $\therefore$ Sum of its opposite angles is $180^{\circ }$ $\therefore \angle \mathrm{A}+\angle \mathrm{C}=180^{\circ }$ But $\angle \mathrm{A}=\angle \mathrm{C}$ From (1) and (2), we have $\angle \mathrm{A}=\angle \mathrm{C}=90^{\circ }$ Similarly, $\angle \mathrm{B}=\angle \mathrm{D}=90^{\circ }$ $\Rightarrow$ Each angle of the parallelogram ABCD is $90^{\circ }$ Thus, ABCD is a rectangle.

6.0Key Features and Benefits: Class 9 Maths Chapter 9 Exercise 9.3

• In-depth Problem Solving: Each response follows a logical sequence of reasoning with proper diagrams as appropriate.
• The concept focus: We emphasize the theorems that form a basis for understanding circular geometry.
• Propositional reasoning: That provides students with an opportunity to engage in more rational thought and logical geometric proof writing.
• Exam focused: Gets students ready to be able to confidently tackle angle based questions in exams.