NCERT Solutions Class 9 Maths Chapter 9 Circles Exercise 9.2

Students engaging with NCERT Solutions Class 9 Maths Chapter 9 Circles Exercise 9.2, will learn some basic properties regarding chords and their locations within a circle.Students will engage with some reasoning and visualization abilities while through theorems and by completing problem-solving questions about chords, distance from the center, and perpendiculars.

1.0Download Class 9 Maths Chapter 9 Ex 9.2 NCERT Solutions PDF

Downloadable PDF for accurate NCERT Solutions for all the problem work corresponding to NCERT Solutions Class 9 Maths Chapter 9 Circles Exercise 9.2 and to get the right understanding using these solutions provided by subject matter experts to prepare for your exams.

2.0Key Concepts of Exercise 9.2

• Equal Chords of a Circle are Equally Distant from the Center.
• Chords that are Equally Distant from the Center are Equal in length.
• The Perpendicular from the Center of a Circle to a Chord Bisects the Chord.
• The Converse of all independently above.
• Thinking, reasoning, and geometry for construction and proof based problems.

3.0CBSE Class 9 Chapter 9 Linear Equations Exercise 9.2 Comprises

1. To Understand Chords and their Properties

To understand what happens with the chords inside of the circle and what is the relationship with the lengths and distance from the center of the circle.

2. Perpendiculars from the Center

To understand what happens when you need to draw a perpendicular from the center to a chord in order to bisect it, I also found using the same distance properties

3. Concept of Equidistant Chords

Understand how chords that are on the same side of the center and are the same distance from the center must be the same length, and how understanding this fact can help with solving proof-based geometry problems.

4. Reason Based Questions

Use deductive reasoning to understand the proof in the shapes involving chords, perpendiculars and centers, and learn to justify where you arrived at each point through logic.

5. Sketching circles to utilize them in proof questions

Practice sketching, annotating, and interpreting circle diagrams correctly to understand and tackle every question properly.

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 9.2 Solutions

• Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm . Find the length of common chord. Sol. We know that if two circles intersect each other at two points, then the line joining their centres is the perpendicular bisector of their common chord.
  
  In $△{\mathrm{PO}}^{\prime }0$ ${\mathrm{OP}}^2={\mathrm{OO}}^{\prime 2}+{\mathrm{PO}}^{\prime 2}$ $5^2=4^2+{\mathrm{PO}}^{\prime 2}$ $\Rightarrow {\mathrm{PO}}^{\prime }=3\mathrm{cm}$ $\therefore$ Length of the common chord $\Rightarrow \mathrm{PQ}=20^{\prime }\mathrm{P}=2\times 3=6\mathrm{cm}$
• If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord. Sol. 0 is the centre of the circle. Chords AB and CD of the circle are equal. $P$ is the point of intersection of AB and CD . Join OP , draw $\mathrm{OL}\perp \mathrm{AB}$ and $\mathrm{OM}\perp \mathrm{OD}$. Here, we find $OL=OM$ $(\because \mathrm{AB}=\mathrm{CD})$ In $△\mathrm{OLP}$ and $△\mathrm{OMP}$,
  
  $\mathrm{OL}=\mathrm{OM}$ (By 1) $\mathrm{OP}=\mathrm{OP}$ (Common hypotenuse) $\angle \mathrm{OLP}=\angle \mathrm{OMP}$ (Each $=90^{\circ }$ ) Then we have $\Delta \mathrm{OLP}\cong \Delta \mathrm{OMP}$ (RHS congruence) By CPCT, PL = PM Now, $\mathrm{AL}=\mathrm{BL}=1/2\mathrm{AB}$ $\begin{array}{rl}& CM=DM=1/2CD\\ \Rightarrow & AL=CM(\because AB=CD)\end{array}$ and $\mathrm{BL}=\mathrm{DM}$ Subtracting (2) from (3), $\begin{array}{rl}& \mathrm{AL}-\mathrm{PL}=\mathrm{CM}-\mathrm{PM}\\ \Rightarrow & \mathrm{AP}=\mathrm{CP}\end{array}$ Adding (2) and (4), $\mathrm{PL}+\mathrm{BL}=\mathrm{PM}+\mathrm{DM}\Rightarrow \mathrm{PB}=\mathrm{PD}$
• If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords. Sol.
  
  Ois the centre of the circle. Chords AB and CD of the circle are equal. $P$ is the point of intersection of AB and CD . Join OP , draw $\mathrm{OL}\perp \mathrm{AB}$ and $\mathrm{OM}\perp \mathrm{CD}$. Here, we find. $\mathrm{OL}=\mathrm{OM}(\because \mathrm{AB}=\mathrm{CD})$ In $△\mathrm{OLP}$ and $△\mathrm{OMP}$, OL $=\mathrm{OM}$ (By 1) $\mathrm{OP}=\mathrm{OP}$ (Common hypotenuse) $\angle \mathrm{OLP}=\angle \mathrm{OMP}$ (Each $=90^{\circ }$ ) Then we have $\Delta \mathrm{OLP}\cong △\mathrm{OMP}$ (RHS congruence) By CPCT, $\angle \mathrm{OPL}=\angle \mathrm{OPM}$
• If a line intersects two concentric circles (circles with the same centre) with centre O at $\mathrm{A},\mathrm{B},\mathrm{C}$ and D , prove that $\mathrm{AB}=\mathrm{CD}$ (see fig).
  
  Sol. Given : Two circles with the common centre 0 . A line " $\ell$ " intersects the outer circle at A and D and the inner circle at B and C .
  
  To prove: $AB=CD$ Construction: Draw $\mathrm{OM}\perp \ell$. Proof: $\mathrm{OM}\perp \ell$ [Construction] For the outer circle, $\therefore AM=MD$ [Perpendicular from the centre bisects the chord] For the inner circle, $\begin{array}{ll}& \mathrm{OM}\perp \ell \text{ [Construction] }\\ \therefore & \mathrm{BM}=\mathrm{MC}\end{array}$ [Perpendicular from the centre to the chord bisects the chord] Subtracting (2) from (1), we have $\Rightarrow \mathrm{AM}-\mathrm{BM}=\mathrm{MD}-\mathrm{MC}$ $\Rightarrow AB=CD$
• Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip? Sol. We draw $\mathrm{SN}\perp \mathrm{RM}$ as shown in figure. Since, $△\mathrm{RSM}$ is an isosceles triangle hence SN bisects RM and also SN (produced) passes through the centre 0 .
  
  Put RN $=\mathrm{x}$ Then $\mathrm{ar}(△\mathrm{ORS})=\frac{1}{2}\times \mathrm{OS}\times \mathrm{RN}$ $=\frac{1}{2}\times 5\times \mathrm{x}(\because \mathrm{OS}=\mathrm{OR}=5\mathrm{m})$ i.e., $\mathrm{ar}(\Delta \mathrm{ORS})=\frac{5}{2}\mathrm{x}$ Now, draw $\mathrm{OP}\perp \mathrm{RS},\mathrm{P}$ is mid-point of RS . $\Rightarrow \mathrm{PR}=\mathrm{PS}=3\mathrm{m}$ $\Rightarrow {\mathrm{OP}}^2=(5{)}^2-(3{)}^2=16$ $\Rightarrow \mathrm{OP}=4\mathrm{m}$ Here, $\mathrm{ar}(△\mathrm{ORS})=\frac{1}{2}\times \mathrm{RS}\times \mathrm{OP}=\frac{1}{2}\times 6\times 4$ i.e., $\mathrm{ar}(△\mathrm{ORS})=12{\mathrm{m}}^2$ From (1) and (2), $\frac{5}{2}x=12\Rightarrow x=4.8\mathrm{m}$ $\Rightarrow \mathrm{RM}=2\mathrm{x}=2\times 4.8\mathrm{m}$ $\Rightarrow \mathrm{RM}=9.6\mathrm{m}$ Thus, distance between Reshma and Mandip is 9.6 m .
• A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone. Sol. Let Ankur, Syed and David are sitting at A, $S$ and $D$ respectively such that $AS=SD=AD$ i.e., $△\mathrm{ASD}$ is an equilateral triangle. Let the length of each side of the equilateral triangle be $2x$ metres.
  
  Draw $\mathrm{AM}\perp \mathrm{SD}$. Since, $△\mathrm{ASD}$ is an equilateral triangle, $\therefore$ So, altitude AM divides SD into two equal parts. Also, $\mathrm{OM}\perp \mathrm{SD}$. $\Rightarrow$ OM bisects SP also. So, we can say AM passes through 0 . $\Rightarrow \mathrm{SM}=\frac{1}{2}\mathrm{SD}=\frac{1}{2}(2\mathrm{x})=\mathrm{x}$ Now, in $△\mathrm{ASM}$, we have ${\mathrm{AM}}^2+{\mathrm{SM}}^2={\mathrm{AS}}^2$ $\Rightarrow {\mathrm{AM}}^2={\mathrm{AS}}^2-{\mathrm{SM}}^2$ $=(2\mathrm{x}{)}^2-{\mathrm{x}}^2=4{\mathrm{x}}^2-{\mathrm{x}}^2=3{\mathrm{x}}^2$ $\Rightarrow \mathrm{AM}=\sqrt{3}\mathrm{x}$ Now, $\mathrm{OM}=\mathrm{AM}-\mathrm{OA}=(\sqrt{3}\mathrm{x}-20)\mathrm{m}$ $\Rightarrow (OS=OA=20\mathrm{cm})$ In $△\mathrm{OSM}$ $\Rightarrow (20{)}^2=x^2+(\sqrt{3}x-20{)}^2$ $\Rightarrow 400=x^2+3x^2-40\sqrt{3}x+400$ $\Rightarrow 4x^2=40\sqrt{3}x$ $\Rightarrow 4\mathrm{x}=40\sqrt{3}$ $\Rightarrow x=10\sqrt{3}m$ Now, SD $=2\mathrm{x}=2\times 10\sqrt{3}\mathrm{m}=20\sqrt{3}\mathrm{m}$ Thus, the length of the string of each phone $=20\sqrt{3}\mathrm{m}$

6.0Key Features and Benefits: Class 9 Maths Chapter 9 Exercise 9.2

• Clarity with well written explanations of solutions with diagrams to visualize information better.
• Lots of logical reasoning used for each step to enhance proof writing capability.
• Following the most recent CBSE syllabus and exam pattern.
• This will help students develop a strong understanding of circle geometry.
• Great for revising a concept, helping with homework and preparation for exams.