CIRCLES — NCERT Solutions | Class 9 Mathematics | Academia Aeternum
Ch 9  ·  Q–
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Class 9 Mathematics Exercise-9.2 NCERT Solutions Olympiad Board Exam
Chapter 9

CIRCLES

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

6 Questions
15–20 min Ideal time
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Q1
NUMERIC3 marks
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 the common chord.
📘 Concept & Theory Theory & Concepts ›
1. Two Intersecting Circles

Two circles can intersect each other in three ways:

  • No intersection: Distance between centres > sum of radii
  • Touch externally: Distance = sum of radii (1 common point)
  • Touch internally: Distance = |difference of radii| (1 common point)
  • Intersect at two points: |r₁ - r₂| < d < (r₁ + r₂) (2 common points)
2. Common Chord of Two Intersecting Circles

The line joining the two points of intersection of two circles is called their >common chord. In the problem, the common chord is \( PQ \).

Key Property: The common chord PQ is always perpendicular to the line joining the centres OO' and is bisected by this line. That is, the perpendicular from the centre of a circle to a chord bisects the chord.

3. Why Circle I Intersects Circle II

Verify the condition for two circles with radii r₁ = 5, r₂ = 3, and d = 4:

  • r₁ - r₂ = 5 - 3 = 2
  • r₁ + r₂ = 5 + 3 = 8
  • Since |r₁ - r₂| < d < (r₁ + r₂), i.e., 2 < 4 < 8, the circles >intersect at two points
4. Right Triangle Setup

When we drop a perpendicular from centre O to the common chord PQ at point M, we get two right triangles: >△OPM and >△O'PM. We apply the Pythagorean theorem in each.

In △OPM: \[ OM^2 + PM^2 = OP^2 \]

In △O'PM: \[ O'M^2 + PM^2 = O'M^2 \]

Since O'M = OO' - OM = d - OM, we get: \[ OM^2 + PM^2 = r_1^2 \] \[ (d - OM)^2 + PM^2 = r_2^2 \]

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Verify Intersection Condition: Check that |r₁ - r₂| < d < (r₁ + r₂) to confirm the circles intersect at two points.

  2. Set Up Diagram: Mark centres O, O', chord PQ, and midpoint M where chord intersects line OO'.

  3. Find O'M Using Formula: O'M = (r₂² - r₁² + d²) / (2d) from subtracting the two equations.

  4. Find OM: OM = d - O'M

  5. Apply Pythagorean Theorem: In △OPM, PM = √(r₁² - OM²)

  6. Find Common Chord: Since M bisects chord PQ, PQ = 2 × P

    M
📊 Graph / Figure Graph / Figure ›
O O' P Q M 4 cm 5 cm 3 cm Common chord PQ = 6 cm
Figure: Two intersecting circles with centres O and O', common chord PQ, and midpoint M where OO' bisects the chord at right angles
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  13 steps
  1. Given
    • Radius of the first (larger) circle, \( r_1 = 5 \text{ cm} \)
    • Radius of the second (smaller) circle, \( r_2 = 3 \text{ cm} \)
    • Distance between the two centres, \( d = OO' = 4 \text{ cm} \)
  2. To Find

    Length of the common chord \( PQ \) of the two intersecting circles.

  3. Step 0: Verify Intersection Condition (Important!)

    Before solving, we must confirm that the circles indeed intersect at two points.

    For two circles to intersect at two points, the following condition must hold: \[ |r_1 - r_2| < d < (r_1 + r_2) \]

    Here, \( r_1 - r_2 = 5 - 3 = 2 \text{ cm} \)

    And \( r_1 + r_2 = 5 + 3 = 8 \text{ cm} \)

    Since \( 2 < 4 < 8 \), we have \( |r_1 - r_2| < d < (r_1 + r_2) \).

    Therefore, the circles >do intersect at exactly two points, so a common chord exists.

  4. Solution

    Set up the diagram and identify the right triangles

    Let the centres of the two circles be \( O \) and \( O' \).

    Let the two intersection points of the circles be \( P \) and \( Q \).

    Since \( PQ \) is the common chord of two intersecting circles, the line joining the centres \( OO' \) perpendicularly bisects the chord \( PQ \) at point \( M \).

    Therefore, \( PM = QM \) (chord bisected by perpendicular from centre) and \( \angle PMO = 90° \) (perpendicular condition).

    This gives us two right triangles: \( \triangle OPM \) and \( \triangle O'M \).

  5. Apply Pythagorean theorem in triangle OPM

    In right triangle \( \triangle OPM \), we have: \[ OP = r_1 = 5 \text{ cm} \] \[ OM = \text{(unknown distance from O to M)} \] \[ PM = \text{(half the chord length, unknown)} \]

    By Pythagorean theorem: \[ OP^2 = OM^2 + PM^2 \] \[ \Rightarrow 5^2 = OM^2 + PM^2 \] \[ \Rightarrow 25 = OM^2 + PM^2 \quad \cdots (1) \]

  6. Apply Pythagorean theorem in triangle O'M

    In right triangle \( \triangle O'M \), we have: \[ O'M = r_2 = 3 \text{ cm} \] \[ O'M = \text{(distance from O' to M)} \] \[ PM = \text{(same as above, since M is midpoint of chord)} \]

    By Pythagorean theorem: \[ O'M^2 = O'M ^2 + PM^2 \] \[ \Rightarrow 3^2 = O'M^2 + PM^2 \] \[ \Rightarrow 9 = O'M^2 + PM^2 \quad \cdots (2) \]

  7. Express O'M in terms of OM

    From the diagram, the distance between the centres is: \[ OO' = d = 4 \text{ cm} \]

    Point M lies on the segment connecting O and O'. Since OM is the distance from O to M (going from O toward O'), and O'M is the remaining distance to O', we have: \[ O'M = OO' - OM \] \[ O'M = d - OM \] \[ O'M = 4 - OM \quad \cdots (3) \]

  8. Find OM by substituting equation (3) into equation (2)

    Substitute \( O'M = 4 - OM \) into equation (2):

    \[ 9 = (4 - OM)^2 + PM^2 \] \[ 9 = (16 - 8 \cdot OM + OM^2) + PM^2 \] \[ 9 = 16 - 8 \cdot OM + OM^2 + PM^2 \] \[ \Rightarrow OM^2 + PM^2 = 9 - 16 + 8 \cdot OM \] \[ \Rightarrow OM^2 + PM^2 = -7 + 8 \cdot OM \quad \cdots (4) \]

  9. Equate equations (1) and (4) to eliminate \( PM^2 \) and solve for OM

    From equation (1): \( OM^2 + PM^2 = 25 \)

    From equation (4): \( OM^2 + PM^2 = -7 + 8 \cdot OM \)

    Equating the two: \[ 25 = -7 + 8 \cdot OM \] \[ \Rightarrow 25 + 7 = 8 \cdot OM \] \[ \Rightarrow 32 = 8 \cdot OM \] \[ \Rightarrow OM = \frac{32}{8} = 4 \text{ cm} \]

  10. Find O'M

    Using equation (3): \[ O'M = d - OM \] \[ O'M = 4 - 4 \] \[ O'M = 0 \text{ cm} \]

    This means point M coincides with centre O'. Note: O'M = 0 implies M is exactly at O'. This is a special case where the perpendicular from M to chord PQ passes through O'. Let's verify:

  11. Calculate PM using Pythagorean theorem in triangle OPM

    From equation (1), we know: \[ OM^2 + PM^2 = 25 \]

    Substitute \( OM = 4 \): \[ (4)^2 + PM^2 = 25 \] \[ 16 + PM^2 = 25 \] \[ PM^2 = 25 - 16 \] \[ PM^2 = 9 \] \[ PM = \sqrt{9} = 3 \text{ cm} \]

    (Taking the positive square root since length is positive)

    Verify with triangle O'M (where O'M = 0): \[ O'M^2 + PM^2 = 0^2 + 3^2 = 0 + 9 = 9 = r_2^2 \quad \checkmark \] The values are consistent.

  12. Since M is the midpoint of PQ (the perpendicular from the centre to a chord bisects it), we have: \[ PQ = PM + QM \] \[ PQ = PM + PM \] \[ PQ = 2 \times PM \]

    Substituting \( PM = 3 \text{ cm} \): \[ PQ = 2 \times 3 \] \[ PQ = 6 \text{ cm} \]

💡 Answer Final Answer ›
Final Answer: The length of the common chord \( PQ \) is 6 cm
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points
  1. Always verify the intersection condition before solving
  2. Chord Perpendicular Property: Common chord is always perpendicular to line joining centres
  3. Chord Bisector Property: This perpendicular also bisects the chord at its midpoint
  4. Pythagorean theorem is the primary tool — apply it in both right triangles
  5. The derivation shows why O'M = (r₂² - r₁² + d²) / (2d) — >know this formula for faster solutions
  6. In this specific problem, O'M = 0 is a >special case where the small circle's centre lies exactly on the common chord's perpendicular bisector line.
🎯 Exam Significance Exam Significance ›
  • 3-4 marks — This is a numerical problem from Exercise 9.2
  • Combines circle chord property with coordinate geometry concepts and Pythagorean theorem
  • Often appears as Application-Based Question in Section C of CBSE exams
  • The intersection condition check is a quick 1-mark verification step students often miss
↑ Top
1 / 6  ·  17%
Q2 →
Q2
NUMERIC3 marks
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to the corresponding segments of the other chord.
📘 Concept & Theory Theory & Concepts ›
  • Equal chords of a circle are equidistant from the centre.
  • The perpendicular drawn from the centre of a circle to a chord bisects the chord.
  • RHS congruence criterion is used when:
    • One angle is \(90^\circ\)
    • Hypotenuse is equal
    • One corresponding side is equal
  • CPCT means “Corresponding Parts of Congruent Triangles”.
  • This question combines circle properties and triangle congruence concepts together.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Draw perpendiculars from the centre to the two equal chords.

  2. Use the property that equal chords are equidistant from the centre.

  3. Consider two right triangles formed inside the circle.

  4. Prove the triangles congruent using RHS congruence criterion.

  5. Use CPCT to prove equality of the corresponding chord segments.

📊 Graph / Figure Graph / Figure ›
O A B C D Q Chord AB S T Chord CD Intersection Q If two equal chords intersect within a circle, their segments are equal.
Figure: Equal chords AB and CD intersect at point P inside the circle. OM and ON are perpendiculars from centre O to the chords. We prove AP = CP and BP = DP.
📐 Proof Proof ›
📌 Given
  • Two equal chords \[ AB = CD \]
  • Chords \(AB\) and \(CD\) intersect at point \(Q\) inside the circle.
🎯 To Prove
\[AQ = CQ\] and \[ BQ = DQ \]
🔧 Construction

Draw perpendiculars from the centre \(O\) to chords \(AB\) and \(CD\).

Let \[ OS \perp AB\] and \[OT \perp CD\] where \(S\) and \(T\) are points on chords \(AB\) and \(CD\) respectively.
✍️ Proof
Step-by-step Proof  ·  19 steps
  1. Since equal chords of a circle are equidistant from the centre,
  2. \[OS = OT\]
  3. Also,\[\angle OSQ = \angle OTQ = 90^\circ\]
  4. In triangles \( \triangle OSQ \) and \( \triangle OTQ \):
  5. \[OQ = OQ\]
    Common Side
  6. \[OS = OT\]
    (Equal chords are equidistant from the centre)
  7. \[\angle OSQ = \angle OTQ = 90^\circ\]
  8. Therefore, by RHS congruence criterion,
  9. \[\triangle OSQ \cong \triangle OTQ\]
  10. By CPCT,\[SQ = TQ\]
  11. Since the perpendicular from the centre bisects the chord,
  12. \[AS = SB\] and \[CT = TD\]
  13. Now, \[ \begin{aligned} AQ &= AS - SQ \\ CQ &= CT - TQ \end{aligned} \]
  14. Since \[AS = CT\] and \[SQ = TQ\]
  15. \[therefore,\]
  16. Similarly,\[ \begin{aligned} BQ &= SB + SQ \\ DQ &= TD + TQ \end{aligned}\]
  17. Since \[SB = TD\] and \[SQ = TQ\]
  18. therefore,\[BQ = DQ\]
  19. Hence, the corresponding segments of the equal chords are equal.
Q.E.D.
🎯 Exam Significance Exam Significance ›
  • 3-4 marks — Proof-based question from Exercise 9.2
  • Tests understanding of three distinct theorems in one problem:
    • Equal chords ↔ Equidistant from centre
    • Perpendicular from centre bisects chord
    • RHS congruence criterion
  • Commonly paired with Q3 or Q4 in a 4-6 mark proof chain
  • Particularly important in geometry construction proofs
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points
  1. Equal chords of a circle are always equidistant from the centre.
  2. The perpendicular from the centre to a chord bisects the chord.
  3. RHS congruence criterion is extremely useful in circle geometry proofs.
  4. CPCT helps establish equality of line segments after proving triangles congruent.
  5. Circle geometry questions often combine:
    Chord properties
    Triangle congruence
    Perpendicular distances
    Symmetry concepts
  6. Such proofs improve mathematical reasoning and stepwise presentation skills needed for board and competitive examinations.
← Q1
2 / 6  ·  33%
Q3 →
Q3
NUMERIC3 marks
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.
📘 Concept & Theory Theory and Concepts ›
  • Equal chords of a circle are equidistant from the centre.
  • The perpendicular from the centre to a chord bisects the chord.
  • If two right triangles have equal hypotenuse and one corresponding side equal, then the triangles are congruent by RHS criterion.
  • Corresponding angles of congruent triangles are equal by CPCT.
  • This question connects chord properties with triangle congruence and angle bisector ideas.
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Draw perpendiculars from the centre to both equal chords.
  2. Use the property that equal chords are equidistant from the centre.
  3. Consider the two right triangles formed using the intersection point and the centre.
  4. Prove the triangles congruent using RHS congruence criterion.
  5. Apply CPCT to prove that the line joining the centre and intersection point makes equal angles with the chords.
📊 Graph / Figure Graph / Figure ›
O A B C D P Q R Chord AB Chord CD Intersection P
The line joining the centre to the intersection point bisects the angle between equal chords.
📐 Proof Proof ›
📌 Given
  • \(AB\) and \(CD\) are equal chords of a circle.
  • The chords intersect at point \(P\).
🎯 To Prove
\[\angle QPO = \angle RPO\]

where \(Q\) and \(R\) are the feet of perpendiculars from the centre \(O\) to chords \(AB\) and \(CD\) respectively.

🔧 Construction

Draw perpendiculars from the centre \(O\) to the chords:

\[ OQ \perp AB \] and \[ OR \perp CD \]

Join \(OP\).

✍️ Proof
Step-by-step Proof  ·  11 steps
  1. Since equal chords of a circle are equidistant from the centre,\[OQ = OR\]
  2. Also, \[\angle OQP = \angle ORP = 90^\circ\]
  3. In triangles \( \triangle PQO \) and \( \triangle PRO \):
  4. \[OP = OP\]
    (Common hypotenuse)
  5. \[OQ = OR\]
    (Equal chords are equidistant from the centre)
  6. \[\angle OQP = \angle ORP = 90^\circ\]
  7. Therefore, by RHS congruence criterion,
  8. \[\triangle PQO \cong \triangle PRO\]
  9. By CPCT,
  10. \[\angle QPO = \angle RPO\]
  11. Hence, the line joining the intersection point \(P\) to the centre \(O\) makes equal angles with the equal chords.
Q.E.D.
🎯 Exam Significance Exam Significance ›
  • This question develops understanding of symmetry properties in circles.
  • It strengthens the application of RHS congruence criterion in geometry proofs.
  • The problem demonstrates how equal chords create equal geometric configurations.
  • Such angle-based proofs are commonly asked in CBSE board examinations.
  • Similar concepts are useful in Olympiads, NTSE, and other competitive entrance examinations.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points
  1. Equal chords of a circle are equidistant from the centre.
  2. The perpendicular from the centre to a chord forms right triangles useful for proofs.
  3. RHS congruence criterion is frequently used in circle geometry problems.
  4. CPCT helps establish equality of corresponding angles and segments.
  5. The line joining the centre to the intersection point behaves like an angle bisector between equal chords.
  6. This problem is an excellent example of combining:
    • Chord properties
    • Triangle congruence
    • Perpendicular distances
    • Angle relationships
← Q2
3 / 6  ·  50%
Q4 →
Q4
NUMERIC3 marks
If a line intersects two concentric circles (circles with the same centre) with centre \( O \) at \( A, B, C \) and \( D \), prove that \( AB = CD \).
📘 Concept & Theory Theory and Concepts ›
  • Concentric circles are circles having the same centre but different radii.
  • The perpendicular drawn from the centre of a circle to a chord bisects the chord.
  • Since the same perpendicular passes through both circles, it bisects both chords formed on the same line.
  • Segment subtraction is used to prove equality of line segments.
  • This question is a direct application of chord properties in concentric circles.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Draw a perpendicular from the common centre \(O\) to the line intersecting the circles.

  2. Use the theorem that the perpendicular from the centre bisects a chord.

  3. Apply the theorem separately to the outer circle and inner circle.

  4. Form equations using equal halves of the chords.

  5. Subtract equal quantities to prove: \[ AB = CD \]

📊 Graph / Figure Graph / Figure ›
A B C D O P
The perpendicular from the common centre bisects chords of both concentric circles.
📐 Proof Proof ›
📌 Given
  • Two concentric circles with common centre \(O\).
  • A line intersects the outer circle at points \(A\) and \(D\).
  • The same line intersects the inner circle at points \(B\) and \(C\).
🎯 To Prove
\[AB = CD\]
🔧 Construction

Draw

\[ OP \perp AD \]

where \(P\) lies on line \(AD\).

✍️ Proof
Step-by-step Proof  ·  10 steps
  1. Since \(OP\) is perpendicular to chord \(AD\) of the outer circle, the perpendicular from the centre bisects the chord.
  2. Therefore,\[AP = PD \tag{1}\]
  3. Also, \(BC\) is a chord of the inner circle and \[OP \perp BC\]
  4. Therefore, \(P\) bisects chord \(BC\).
  5. Hence,\[BP = PC \tag{2}\]
  6. Now,\[AB = AP - BP\] and \[CD = PD - PC\]
  7. From equations \((1)\) and \((2)\),\[AP = PD\] and \[BP = PC\]
  8. Subtracting equal quantities from equal quantities:\[AP - BP = PD - PC\]
  9. Therefore,\[AB = CD\]
  10. Hence proved.
Q.E.D.
🎯 Exam Significance Exam Significance ›
  • This question develops understanding of concentric circles and chord properties.
  • It strengthens theorem-based proof writing required in board examinations.
  • The problem demonstrates practical use of perpendicular bisector properties in circles.
  • Similar problems are frequently asked in CBSE, NTSE, and Olympiad examinations.
  • The concept is useful in advanced geometry involving symmetry and equal segments.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points
  1. Concentric circles have the same centre but different radii.
  2. The perpendicular from the centre to a chord always bisects the chord.
  3. A single perpendicular can bisect chords of both concentric circles simultaneously.
  4. Subtracting equal quantities from equal quantities preserves equality.
  5. Geometry proofs often combine:
    • Chord properties
    • Perpendicular theorems
    • Segment relationships
    • Algebraic subtraction
  6. Such theorem-based proofs are important for board exams and competitive entrance tests.
← Q3
4 / 6  ·  67%
Q5 →
Q5
NUMERIC3 marks
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius \(5\text{ m}\) drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, and Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is \(6\text{ m}\) each, what is the distance between Reshma and Mandip?
📘 Concept & Theory Concept and theory ›
  • Equal chords of a circle subtend equal angles at the centre.
  • The perpendicular from the centre to a chord bisects the chord.
  • Pythagoras Theorem is used to find the perpendicular distance from the centre to the chord.
  • The chord joining two points farther apart on the circle can be determined using properties of isosceles triangles and equal central angles.
  • This problem combines circle geometry with triangle properties and algebraic calculations.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Draw radii from the centre to the endpoints of chord \(RS\).

  2. Draw a perpendicular from the centre to chord \(RS\), which bisects the chord.

  3. Use Pythagoras Theorem to calculate the perpendicular distance from the centre.

  4. Since chords \(RS\) and \(SM\) are equal, the corresponding central angles are equal.

  5. Use symmetry to determine the length of chord \(RM\).

📊 Graph / Figure Graph / Figure ›
O R P S M 6m 6m ? (to find) r=5m RS = SM = 6 m (Given)
Figure: Three points R, S, M on a circle of radius 5 m. Reshma (R) and Salma (S) are 6 m apart; Salma (S) and Mandip (M) are 6 m apart. We find the distance RM between Reshma and Mandip
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  21 steps

  1. Given:

    • Radius of the circle: \[ OR = OS = OM = 5\text{ m} \]
    • Distance between Reshma and Salma: \[ RS = 6\text{ m} \]
    • Distance between Salma and Mandip: \[ SM = 6\text{ m} \]

  2. To Find:

    \[RM\]

  3. Construction:

  4. Draw perpendicular \(OP\) from the centre \(O\) to chord \(RS\).

  5. Since the perpendicular from the centre to a chord bisects the chord,
  6. \[\begin{aligned}RP = PS &= \frac{RS}{2}\\RP = PS &= \frac{6}{2} = 3\text{ m}\end{aligned}\]

  7. Apply Pythagoras Theorem in right triangle \(ORP\):

    \[OR^2 = OP^2 + RP^2\]
  8. Substituting the values:\[ \begin{aligned} 5^2 &= OP^2 + 3^2\\ 25 &= OP^2 + 9\\ OP^2 &= 25 - 9\\ OP^2 &= 16\\ OP &= 4\text{ m} \end{aligned} \]
  9. Therefore, the perpendicular distance from the centre to chord \(RS\) is \[OP = 4\text{ m}\]
  10. Since\[RS = SM\]
  11. equal chords subtend equal angles at the centre.
  12. Therefore, points \(R\) and \(M\) are symmetrically placed about line \(OS\).
  13. Hence, the perpendicular distance from \(O\) to chord \(RM\) is equal to\[1\text{ m}\]
  14. Let \(Q\) be the midpoint of chord \(RM\).
  15. Then,\[OQ = 1\text{ m}\]
  16. In right triangle \(ORQ\):\[OR^2 = OQ^2 + RQ^2\]
  17. Substituting the values:\[ \begin{aligned} 5^2 &= 1^2 + RQ^2 \\ 25 &= 1 + RQ^2 \\ RQ^2 &= 24 \\ RQ &= \sqrt{24} \\ &= 2\sqrt{6} \end{aligned} \]
  18. Therefore,\[ \begin{aligned} RM &= 2 \times RQ \\ &= 2 \times 2\sqrt{6} \\ &= 4\sqrt{6} \end{aligned} \]
  19. Taking
  20. \[\sqrt{6} \approx 2.45\]
  21. \[ \begin{aligned} RM &\approx 4 \times 2.45 \\ &\approx 9.8\text{ m} \end{aligned} \]
💡 Answer Final Answer ›
Final Answer: Therefore, the distance between Reshma and Mandip is approximately \[ \boxed{9.8\text{ m}} \]
🎯 Exam Significance Exam Significance ›
  • This question develops understanding of chord properties in circles.
  • It strengthens the application of the Pythagoras Theorem in geometry.
  • The problem demonstrates how equal chords create symmetrical geometric configurations.
  • Similar questions are frequently asked in CBSE board examinations and scholarship tests.
  • The concepts are useful in Olympiad-level geometry and entrance examination preparation.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points
  1. Equal chords subtend equal angles at the centre of a circle.
  2. The perpendicular from the centre to a chord bisects the chord.
  3. Pythagoras Theorem is very useful in finding distances related to chords.
  4. Symmetry in circles simplifies many geometry problems.
  5. Circle geometry problems often combine:
    • Chord properties
    • Symmetry
    • Right triangles
    • Pythagoras Theorem
  6. Such problems are important for board examinations, NTSE, Olympiads, and other competitive entrance examinations.
← Q4
5 / 6  ·  83%
Q6 →
Q6
NUMERIC3 marks
A circular park of radius \(20\text{ m}\) 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 to each other. Find the length of the string of each phone.
📘 Concept & Theory Theory and Concepts ›
  • Three points equally spaced on a circle form an equilateral triangle.
  • In an equilateral triangle:
    • All sides are equal
    • All angles are \(60^\circ\)
    • Median, altitude, and perpendicular bisector coincide
  • The centre of the circumscribed circle of an equilateral triangle is also its centroid.
  • The centroid divides the median in the ratio: \[ 2:1 \]
  • Pythagoras Theorem is used to determine the side length of the equilateral triangle.
🗺️ Solution Roadmap Step-by-step Plan ›
  1. Recognize that the boys form an equilateral triangle on the circle.
  2. Draw the median from one vertex to the opposite side.
  3. Use the centroid property of the equilateral triangle.
  4. Find the full height of the triangle using the radius.
  5. Apply Pythagoras Theorem to calculate the side length of the equilateral triangle.
📊 Graph / Figure Graph / Figure ›
A S D O P x x x 20 m

Three equally spaced points on a circle form an equilateral triangle.

Three equally spaced points on a circle form an equilateral triangle.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  20 steps

  1. Given:

    • Radius of the circular park: \[ OA = OS = OD = 20\text{ m} \]
    • The three boys sit at equal distances on the boundary.
  3. Therefore, the points \(A\), \(S\), and \(D\) form an equilateral triangle.

  4. To Find:

  5. Length of the string between each pair of boys.
  6. Let each side of the equilateral triangle be\[x\text{ m}\]

  7. Construction:

  8. Draw median \(AP\) to side \(SD\).
  9. In an equilateral triangle, the median is also perpendicular to the side.
  10. Therefore,\[SP = PD = \frac{x}{2}\]
  11. Also, the centre \(O\) lies on the median.
  12. The centroid divides the median in the ratio:\[AO : OP = 2 : 1\]
  13. Since, \[AO = 20\text{ m}\]
  14. therefore,\[OP = 10\text{ m}\]
  15. Hence, the total height of the triangle is:\[ \begin{aligned}AP &= AO + OP\\AP &= 20 + 10\\&=30\text{ m}\end{aligned}\]

  16. Apply Pythagoras Theorem in right triangle \(APS\):

    \[AS^2 = AP^2 + PS^2\]
  17. Substitute the values: \[ \begin{aligned} x^2 &= 30^2 + \left(\frac{x}{2}\right)^2 \\ x^2 &= 900 + \frac{x^2}{4} \end{aligned} \]
  18. Multiply both sides by \(4\): \[ \begin{aligned} 4x^2 &= 3600 + x^2 \\ 4x^2 - x^2 &= 3600 \\ 3x^2 &= 3600 \\ x^2 &= 1200 \\ x &= \sqrt{1200} \\ x &= 20\sqrt{3} \end{aligned} \]
  19. Taking \[\sqrt{3} \approx 1.732\]
  20. \[ \begin{aligned} x &\approx 20 \times 1.732 \\ &\approx 34.64\text{ m} \end{aligned} \]
💡 Answer Final Answer ›
Therefore, the length of the string of each telephone is \[ \boxed{20\sqrt{3}\text{ m}} \] or approximately \[ \boxed{34.64\text{ m}} \]
🎯 Exam Significance Exam Significance ›
  • This question develops understanding of equilateral triangles inscribed in circles.
  • It demonstrates the relationship between centroid, median, and circumcentre.
  • The problem strengthens the application of Pythagoras Theorem in geometry.
  • Similar geometry-based applications are frequently asked in board examinations and scholarship tests.
  • The concepts are useful in Olympiads, NTSE, and foundation-level entrance examinations.
🔑 Key Takeaways Key Takeaways ›
Key Takeaways  ·  6 points
  1. Equal spacing on a circle forms an equilateral triangle.
  2. In an equilateral triangle:
    • All sides are equal
    • All medians are equal
    • The centroid and circumcentre coincide
  3. The centroid divides the median in the ratio: \[ 2:1 \]
  4. Pythagoras Theorem helps determine unknown side lengths efficiently.
  5. Circle geometry often combines:
    • Triangle properties
    • Symmetry
    • Centroid concepts
    • Pythagoras Theorem
  6. Such integrated geometry problems are important for board exams and competitive entrance examinations.
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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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NCERT Class 9 Circles Exercise 9.2 Solutions
NCERT Class 9 Circles Exercise 9.2 Solutions — Complete Notes & Solutions · academia-aeternum.com
NCERT Mathematics Class IX Chapter 9 "Circles" Textbook Exercise Solutions provide detailed, step-by-step answers to all exercise problems, helping students master crucial concepts about circles, their properties, theorems, and applications. This resource offers clear explanations, diagrams, and exam-focused guidance, making it an ideal companion for scoring high in school exams and competitive tests. From basics to challenging problems, this collection ensures conceptual clarity and boosts…
🎓 Class 9 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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