Class 9 Maths Ganita Manjari Chapter 5 End of Chapter Exercises

Chord, arc and cyclic quadrilateral questions in Class 9 Maths Ganita Manjari Chapter 5 End of Chapter Exercises revise the full circle chapter. These solutions connect radius, perpendicular distance, chord length, angle in a semicircle and opposite angles of a cyclic quadrilateral.

The End of Chapter Exercises from Chapter 5, I’m Up and Down and Round and Round Class 9, bring together all major circle results from the chapter. Students solve numerical questions on chord length, radius and distance from the centre, then move to proof questions on perpendicular bisectors, cyclic quadrilaterals, rectangles in circles and chords through an interior point. 

Class 9 Maths Ganita Manjari Chapter 5 End of Chapter Exercises contain 26 questions, including starred higher-order questions. These Class 9 Maths Chapter 5 End of Chapter Exercise Solutions follow the textbook order and keep equations in copy-friendly form. The textbook places these exercises after the chapter’s results on chords, arcs, cyclic quadrilaterals and angle properties.

Key Takeaways

• Chord Formula: Chord length is found using radius, perpendicular distance and half-chord.
• Arc Angle: An arc subtends half its central angle at any point on the remaining circle.
• Cyclic Quadrilateral: Opposite angles of a cyclic quadrilateral add up to 180°.
• Diameter: A diameter is the longest chord of a circle.

Class 9 Maths Ganita Manjari Chapter 5 End of Chapter Exercises Structure 2026

Class 9 Maths Ganita Manjari Chapter 5 End of Chapter Exercises Solutions for Chords

The first group of questions uses one right triangle inside a circle. The radius is the hypotenuse, and the perpendicular from the centre bisects the chord.

Q1. In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?

The length of the chord is 24 cm.

Given:

Radius = 13 cm

Distance from centre to chord = 5 cm

Let half of the chord be x.

Using Baudhāyana-Pythagoras theorem:

13² = 5² + x²

169 = 25 + x²

x² = 169 - 25

x² = 144

x = 12 cm

Full chord = 2x

Full chord = 2 × 12

Full chord = 24 cm

Answer:

The length of the chord is 24 cm.

Q2. An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?

The angle subtended by the arc at a point on the circle is 35°.

Given:

Angle at centre = 70°

Rule:

Angle at centre = 2 × angle at circle

So:

70° = 2 × angle at circle

Angle at circle = 70° ÷ 2

Angle at circle = 35°

Answer:

The measure of the angle is 35°.

Q3. The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.

The distance from the centre to the chord is 5 cm.

Given:

Diameter = 26 cm

Radius = 26 ÷ 2

Radius = 13 cm

Chord length = 24 cm

Half chord = 24 ÷ 2

Half chord = 12 cm

Let distance from centre to chord be d.

Using Baudhāyana-Pythagoras theorem:

13² = d² + 12²

169 = d² + 144

d² = 169 - 144

d² = 25

d = 5 cm

Answer:

The distance from the centre to the chord is 5 cm.

Q4. A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?

The length of the chord is 24 cm.

Given:

Radius = 15 cm

Distance from centre to chord = 9 cm

Let half of the chord be x.

Using Baudhāyana-Pythagoras theorem:

15² = 9² + x²

225 = 81 + x²

x² = 225 - 81

x² = 144

x = 12 cm

Full chord = 2x

Full chord = 2 × 12

Full chord = 24 cm

Answer:

The length of the chord is 24 cm.

Ganita Manjari Class 9 Chapter 5 End of Chapter Exercises: Chord Proofs

The perpendicular bisector of a chord passes through the centre because both chord endpoints are equally distant from the centre. This proof uses the locus idea from the chapter.

Q5. Prove that the perpendicular bisector of a chord passes through the centre of the circle.

The perpendicular bisector of a chord passes through the centre because the centre is equidistant from both endpoints of the chord.

Given:

AB is a chord of a circle with centre O.

Let l be the perpendicular bisector of AB.

To prove:

O lies on l.

Proof:

1. OA and OB are radii of the same circle.

Equation:

OA = OB

1. Any point equidistant from A and B lies on the perpendicular bisector of AB.
2. O is equidistant from A and B.
3. Therefore, O lies on the perpendicular bisector of AB.

Answer:

The perpendicular bisector of a chord passes through the centre of the circle.

Q6. The diameter of a circle is AB. Point C is on the circumference. What is the measure of ∠ACB? Explain your reasoning.

The measure of ∠ACB is 90°.

Given:

AB is a diameter.

C lies on the circumference.

Rule:

The angle subtended by a diameter at any point on the circle is 90°.

So:

∠ACB = 90°

Answer:

∠ACB = 90°.

Class 9 Maths Chapter 5 End of Chapter Exercise Solutions for Cyclic Quadrilaterals

A cyclic quadrilateral has all four vertices on one circle. The textbook summary states that opposite angle pairs in a cyclic quadrilateral add up to 180°.

Q7. ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is ∠C? If ∠B measures 110°, what is ∠D?

The values are ∠C = 105° and ∠D = 70°.

In a cyclic quadrilateral:

∠A + ∠C = 180°

∠B + ∠D = 180°

For ∠C:

75° + ∠C = 180°

∠C = 180° - 75°

∠C = 105°

For ∠D:

110° + ∠D = 180°

∠D = 180° - 110°

∠D = 70°

Answer:

∠C = 105° and ∠D = 70°.

Q8. Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find x, ∠P and ∠R.

The value of x is 38, ∠P is 86° and ∠R is 94°.

Since PQRS is cyclic:

∠P + ∠R = 180°

Substitute values:

(2x + 10)° + (3x - 20)° = 180°

5x - 10 = 180

5x = 190

x = 38

Now:

∠P = 2x + 10

∠P = 2 × 38 + 10

∠P = 76 + 10

∠P = 86°

Also:

∠R = 3x - 20

∠R = 3 × 38 - 20

∠R = 114 - 20

∠R = 94°

Answer:

x = 38, ∠P = 86° and ∠R = 94°.

Q9. The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.

The radius of the circle is 10 cm.

Given:

Chord length = 16 cm

Half chord = 8 cm

Distance from centre = 6 cm

Let radius be r.

Using Baudhāyana-Pythagoras theorem:

r² = 6² + 8²

r² = 36 + 64

r² = 100

r = 10 cm

Answer:

The radius of the circle is 10 cm.

Q10. A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.

The area of the cyclic quadrilateral is 60 square units.

Given sides:

5, 5, 12, 12

Semi-perimeter:

s = (5 + 5 + 12 + 12) ÷ 2

s = 34 ÷ 2

s = 17

Using Brahmagupta’s formula:

Area = √[(s - a)(s - b)(s - c)(s - d)]

Area = √[(17 - 5)(17 - 5)(17 - 12)(17 - 12)]

Area = √[12 × 12 × 5 × 5]

Area = √3600

Area = 60 square units

Answer:

The area of the cyclic quadrilateral is 60 square units.

I’m Up and Down and Round and Round Class 9: Higher-Order End Exercises

The starred questions test whether students can extend the chapter results without relying only on fixed diagrams. These questions use perpendicular bisectors, chord symmetry, cyclic quadrilaterals and distance from the centre.

Q11. Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find whether the centre of the circumcircle lies inside or outside?

The best way is to draw perpendicular bisectors of two sides and locate their intersection.

Steps:

1. Take any two sides of the cyclic quadrilateral.
2. Draw the perpendicular bisector of the first side.
3. Draw the perpendicular bisector of the second side.
4. Let both perpendicular bisectors meet at O.
5. O is the circumcentre.
6. Check whether O lies inside or outside the quadrilateral.

Answer:

The centre is found by intersecting perpendicular bisectors of two sides, then checking its position.

Q12. When two chords intersect, show that if the intersecting chords are equal in length, then the line segments of one chord are equal to the corresponding line segments of the other chord.

The corresponding line segments are equal because equal intersecting chords are symmetric around their common intersection.

Let equal chords AB and CD intersect at P.

Given:

AB = CD

So:

AP + PB = CP + PD

For equal intersecting chords, corresponding parts match.

Therefore:

AP = CP

PB = PD

or, according to the labelling of the diagram:

AP = PD

PB = CP

Answer:

The two parts of one chord are equal to the corresponding two parts of the other chord.

Q13. Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.

The required circle has radius 3√2 cm.

Given:

Chord length = 6 cm

Half chord = 3 cm

Distance from centre = 3 cm

Let radius be r.

Using Baudhāyana-Pythagoras theorem:

r² = 3² + 3²

r² = 9 + 9

r² = 18

r = √18

r = 3√2 cm

Construction:

1. Draw OM = 3 cm.
2. At M, draw a line perpendicular to OM.
3. Mark A and B on this line such that AM = 3 cm and MB = 3 cm.
4. Join OA.
5. Draw a circle with centre O and radius OA.

Answer:

The required circle has centre O, chord AB = 6 cm and radius 3√2 cm.

Q14. Show that rectangle is the only parallelogram that can be inscribed in a circle.

A rectangle is the only parallelogram that can be cyclic because each angle becomes 90°.

Given:

ABCD is a parallelogram inscribed in a circle.

In a parallelogram:

∠A = ∠C

In a cyclic quadrilateral:

∠A + ∠C = 180°

So:

∠A + ∠A = 180°

2∠A = 180°

∠A = 90°

Similarly:

∠B = 90°

Answer:

The parallelogram is a rectangle.

Q15. Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals lies at the centre of the circle.

The intersection of the diagonals is the centre because it is equidistant from all four vertices.

Given:

ABCD is a rectangle inscribed in a circle.

Let diagonals AC and BD meet at O.

In a rectangle:

AC = BD

Diagonals bisect each other, so:

OA = OC

OB = OD

Since AC = BD:

OA = OB = OC = OD

Answer:

O is equidistant from A, B, C and D, so O is the centre of the circle.

Q16. Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?

The midpoints form a circle with the same centre as the original circle.

Let original radius be r.

Let fixed chord length be l.

Half chord = l ÷ 2

Let distance of each chord midpoint from centre be d.

Using Baudhāyana-Pythagoras theorem:

d² + (l/2)² = r²

d² = r² - (l/2)²

d = √[r² - (l/2)²]

Since r and l are fixed, d is fixed.

Answer:

The midpoints form a concentric circle.

Q17. In a circle with centre O, chords AB and AC are congruent. Explain why the centre lies on the angle bisector of ∠BAC.

The centre lies on the angle bisector because ΔAOB and ΔAOC are congruent.

Given:

AB = AC

OB and OC are radii.

So:

OB = OC

AO is common.

So:

AO = AO

Now:

AB = AC

OB = OC

AO = AO

Therefore:

ΔAOB ≅ ΔAOC by SSS congruence.

So:

∠BAO = ∠OAC

Answer:

AO bisects ∠BAC, so the centre O lies on the angle bisector of ∠BAC.

Class 9 Maths Circles Solutions: Parallel Chords and Regular Hexagon

Parallel chord questions use half-chord lengths and distance from centre. A regular hexagon in a circle uses six equilateral triangles.

Q18. Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre. The distance between the chords is 7 cm. Find the radius.

The radius of the circle is 13 cm.

Let the distance of the 24 cm chord from the centre be a.

The 10 cm chord is farther from the centre.

So:

Distance of 10 cm chord = a + 7

For the 24 cm chord:

Half chord = 12 cm

r² = a² + 12²

r² = a² + 144

For the 10 cm chord:

Half chord = 5 cm

r² = (a + 7)² + 5²

r² = (a + 7)² + 25

Equate both:

a² + 144 = (a + 7)² + 25

a² + 144 = a² + 14a + 49 + 25

144 = 14a + 74

70 = 14a

a = 5

Now:

r² = a² + 144

r² = 5² + 144

r² = 25 + 144

r² = 169

r = 13 cm

Answer:

The radius of the circle is 13 cm.

Q19. A regular hexagon is inscribed in a circle of radius r. Find the side length and distance of each side from the centre.

The side length of the regular hexagon is r, and the distance of each side from the centre is r√3/2.

In a regular hexagon, the centre joins all vertices.

Each central angle is:

360° ÷ 6 = 60°

Each triangle formed has two radii and included angle 60°.

So each triangle is equilateral.

Therefore:

Side length = r

Distance from centre to each side is the altitude of an equilateral triangle of side r.

Distance = r√3/2

Answer:

Side length = r

Distance of each side from centre = r√3/2

Q20. A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP?

∠MOP and ∠MNP are supplementary angles.

Since MNOP is cyclic, opposite angles add up to 180°.

Here, ∠MOP and ∠MNP are opposite angles.

So:

∠MOP + ∠MNP = 180°

Answer:

∠MOP and ∠MNP are supplementary angles.

Q21. Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle.

The exterior angle equals the interior opposite angle because opposite angles of a cyclic quadrilateral add up to 180°.

Let side CD be extended to E.

In cyclic quadrilateral ABCD:

∠CDA + ∠ABC = 180°

The exterior angle is:

∠CDE = 180° - ∠CDA

So:

∠CDE = ∠ABC

Answer:

The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Class 9 Maths Ganita Manjari Chapter 5 Solutions: Final Proof Questions

The last questions use the chapter’s biggest ideas together: diameter as the longest chord, shortest chord through an interior point, and angle relations in cyclic figures. The chapter summary confirms these results for chords, arcs, diameters and cyclic quadrilaterals.

Q22. “There is no chord of a circle that is longer than its diameter.” How do you justify this statement?

No chord can be longer than the diameter because the diameter passes through the centre.

Let radius be r.

Diameter = 2r

For any chord at distance d from the centre:

Chord length = 2√(r² - d²)

Since d² ≥ 0:

r² - d² ≤ r²

So:

√(r² - d²) ≤ r

Therefore:

Chord length ≤ 2r

Answer:

The diameter is the longest chord of a circle.

Q23. Let A be any point within a given circle with centre O. Show that the shortest chord through A is perpendicular to OA.

The shortest chord through A is perpendicular to OA because it has the greatest distance from the centre.

For any chord through A, its length decreases as its distance from O increases.

The greatest possible perpendicular distance from O to a line through A is OA.

This happens when the chord is perpendicular to OA at A.

Therefore:

Shortest chord through A ⟂ OA

Answer:

The shortest chord through A is the chord perpendicular to OA.

Q24. How would you use Fig. 5.30 to justify that the angle in a semicircle is 90°?

The angle in a semicircle is 90° because the diameter subtends a straight angle at the centre.

Let AB be the diameter.

Angle at centre = 180°

Angle at circle = half of angle at centre

So:

Angle at circle = 180° ÷ 2

Angle at circle = 90°

Answer:

The angle in a semicircle is 90°.

Q25. In a circle, two chords CC' and DD' are drawn perpendicular to a diameter AB. Prove that MM' joining the midpoints of CD and C'D' is perpendicular to AB.

MM' is perpendicular to AB because points on the two sides of AB are mirror images.

Since CC' and DD' are perpendicular to diameter AB, the diameter AB acts as a line of symmetry.

So C and C' are symmetric about AB.

Also:

D and D' are symmetric about AB.

M is the midpoint of CD.

M' is the midpoint of C'D'.

Therefore, M and M' are also symmetric about AB.

The line joining symmetric points is perpendicular to the line of symmetry.

Answer:

MM' ⟂ AB.

Q26. How would you use Fig. 5.31 to justify that the sum of opposite angles of a cyclic quadrilateral is 180°?

Opposite angles of a cyclic quadrilateral add up to 180° because they are half of arcs that together make a full circle.

Let the central angles around O be p, q, u and v.

Full angle around centre:

p + q + u + v = 360°

Each angle of the cyclic quadrilateral is half of the opposite arc.

So the two opposite angles together equal:

1/2 × 360°

= 180°

Answer:

The sum of opposite angles of a cyclic quadrilateral is 180°.

Quick Formula Table for Class 9 Maths Ganita Manjari Chapter 5 End of Chapter Exercises

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5