Class 9 Maths Ganita Manjari Chapter 5 End of Chapter Exercises – I’m Up and Down, and Round and Round

The NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 End of Chapter Exercises – I’m Up and Down, and Round and Round provide detailed, step-by-step solutions to all the questions given at the end of the chapter. These solutions help students revise the concepts covered in the chapter and develop a clear understanding of different mathematical problems.

Students can use these solutions to practise important questions, verify their answers, and understand the correct approach to solving each problem. The explanations are presented in a simple and easy-to-follow manner, making them useful for homework, revision, and exam preparation. A printable PDF of the solutions is also available for convenient offline practice and revision.

Class 9 Maths Ganita Manjari Chapter 5 End of Chapter Exercises - I’m Up and Down, and Round and Round

NCERT Solutions for Class 9 Maths Chapter 5 End-of-Chapter Exercises

The end-of-chapter exercises include chord-length questions, arc-angle questions, cyclic quadrilateral problems, proof-based questions and construction problems.

Question 1. In a circle, a chord is 5 cm away from the centre. If the radius is 13 cm, what is the length of the chord?

Answer:
Radius = 13 cm
Distance from centre = 5 cm

Half chord = √(13² - 5²)
= √(169 - 25)
= √144
= 12 cm

Chord length = 2 × 12
= 24 cm

Final answer:
24 cm

Question 2. 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?

Answer:
Angle subtended by an arc at the centre is twice the angle subtended at a point on the circle.

So:

Angle at point on circle = 70°/2
= 35°

Final answer:
35°

Question 3. The diameter of a circle is 26 cm. A chord of length 24 cm is drawn. Find the distance from the centre of the circle to the chord.

Answer:
Diameter = 26 cm
Radius = 13 cm
Chord = 24 cm

Half chord = 12 cm

Distance from centre:

d = √(13² - 12²)
= √(169 - 144)
= √25
= 5 cm

Final answer:
5 cm

Question 4. A circle has radius 15 cm. A chord is drawn. The distance from the centre to the chord is 9 cm. What is the length of the chord?

Answer:
Radius = 15 cm
Distance from centre = 9 cm

Half chord = √(15² - 9²)
= √(225 - 81)
= √144
= 12 cm

Chord length = 2 × 12
= 24 cm

Final answer:
24 cm

Question 5. Prove that the perpendicular bisector of a chord passes through the centre of the circle.

Answer:
Let AB be a chord and let its perpendicular bisector meet AB at M.

Every point on the perpendicular bisector of AB is equidistant from A and B.

The centre O of the circle is also equidistant from A and B because:

OA = OB = radius

Therefore, O lies on the perpendicular bisector of AB.

Final answer:
The perpendicular bisector of a chord passes through the centre.

Question 6. The diameter of a circle is AB. Point C is on the circumference. What is the measure of ∠ACB?

Answer:
A diameter subtends a right angle at any point on the circle.

Therefore:

∠ACB = 90°

Final answer:
∠ACB = 90°

Question 7. ABCD is a cyclic quadrilateral. If ∠A = 75°, find ∠C. If ∠B = 110°, find ∠D.

Answer:
Opposite angles of a cyclic quadrilateral add up to 180°.

∠A + ∠C = 180°
75° + ∠C = 180°
∠C = 105°

Also:

∠B + ∠D = 180°
110° + ∠D = 180°
∠D = 70°

Final answer:
∠C = 105°, ∠D = 70°

Question 8. Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x - 20)°, find x and the measures of ∠P and ∠R.

Answer:
Opposite angles of a cyclic quadrilateral add up to 180°.

∠P + ∠R = 180°

(2x + 10) + (3x - 20) = 180
5x - 10 = 180
5x = 190
x = 38

Now:

∠P = 2(38) + 10 = 86°
∠R = 3(38) - 20 = 94°

Final answer:
x = 38, ∠P = 86°, ∠R = 94°

Question 9. The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius.

Answer:
Chord = 16 cm
Half chord = 8 cm
Distance from centre = 6 cm

Radius² = 8² + 6²
= 64 + 36
= 100

Radius = 10 cm

Final answer:
10 cm

Question 10. A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.

Answer:
A cyclic quadrilateral with sides 5, 5, 12, 12 can be treated as an isosceles trapezium.

Let the parallel sides be 12 and 12, and the other two equal sides be 5 and 5. In this case, it forms a rectangle-like cyclic case with height 5.

Area = 12 × 5
= 60 square units

Final answer:
60 square units

Question 11. Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside or outside the quadrilateral?

Answer:
Use the angles of the cyclic quadrilateral.

If the cyclic quadrilateral is acute in a suitable sense and its diagonals are positioned so that the perpendicular bisectors meet inside, the circumcentre lies inside.

If the quadrilateral contains an obtuse angle or the perpendicular bisectors of two sides meet outside the quadrilateral, the circumcentre lies outside.

Best method:

  1. Draw perpendicular bisectors of two sides.
  2. Their point of intersection is the circumcentre.
  3. Check whether this point lies inside or outside the quadrilateral.

Final answer:
Find the intersection of perpendicular bisectors of two sides and check its position.

Question 12. When two chords intersect, each is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.

Answer:
Let equal chords AB and CD intersect at P.

Since AB = CD:

AP + PB = CP + PD

For intersecting chords in a circle:

AP × PB = CP × PD

If two positive pairs have the same sum and same product, then the pairs are equal in corresponding order.

Therefore:

AP = CP and PB = PD

or

AP = PD and PB = CP

Final answer:
Equal intersecting chords are divided into equal corresponding segments.

Question 13. Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.

Answer:
Steps of construction:

  1. Draw a line segment AB = 6 cm.
  2. Find its midpoint M.
  3. Draw a perpendicular line through M.
  4. Mark O on this perpendicular such that OM = 3 cm.
  5. Join OA.
  6. With O as centre and OA as radius, draw a circle.

Since OM is perpendicular to AB and passes through the midpoint of AB, AB is a chord at distance 3 cm from centre O.

Final answer:
The required circle is obtained with centre O and radius OA.

Question 14. Show that rectangle is the only parallelogram that can be inscribed in a circle.

Answer:
In a cyclic quadrilateral, opposite angles add up to 180°.

In a parallelogram, opposite angles are equal.

Let one pair of opposite angles be A and C.

Since it is cyclic:

A + C = 180°

Since it is a parallelogram:

A = C

Therefore:

2A = 180°
A = 90°

So all angles of the parallelogram are 90°.

Final answer:
A cyclic parallelogram must be a rectangle.

Question 15. Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.

Answer:
Let ABCD be a rectangle inscribed in a circle.

The diagonals AC and BD of a rectangle are equal and bisect each other.

Let their point of intersection be O.

Then:

OA = OB = OC = OD

So O is equidistant from all four vertices.

The centre of the circle is also the point equidistant from all points on the circle.

Final answer:
The intersection of the diagonals is the centre of the circle.

Question 16. Consider all chords of a circle of a fixed length. What shape is formed by the midpoints of all these chords?

Answer:
Equal chords are equidistant from the centre.

So, the midpoint of every chord of fixed length is at the same distance from the centre.

The locus of points at a fixed distance from a fixed point is a circle.

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

Question 17. In a circle with centre O, chords AB and AC are congruent. Explain why the centre lies on the angle bisector of ∠BAC.

Answer:
Given:

AB = AC

Also:

OB = OC because both are radii.

OA is common.

So triangles OAB and OAC are congruent by SSS.

Therefore:

∠BAO = ∠OAC

This means AO bisects ∠BAC.

Final answer:
The centre O lies on the angle bisector of ∠BAC.

Question 18. 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.

Answer:
Let radius be r.

For chord 10 cm:

Half chord = 5 cm
Distance from centre = √(r² - 25)

For chord 24 cm:

Half chord = 12 cm
Distance from centre = √(r² - 144)

Since the longer chord is closer to the centre, the distance between the chords is:

√(r² - 25) - √(r² - 144) = 7

Try r = 13:

√(169 - 25) - √(169 - 144)
= √144 - √25
= 12 - 5
= 7

Final answer:
Radius = 13 cm

Question 19. A regular hexagon is inscribed in a circle of radius r. Find the length of each side of the hexagon and the distance of each side from the centre.

Answer:
A regular hexagon divides the circle into 6 equal central angles.

Each central angle = 360°/6 = 60°

Each side of the hexagon is a chord subtending 60° at the centre.

The triangle formed by two radii and one side is equilateral.

So:

Side length = r

Distance from centre to each side:

Draw perpendicular from centre to a side. It bisects the side.

Half side = r/2

Distance d:

d² + (r/2)² = r²
d² = r² - r²/4
= 3r²/4

d = (√3/2)r

Final answer:
Side length = r, distance of each side from centre = (√3/2)r.

Question 20. A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP?

Answer:
Since MN is a diameter, the angle subtended by MN at any point on the circle is 90°.

If O lies on the circle, then:

∠MON or ∠MPN-type angles subtended by diameter MN are 90° depending on the vertex.

For the given angles, both are angles in the cyclic quadrilateral related to arcs and opposite angles.

Since MNOP is cyclic:

∠MOP + ∠MNP = 180°

Final answer:
∠MOP and ∠MNP are supplementary. Their sum is 180°.

Question 21. Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle.

Answer:
Let exterior angle at D be ∠CDE, where E lies on extension of CD.

Since ABCD is cyclic:

∠ADC + ∠ABC = 180°

Also, ∠ADC + ∠CDE = 180° because they form a linear pair.

Therefore:

∠CDE = ∠ABC

Final answer:
The exterior angle of a cyclic quadrilateral equals the interior opposite angle.

Question 22. There is no chord of a circle longer than its diameter. Justify this statement.

Answer:
The diameter passes through the centre and is the longest possible chord.

Any other chord lies at a positive distance from the centre. Its length is:

2√(r² - d²)

where d > 0.

Since r² - d² < r²:

2√(r² - d²) < 2r

But diameter = 2r.

Final answer:
No chord can be longer than the diameter.

Question 23. Let A be any point within a given circle with centre O. Show that the shortest chord passing through A is perpendicular to OA.

Answer:
All chords passing through A have different distances from O.

The chord perpendicular to OA at A has distance OA from the centre. This is the greatest possible perpendicular distance among chords through A.

The farther a chord is from the centre, the shorter it is.

Therefore, the chord through A perpendicular to OA is the shortest.

Final answer:
The shortest chord through A is perpendicular to OA.

Question 24. How would you use the given figure to justify that the angle in a semicircle is 90°?

Answer:
Let AB be a diameter and O be the centre.

The arc AB is a semicircle, so the angle subtended by arc AB at the centre is 180°.

The angle subtended by the same arc at any point on the circle is half the central angle.

So:

Angle in semicircle = 180°/2
= 90°

Final answer:
The angle in a semicircle is 90°.

Question 25. In a circle, two chords CC′ and DD′ are drawn perpendicular to a diameter AB. Prove that the segment MM′ joining the midpoints of CD and C′D′ is perpendicular to AB.

Answer:
Since CC′ and DD′ are perpendicular to diameter AB, they are parallel chords.

The midpoints of corresponding segments formed symmetrically about AB lie on a line parallel to the chords and perpendicular to AB.

By symmetry of the circle about the diameter AB, the midpoint M of CD and midpoint M′ of C′D′ are mirror-related across the diameter.

Therefore, MM′ is perpendicular to AB.

Final answer:
MM′ is perpendicular to AB.

Question 26. How would you use the figure to justify that the sum of opposite angles of a cyclic quadrilateral is 180°?

Answer:
Let ABCD be a cyclic quadrilateral with centre O.

Angle ∠A subtends arc BCD.
Angle ∠C subtends arc BAD.

The two arcs together make the full circle, whose angle at the centre is 360°.

Since angle at the circle is half the corresponding angle at the centre:

∠A + ∠C = 1/2 × 360°
= 180°

Similarly:

∠B + ∠D = 180°

Final answer:
Opposite angles of a cyclic quadrilateral add up to 180°.

NCERT Solutions for Class 9 Maths Chapter 5 - Related Links