I’m Up and Down, and Round and Round for Class 9: Important Questions for Complete Revision

Important questions on ‘I’m Up and Down, and Round and Round’ for Class 9 are available in this Maths article to help students understand key concepts related to vertical and circular motion in a simple way. These questions focus on important ideas such as up-and-down movements, round-and-round motion, and their real-life applications with solved examples and exam-oriented explanations. Whether you follow the standard NCERT textbook or Ganita Manjari, this material supports quick revision and better exam preparation

Table of Contents

• Key Concepts of Important Questions on I’m Up and Down, and Round and Round for Class 9
• Important Questions on I’m Up and Down, and Round and Round for Class 9

Key Concepts of Important Questions on I’m Up and Down, and Round and Round for Class 9

This chapter has 12 theorems.

T1: Perpendicular from Centre to Chord

The perpendicular from the centre of a circle to a chord bisects the chord. Converse: the line from the centre bisecting a chord is perpendicular to it.

T2: Equal Chords are equidistant.

Equal chords of a circle (or congruent circles) are equidistant from the centre. And vice versa, chords equidistant from the centre are equal.

T3: Unique Circle through Three Non-Collinear Points

There is one and only one circle passing through three non-collinear points.

T4: Equal Chords ↔ Equal Arcs

In the same circle (or equal circles), equal chords subtend equal arcs; and equal arcs correspond to equal chords.

T5: Central Angle = 2 × Inscribed Angle

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.

T6: Angles in the Same Segment are Equal

Angles in the same segment of a circle (subtended by the same arc) are equal.

T7: Angle in a Semicircle = 90°

The angle subtended by a diameter (semicircle) at any point on the circle is a right angle (90°). Converse is also true.

T8: Converse of T6

If a line segment subtends equal angles at two points on the same side, those two points and the endpoints of the segment are concyclic.

T9: Sum of Angles in a Cyclic Quadrilateral

The sum of opposite angles in a cyclic quadrilateral is 180°. i.e., ∠A + ∠C = 180° and ∠B + ∠D = 180°.

T10: Condition for a Cyclic Quadrilateral

If the sum of opposite angles of a quadrilateral is 180°, it is cyclic (all four vertices lie on a circle).

T11: Exterior Angle of a Cyclic Quadrilateral

An exterior angle of a cyclic quadrilateral equals the interior opposite angle.

T12: Converse of T10

If the exterior angle of a quadrilateral equals the interior opposite angle, the quadrilateral is cyclic.

Important Questions on I’m Up and Down, and Round and Round for Class 9

Q1: Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?

Solution:

Steps of Construction:

Draw the line segment AB = 5 cm.

At point A, construct ∠BAC = 100°.

On the ray AX, mark point C such that AC = 4 cm.

Join C to B. Then ΔABC is formed.

Draw the perpendicular bisector of AB.

Draw the perpendicular bisector of AC and BC.

Let these bisectors meet at O. This point O is the circumcentre.

With centre O and radius OA, draw the circle through A, B and C. This is the circumcircle.

 

Q2: How many circles can pass through three non-collinear points?

Solution: Exactly one. Three non-collinear points determine a unique circle, the circumcircle of the triangle formed by those three points. If the three points were collinear (on a straight line), no circle could pass through all three.

Q3: ABCD is a cyclic quadrilateral. If ∠A = 70°, find ∠C.

Solution: ∠A + ∠C = 180° (opposite angles of cyclic quadrilateral)

70° + ∠C = 180° 

⇒ ∠C = 110°

∴ ∠C = 110°

Q4: The perpendicular from the centre O of a circle bisects a chord AB at M. If OA = 5 cm and OM = 3 cm, find the length of chord AB.

Solution: By Pythagoras: AM² = OA² − OM² = 25 − 9 = 16

AM = √16 = 4 cm

AB = 2 × AM = 8 cm (perpendicular from centre bisects chord)

∴ AB = 8 cm

Q5: Two equal chords of a circle are at distances 4 cm and d cm from the centre. What is the value of d?

Solution: d = 4 cm. Equal chords are equidistant from the centre. So if one equal chord is 4 cm from the centre, the other must also be 4 cm away.

Q6: Prove that angles in the same segment of a circle are equal.

Solution: Given: A circle with centre O. Chord AB. Points P and Q on the same (major) segment. To prove: ∠APB = ∠AQB.

∠AOB = 2∠APB (∵ angle at centre = 2 × angle at circumference, arc AB)

Similarly: ∠AOB = 2∠AQB (same arc AB, point Q also on major arc)

∴ 2∠APB = 2∠AQB 

∴ ∠APB = ∠AQB   (Hence proved)

Q7: In the figure, O is the centre of a circle and ∠BOC = 100°. Find ∠BAC, where A is a point on the major arc.

Solution: Given: ∠BOC = 100° (angle at centre, subtended by arc BC)

Reflex ∠BOC: 360° − 100° = 260° (angle at centre for major arc BC)

∠BAC = (1/2) × ∠BOC (∵ Central Angle = 2 × Inscribed Angle)

∠BAC = 100° ÷ 2 = 50°

∴ ∠BAC = 50°

Q8:ABCD is a cyclic quadrilateral in which ∠A = 3x and ∠C = x. Find both angles.

Solution: ∠A + ∠C = 180°(∵ sum of opposite angles in a cyclic quadrilateral is 180°)

3x + x = 180° 

⇒ 4x = 180° 

⇒ x = 45°

∠A = 3 × 45° = 135°

∠C = 45°

∴ ∠A = 135° and ∠C = 45°

Q9: A chord of a circle of radius 10 cm is at a distance of 6 cm from the centre. Find the length of the chord.

Solution: Radius (OA) = 10 cm

Distance from centre (OM) = 6 cm

By Pythagoras: AM² = OA² − OM² = 100 − 36 = 64 

⇒ AM = 8 cm

Chord AB = 2 × AM = 16 cm (perpendicular bisects chord, Theorem 1)

∴ Length of chord = 16 cm

Q10: In a circle with centre O, ∠AOB = 60° (where A and B are on the circle). If AB = 7 cm, find OA.

Solution: Given ∠AOB = 60° and OA = OB (radii)

Triangle OAB is equilateral (Isosceles + 60° at apex ⇒ all angles 60°)

OA = AB = OB

∴ OA = 7 cm

Q 11: Prove that 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. 

Solution: Given: Arc PQ of a circle with centre O. Point R on the remaining arc. To Prove: ∠POQ = 2∠PRQ.

 

Construction: Join RO and extend it to a point A (outside the circle on the other side).

In triangle OPR: OP = OR (radii of same circle, radii are equal)

∴ ∠OPR = ∠ORP (angles opposite equal sides)

∠POA = ∠OPR + ∠ORP = 2∠ORP (exterior angle of triangle = sum of interior opposite angles)

Similarly, in triangle OQR: ∠QOA = 2∠ORQ

∴ ∠POA + ∠QOA = 2∠ORP + 2∠ORQ, adding both results

∴ ∠POQ = 2∠PRQ   (Hence proved)

Q12: Prove that opposite angles of a cyclic quadrilateral sum to 180°.

Solution: Given: Cyclic quadrilateral ABCD with centre O. 

To Prove: ∠BAD + ∠BCD = 180° and ∠ABC + ∠ADC = 180°.

Arc BCD subtends ∠BAD at point A on the remaining arc (inscribed angle on major arc)

The same arc BCD subtends angle ∠BOD at centre O (central angle on same arc)

∴ ∠BOD = 2∠BAD   … (i)

Arc BAD subtends reflex ∠BOD at centre and ∠BCD at circumference (inscribed angle theorem)

∴ Reflex ∠BOD = 2∠BCD  

∠BOD + Reflex ∠BOD = 360°

From (i) and (ii): 2∠BAD + 2∠BCD = 360°

∴ ∠BAD + ∠BCD = 180°   (Hence proved)

Q13: In a circle with centre O, chords AB and CD are equal. Prove that the arcs AB and CD are also equal.

Solution: Given: Circle with centre O. AB = CD (equal chords). To Prove: Arc AB = Arc CD.

In triangles AOB and COD: OA = OC (radii), OB = OD (radii), AB = CD (given)

∴ △AOB ≅ △COD (SSS congruence rule)

∴ ∠AOB = ∠COD (corresponding parts of congruent triangles)

Equal central angles correspond to equal arcs in the same circle

∴ Arc AB = Arc CD   (Hence proved)

Q14: ABCD is a cyclic quadrilateral. ∠A = 5x + 10° and ∠C = 3x + 20°. ∠B = 4y − 10° and ∠D = 2y + 30°. Find the values of x and y.

Solution: ∠A + ∠C = 180°

(5x + 10) + (3x + 20) = 180

8x + 30 = 180 

⇒ 8x = 150 

⇒ x = 18.75°

∠B + ∠D = 180°

⇒ (4y − 10) + (2y + 30) = 180

6y + 20 = 180 ⇒ 6y = 160 ⇒ y = 26.67°

∴ x = 18.75° and y ≈ 26.67°