Two chords of a circle are equal in length. What can be said about their distance from the centre?

They are equidistant from the centre.

The longer chord is closer to the centre.

They are equidistant from the centre.

The shorter chord is closer to the centre.

Their distances from the centre are unequal.

Which of the following statements is true regarding the relationship between a chord and its distance from the centre?

The longest chord is closest to the centre.

Longer chords are farther from the centre.

Shorter chords are closer to the centre.

The longest chord is closest to the centre.

All chords are equidistant from the centre.

Assertion (A): Infinitely many circles can be drawn through two given points. Reason (R): The centres of all circles passing through two given points lie on the perpendicular bisector of the line segment joining the two points.

A and R both correct, R explains A

Is this Class 9 Maths Chapter 5 quiz free?

Yes, this quiz is completely free for all students. You can attempt it as many times as you like without any charges or sign-ups.

How many questions are in the quiz?

The main 'Game Mode' quiz contains 40 questions covering various topics from Chapter 5, including easy, medium, and hard difficulty levels.

Is this quiz aligned with the NCERT syllabus?

Absolutely! All questions are carefully designed based on the CBSE Class 9 Maths NCERT textbook, Chapter 5: I’m Up and Down, and Round and Round.

Can I find solutions to NCERT textbook questions here?

Yes, the 'Book Questions' section provides solved examples and explanations for selected questions directly from your NCERT textbook exercises for Chapter 5.

What topics does this quiz cover?

This quiz covers definitions of circles and related terms, symmetries, properties of chords, angles subtended by arcs, and concyclicity of points.

How can this quiz help me prepare for exams?

By attempting this quiz, you can test your understanding of key concepts, identify areas where you need more practice, and learn from detailed explanations for each question, helping you to score better in exams.

Home › CBSE › Class 9 › Maths › I’m Up and Down, and Round and Round

CBSE · Class 9 · 🧮 Maths · Chapter 5

I’m Up and Down, and Round and Round

Definition of a circleSymmetries of a circleChords and their propertiesArcs and angles subtendedConcyclicity of pointsCyclic quadrilaterals

Chapter 5 delves into the fascinating world of circles, defining key terms such as centre, radius, chord, diameter, arc, and segment. It explores the symmetries of a circle, the relationship between chords and their distance from the centre, and the angles subtended by arcs. Understanding these concepts is crucial for building a strong foundation in geometry and preparing for higher-level mathematics.

Circle aur Uske Related Terms

Circle ek closed plane figure hai. Is chapter mein hum circles aur unse related concepts ko detail mein padhenge.

Circle: Plane mein un sabhi points ka collection jo ek fixed point (centre) se fixed distance (radius) par hote hain.
Centre (O): Circle ka fixed point. Sabhi points isse equidistant hote hain.
Radius (r): Centre se circle par kisi bhi point tak ka distance. Sabhi radii ki length equal hoti hai.
Chord: Circle par do points ko join karne wala line segment. Jaise AB.
Diameter (d): Sabse lambi chord jo centre se pass hoti hai. Diameter = 2 × Radius.
Arc: Circle ka ek part ya piece. Do points ke beech ka circle ka portion.
Minor Arc: Chhota arc.
Major Arc: Bada arc.
Circumference: Circle ki total length ya perimeter.
Segment: Chord aur arc ke beech ka region.
Minor Segment: Chord aur minor arc ke beech ka region.
Major Segment: Chord aur major arc ke beech ka region.
Sector: Do radii aur arc ke beech ka region.
Minor Sector: Do radii aur minor arc ke beech ka region.
Major Sector: Do radii aur major arc ke beech ka region.
Interior of a Circle: Circle ke andar ka region.
Exterior of a Circle: Circle ke bahar ka region.
Circular Region: Interior of a circle aur circle ko mila kar.

Locus of points: Points ka set jo ek given condition ko satisfy karta hai. Circle, un points ka locus hai jo ek fixed point se equidistant hain.

📖Definition

Circle: Ek plane mein un sabhi points ka collection jo ek fixed point se fixed distance par hote hain.

❗Important

Diameter circle ki sabse lambi chord hoti hai.

Circle ki Symmetries

Circles mein bahut high degree ki symmetry hoti hai.

Reflection Symmetry:
Circle ke kisi bhi diameter ke across reflection symmetry hoti hai.
Agar aap circle ko kisi diameter ke along fold karein, toh dono halves perfectly overlap karenge.
Iska matlab hai ki har diameter circle ke liye ek line of symmetry hai.
Rotational Symmetry:
Circle mein complete rotational symmetry hoti hai.
Agar aap circle ko uske centre ke around kisi bhi angle se rotate karein, toh woh exactly same dikhega.
Is property ki wajah se wheel jaise objects rotate hone par bhi same dikhte hain.

Example: Ek gadi ka wheel. Rotate hone par bhi woh hamesha same dikhta hai, kyunki circle mein rotational symmetry hoti hai.

❗Important

Har diameter circle ki line of symmetry hoti hai.

Points se Pass Hone Wale Circles

Hum dekhenge ki kitne circles ek ya do ya teen points se pass ho sakte hain.

Ek point se:
Ek single point (jaise A) se infinitely many circles pass ho sakte hain.
Aap us point ko centre maan kar koi bhi radius le sakte hain, ya us point ko circumference par rakh kar centre kahin aur bhi le sakte hain.
Do points se:
Do given points (jaise A aur B) se bhi infinitely many circles pass ho sakte hain.
In sabhi circles ka centre line segment AB ke perpendicular bisector par lie karta hai.
Perpendicular Bisector: Ek line jo segment ko do equal parts mein divide karti hai aur us par perpendicular hoti hai.
Perpendicular bisector par har point A aur B se equidistant hota hai. Yahi distance circle ka radius banega.
Smallest Circle: Sabse chhota circle woh hoga jiska diameter AB ho. Iska centre AB ka midpoint hoga aur radius \( \frac{AB}{2} \) hoga.
Teen non-collinear points se:
Teen non-collinear points (jo ek hi line par nahi hain) se exactly one unique circle pass hota hai.
Is circle ko circumcircle kehte hain aur iske centre ko circumcentre.
Circumcentre, triangle ke sides ke perpendicular bisectors ka intersection point hota hai.
Teen collinear points se:
Agar teen points collinear (ek hi line par) hain, toh unse koi circle pass nahi ho sakta.

Construction Steps for Circumcircle:

Teen non-collinear points A, B, C lo.
Line segments AB aur BC ko join karo.
AB ka perpendicular bisector draw karo.
BC ka perpendicular bisector draw karo.
Jahan ye dono perpendicular bisectors intersect karte hain, woh point O (circumcentre) hoga.
O ko centre maan kar aur OA (ya OB ya OC) ko radius maan kar circle draw karo. Ye circle A, B, C se pass hoga.

❗Important

Teen non-collinear points se ek unique circle pass hota hai.

💡Tip

Circumcentre hamesha perpendicular bisectors ke intersection par hota hai. Is concept par construction questions aate hain.

Chords aur Angles jo Wo Centre par Banate Hain

Chords circle ke centre par angles subtend karte hain. In angles ki properties bahut important hain.

Angle Subtended by a Chord at the Centre:
Jab ek chord AB centre O par angle banati hai, toh use \( \angle AOB \) kehte hain.
Agar do chords equal hain, toh woh centre par equal angles subtend karte hain.
Theorem 1: Equal chords of a circle subtend equal angles at the centre.
Given: Circle with centre O, chords AB = CD.
To Prove: \( \angle AOB = \angle COD \).
Proof: \( \triangle AOB \) aur \( \triangle COD \) mein:
OA = OC (Radii)
OB = OD (Radii)
AB = CD (Given)
By SSS congruence, \( \triangle AOB \cong \triangle COD \).
Isliye, \( \angle AOB = \angle COD \) (CPCTC).
Converse of Theorem 1: Agar do chords centre par equal angles subtend karte hain, toh woh chords equal hote hain.
Given: Circle with centre O, \( \angle AOB = \angle COD \).
To Prove: AB = CD.
Proof: \( \triangle AOB \) aur \( \triangle COD \) mein:
OA = OC (Radii)
\( \angle AOB = \angle COD \) (Given)
OB = OD (Radii)
By SAS congruence, \( \triangle AOB \cong \triangle COD \).
Isliye, AB = CD (CPCTC).

Application: Is property ka use karke hum unknown chord lengths ya angles nikal sakte hain.

📐Theorem

Theorem: Equal chords of a circle subtend equal angles at the centre. (Aur iska converse bhi true hai).

Centre se Chord par Perpendicular

Centre se chord par draw kiya gaya perpendicular chord ki properties ko affect karta hai.

Theorem 2: The perpendicular from the centre of a circle to a chord bisects the chord.
Given: Circle with centre O, chord AB, \( OM \perp AB \).
To Prove: AM = MB.
Proof: \( \triangle OMA \) aur \( \triangle OMB \) mein:
OA = OB (Radii)
OM = OM (Common side)
\( \angle OMA = \angle OMB = 90^\circ \) (Given)
By RHS congruence, \( \triangle OMA \cong \triangle OMB \).
Isliye, AM = MB (CPCTC).
Converse of Theorem 2: The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
Given: Circle with centre O, chord AB, M is the midpoint of AB (AM = MB).
To Prove: \( OM \perp AB \).
Proof: \( \triangle OMA \) aur \( \triangle OMB \) mein:
OA = OB (Radii)
AM = MB (Given)
OM = OM (Common side)
By SSS congruence, \( \triangle OMA \cong \triangle OMB \).
Isliye, \( \angle OMA = \angle OMB \) (CPCTC).
Since \( \angle OMA + \angle OMB = 180^\circ \) (Linear Pair), toh \( 2 \angle OMA = 180^\circ \).
\( \angle OMA = 90^\circ \). Hence, \( OM \perp AB \).

Key Takeaway: Ye dono theorems ek dusre ke converse hain aur bahut important hain problems solve karne ke liye.

📐Theorem

Centre se chord par perpendicular chord ko bisect karta hai. Aur iska converse bhi true hai.

💡Tip

Is property ka use karke chord ki length, centre se distance, ya radius nikalne wale questions commonly puche jaate hain. Often Pythagoras theorem use hota hai.

Equal Chords aur Centre se Unki Distance

Chords ki length aur centre se unki distance ke beech ek direct relationship hota hai.

Theorem 3: Equal chords of a circle are equidistant from the centre.
Given: Circle with centre O, chords AB = CD. \( OM \perp AB \) aur \( ON \perp CD \).
To Prove: OM = ON.
Proof: Since \( OM \perp AB \) aur \( ON \perp CD \), toh M aur N midpoints hain AB aur CD ke (Theorem 2).
Toh AM = MB = \( \frac{1}{2} AB \) aur CN = ND = \( \frac{1}{2} CD \).
Since AB = CD, toh AM = CN.
\( \triangle OMA \) aur \( \triangle ONC \) mein:
OA = OC (Radii)
AM = CN (Proved above)
\( \angle OMA = \angle ONC = 90^\circ \) (By construction)
By RHS congruence, \( \triangle OMA \cong \triangle ONC \).
Isliye, OM = ON (CPCTC).
Converse of Theorem 3: Chords equidistant from the centre of a circle are equal in length.
Given: Circle with centre O, \( OM \perp AB \), \( ON \perp CD \) aur OM = ON.
To Prove: AB = CD.
Proof: \( \triangle OMA \) aur \( \triangle ONC \) mein:
OA = OC (Radii)
OM = ON (Given)
\( \angle OMA = \angle ONC = 90^\circ \) (Given)
By RHS congruence, \( \triangle OMA \cong \triangle ONC \).
Isliye, AM = CN (CPCTC).
Since M aur N midpoints hain (Theorem 2), toh AB = 2AM aur CD = 2CN.
Since AM = CN, toh AB = CD.

Unequal Chords:
Agar do chords unequal hain, toh longer chord centre ke zyada paas hoti hai.
Conversely, jo chord centre ke zyada paas hoti hai, woh zyada lambi hoti hai.
Example: Agar AB > CD, toh OM < ON (jahan OM aur ON centre se perpendicular distances hain).

Summary Table:

📐Theorem

Theorem: Equal chords of a circle are equidistant from the centre. (Aur iska converse bhi true hai).

⭐Remember

Chord ki length aur centre se distance ka inverse relation hai: jitni lambi chord, utni kam distance centre se.

Arc dwara Subtended Angle

Arcs bhi circle ke centre par aur circumference par angles subtend karte hain.

Angle Subtended by an Arc at the Centre:
Minor arc AB centre O par \( \angle AOB \) subtend karta hai.
Major arc AB centre par reflex \( \angle AOB \) subtend karta hai.
Angle Subtended by an Arc at any point on the Remaining Part of the Circle:
Arc AB circle ke remaining part par kisi bhi point C par \( \angle ACB \) subtend karta hai.
Theorem 4: 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.
Given: Arc AB, centre O, point C on remaining part of circle.
To Prove: \( \angle AOB = 2 \angle ACB \).
Proof: (Teen cases possible hain: arc minor, major, ya semicircle. Sabhi cases mein proof similar hai, auxiliary construction OD join karke.)
OD ko join karo.
\( \triangle AOC \) mein, OA = OC (Radii), toh \( \angle OAC = \angle OCA \).
Exterior angle \( \angle AOD = \angle OAC + \angle OCA = 2 \angle OCA \).
Similarly, \( \triangle BOC \) mein, OB = OC (Radii), toh \( \angle OBC = \angle OCB \).
Exterior angle \( \angle BOD = \angle OBC + \angle OCB = 2 \angle OCB \).
\( \angle AOB = \angle AOD + \angle BOD = 2 \angle OCA + 2 \angle OCB = 2 (\angle OCA + \angle OCB) = 2 \angle ACB \).
Agar major arc hai, toh reflex angle use hoga.

Corollaries of Theorem 4:
Angles in the same segment are equal: Agar do points C aur D same segment mein hain, toh \( \angle ACB = \angle ADB \).
Proof: \( \angle AOB = 2 \angle ACB \) aur \( \angle AOB = 2 \angle ADB \).
Isliye, \( \angle ACB = \angle ADB \).
Angle in a semicircle is a right angle: Diameter AB circle ke kisi bhi point C par \( 90^\circ \) ka angle subtend karta hai (\( \angle ACB = 90^\circ \)).
Proof: Diameter AB centre par \( 180^\circ \) ka angle subtend karta hai (straight line).
Toh \( \angle AOB = 180^\circ \).
By Theorem 4, \( \angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} (180^\circ) = 90^\circ \).

Concyclic Points: Agar ek line segment do other points par same side mein equal angles subtend karta hai, toh woh charo points concyclic hote hain (ek hi circle par lie karte hain).

📐Theorem

Theorem: Centre par bana angle, remaining part par bane angle ka double hota hai.

❗Important

Angles in the same segment are equal aur Angle in a semicircle is 90°, ye dono Theorem 4 ke direct results hain.

Cyclic Quadrilaterals

Jab ek quadrilateral ke sabhi vertices ek circle par lie karte hain, toh use cyclic quadrilateral kehte hain.

Definition: A quadrilateral ABCD whose all four vertices lie on a circle is called a cyclic quadrilateral.
Theorem 5: The sum of either pair of opposite angles of a cyclic quadrilateral is \( 180^\circ \).
Given: Cyclic quadrilateral ABCD.
To Prove: \( \angle A + \angle C = 180^\circ \) aur \( \angle B + \angle D = 180^\circ \).
Proof: Centre O se A aur C ko join karo.
Arc ADC centre par \( \angle AOC \) subtend karta hai aur remaining part B par \( \angle ABC \) subtend karta hai.
Toh \( \angle AOC = 2 \angle ABC \) (Theorem 4).
Similarly, Arc ABC centre par reflex \( \angle AOC \) subtend karta hai aur remaining part D par \( \angle ADC \) subtend karta hai.
Toh reflex \( \angle AOC = 2 \angle ADC \).
\( \angle AOC + \text{reflex } \angle AOC = 360^\circ \).
\( 2 \angle ABC + 2 \angle ADC = 360^\circ \).
\( 2 (\angle ABC + \angle ADC) = 360^\circ \).
\( \angle ABC + \angle ADC = 180^\circ \) (i.e., \( \angle B + \angle D = 180^\circ \)).
Similarly, \( \angle A + \angle C = 180^\circ \) prove kar sakte hain.
Converse of Theorem 5: If the sum of a pair of opposite angles of a quadrilateral is \( 180^\circ \), then the quadrilateral is cyclic.
Given: Quadrilateral ABCD jismein \( \angle B + \angle D = 180^\circ \).
To Prove: ABCD ek cyclic quadrilateral hai.
Proof (By contradiction): Maan lo ABCD cyclic nahi hai. Toh point D circle par nahi hai. Maan lo D' circle par hai.
Toh ABCD' ek cyclic quadrilateral hai. Isliye \( \angle B + \angle D' = 180^\circ \).
Lekin humein diya hai \( \angle B + \angle D = 180^\circ \).
Toh \( \angle D = \angle D' \). Ye tabhi possible hai jab D aur D' coincide karein. Iska matlab D circle par hai. Toh ABCD cyclic hai.

Exterior Angle of a Cyclic Quadrilateral:

Ek cyclic quadrilateral ka exterior angle uske interior opposite angle ke equal hota hai.
\( \angle ADE = \angle ABC \) (jahan E, CD ko extend karne par point hai).
Proof: \( \angle ADC + \angle ABC = 180^\circ \) (Cyclic Quad).
\( \angle ADC + \angle ADE = 180^\circ \) (Linear Pair).
Toh \( \angle ABC = \angle ADE \).

Concyclic Points: Char points tab concyclic hote hain jab unse ek circle pass ho sake. Cyclic quadrilateral ki property concyclic points ko determine karne mein help karti hai.

📐Theorem

Theorem: Cyclic quadrilateral ke opposite angles ka sum \( 180^\circ \) hota hai. (Aur iska converse bhi true hai).

❗Important

Cyclic quadrilateral ka exterior angle uske interior opposite angle ke equal hota hai. Ye property bhi bahut use hoti hai.

Easy (15)

What is the set of all points on a plane that are equidistant from a given point on that plane called?
- Square
- Triangle
- Circle (correct)
- Rectangle
Why: A circle is defined as the locus of points equidistant from a central point.
The distance from the centre to any point on the circle is known as its:
- Diameter
- Chord
- Radius (correct)
- Arc
Why: The radius is the distance from the centre to any point on the circle.
A line segment connecting two points on a circle is called a:
- Tangent
- Secant
- Chord (correct)
- Radius
Why: A chord connects two points on the circumference of a circle.
Which of these is a line of reflection symmetry for a circle?
- Any chord
- Any tangent
- Any diameter (correct)
- Any radius
Why: A diameter acts as a line of reflection symmetry for a circle.
The longest chord of a circle is its:
- Radius
- Arc
- Diameter (correct)
- Segment
Why: The diameter is the longest chord, passing through the centre.
How many circles can pass through two given points A and B?
- One
- Two
- Finite number
- Infinitely many (correct)
Why: Infinitely many circles can pass through two given points.
What is a connected portion of a circle called?
- Segment
- Sector
- Arc (correct)
- Chord
Why: An arc is a connected part of the circumference of a circle.
Points that lie on the same circle are called:
- Collinear
- Concyclic (correct)
- Concurrent
- Coplanar
Why: Points lying on the same circle are referred to as concyclic points.
A circle has complete rotational symmetry.
Why: A circle looks the same after rotation by any angle about its centre.
The perpendicular bisector of a chord always passes through the centre of the circle.
Why: This is a fundamental property: the perpendicular bisector of any chord passes through the circle's centre.
The radius of the smallest circle passing through two points A and B is half the length of AB.
Why: The smallest circle has AB as its diameter, so its radius is AB/2.
A chord passing through the centre of a circle is called a __________.
Why: The diameter is the special chord that passes through the centre.
The set of points that satisfy a given condition is also called the __________ of points that satisfy the condition.
Why: Locus refers to the path traced by a point satisfying a given condition.
The bigger of the two arcs between two points on a circle is called the __________ arc.
Why: The larger arc connecting two points is called the major arc.
A quadrilateral whose vertices are concyclic is called a __________ quadrilateral.
Why: A quadrilateral with all its vertices on a circle is a cyclic quadrilateral.

Medium (15)

If the radius of a circle is 7 cm, what is the length of its diameter?
- 3.5 cm
- 7 cm
- 14 cm (correct)
- 21 cm
Why: Diameter is twice the radius, so 2 * 7 cm = 14 cm.
Two chords of a circle are equal in length. What can be said about their distance from the centre?
- The longer chord is closer to the centre.
- They are equidistant from the centre. (correct)
- The shorter chord is closer to the centre.
- Their distances from the centre are unequal.
Why: Equal chords in a circle are always equidistant from the centre.
If an arc subtends an angle of 60° at the centre of a circle, what angle does it subtend at any point on the remaining part of the circle?
- 30° (correct)
- 60°
- 90°
- 120°
Why: The angle subtended by an arc at the centre is twice the angle subtended at any point on the remaining part of the circle.
Which of the following statements is true regarding the relationship between a chord and its distance from the centre?
- Longer chords are farther from the centre.
- Shorter chords are closer to the centre.
- The longest chord is closest to the centre. (correct)
- All chords are equidistant from the centre.
Why: The longest chord (diameter) passes through the centre, making its distance from the centre zero.
If a line segment AB subtends equal angles at two other points C and D on the same side of AB, what can be concluded about points A, B, C, D?
- They form a square.
- They are collinear.
- They are concyclic. (correct)
- They form a parallelogram.
Why: If a segment subtends equal angles at two points on the same side, those four points are concyclic.
Match the following terms with their definitions:
Why: These are the standard definitions for basic circle components.
Match the following properties with their corresponding terms:
Why: This matches key properties and definitions from the chapter.
Match the following arc types with their descriptions:
Why: These terms describe different parts of a circle's circumference and associated angles.
What is the term for the set of points equidistant from a given point?
Why: A circle is defined by points equidistant from a central point.
What geometric figure is formed by the locus of points equidistant from two given points?
Why: The perpendicular bisector contains all points equidistant from two given points.
Arrange the following terms in increasing order of their typical length in a given circle: Radius, Diameter, Chord (not diameter).
Why: Radius is half the diameter, and a non-diameter chord is shorter than the diameter.
Arrange the following steps to find the centre of a circle using paper folding:
Why: The intersection of perpendicular bisectors of two chords gives the centre.
In the given figure, O is the centre of the circle. If AB is a chord and OM ⊥ AB, what is the relationship between AM and MB?
- AM > MB
- AM < MB
- AM = MB (correct)
- No relation can be determined
Why: A perpendicular from the centre to a chord bisects the chord.
Observe the given figure. If chord PQ is longer than chord RS, which chord is closer to the centre O?
- Chord RS
- Chord PQ (correct)
- Both are equidistant
- Cannot be determined
Why: The longer chord is always closer to the centre of the circle.
In the given circle with centre O, if ∠AOB = 100°, what is the measure of ∠ACB?
- 50° (correct)
- 100°
- 200°
- 80°
Why: The angle at the circumference is half the angle at the centre subtended by the same arc.

Hard (10)

A chord of length 16 cm is drawn in a circle of radius 10 cm. What is the distance of the chord from the centre of the circle (in cm)?
Why: Use Pythagoras theorem: (radius)^2 = (half chord)^2 + (distance from center)^2.
In a circle, two chords AB and CD are equal and intersect at point P. If the radius of the circle is 5 cm and the length of each chord is 8 cm, what is the distance of each chord from the centre (in cm)?
Why: Equal chords are equidistant from the centre. Use Pythagoras theorem with half chord and radius.
If a cyclic quadrilateral ABCD has ∠A = 70°, what is the measure of ∠C (in degrees)?
Why: Opposite angles of a cyclic quadrilateral are supplementary (sum to 180°).
A circular park has a radius of 13 meters. A straight path of length 24 meters is constructed within the park. What is the shortest distance (in meters) from the centre of the park to this path?
Why: The path is a chord. Use Pythagoras theorem with radius and half-chord length.
Assertion (A): Infinitely many circles can be drawn through two given points. Reason (R): The centres of all circles passing through two given points lie on the perpendicular bisector of the line segment joining the two points.
Why: The reason correctly explains why infinite circles can pass through two points.
Assertion (A): A diameter is the longest chord of a circle. Reason (R): All chords of a circle are equidistant from the centre.
Why: Assertion is true, but the reason is false; only equal chords are equidistant from the centre.
A circular clock face has its minute hand as a radius of 10 cm. At 3:00 PM, the minute hand points to 12 and the hour hand points to 3. What is the length of the chord connecting the tips of the hour and minute hands (in cm)? (Assume the hour hand is 7 cm long and the clock is a perfect circle.)
- √149 (correct)
- √100
- √49
- 17
Why: The angle between hands at 3 PM is 90°. Use Pythagoras theorem with the lengths of the hands.
A circular garden has three points A, B, and C on its boundary, forming an equilateral triangle. If the side length of the triangle is 6√3 meters, what is the radius of the circular garden (in meters)?
- 6 (correct)
- 9
- 12
- 3√3
Why: For an equilateral triangle inscribed in a circle, Radius = side / √3.
In the given figure, O is the centre of the circle. If ∠OAB = 30°, what is the measure of ∠ACB?
- 30°
- 60° (correct)
- 90°
- 120°
Why: Triangle OAB is isosceles. Find ∠AOB, then ∠ACB is half of ∠AOB.
In the given figure, AB is a diameter of the circle with centre O. If ∠AOC = 130°, find ∠ABC.
- 25° (correct)
- 50°
- 65°
- 130°
Why: ∠BOC = 180° - ∠AOC. ∠ABC is half of ∠AOC.