Circles Class 9 Worksheet (Ganita Manjari Ch 5) with Answers

Class 9 Circles worksheet — "I'm Up and Down, and Round and Round" (Ganita Manjari Ch 5). 75 questions with step-by-step solutions. Free PDF.

This free Worksheet for CBSE Class IX Maths, Chapter 5: I'm Up and Down, and Round and Round, contains a structured worksheet with MCQs, short answer, case-based and HOTS questions in one place. It has been prepared by Sumeet Sahu at Unique Study Point, Indore, strictly following the latest NCERT syllabus for Session 2026-27.

📌 How to use this Worksheet

First revise the chapter — I'm Up and Down, and Round and Round from your notes or textbook.
Attempt every question on your own before checking answers — this is how marks actually improve.
Mark the questions you got wrong and re-attempt them after 2–3 days.

"I'm Up and Down, and Round and Round" — Class 09

UNIQUE STUDY POINT BY SUMEET SAHU

"I'm Up and Down, and Round and Round"

Class 09 · Maths (Ganita Manjari) · Practice Worksheet with Solutions

75 Questions

Tap any question's "Show Answer" button to reveal the full step-by-step solution.

Section A · Multiple Choice / Fill in the Blanks (Q1–Q20)

1 Mark each

The length of an arc of a circle with radius 12 cm is 10π cm. The central angle subtended by this arc at the centre, is:MCQ

a) 120°
b) 150°
c) 6°
d) 75°

✓ Correct Answer: (b) 150°

Radius of circle = 12 cm, Arc length = 10π cm.
Arc length = θ360° × 2πr
10π = θ360° × 2 × π × 12 ⇒ θ = 150°.

Every cyclic parallelogram is a:MCQ

a) square
b) rectangle
c) rhombus
d) isosceles trapezium

✓ Correct Answer: (b) rectangle

Let ABCD be a cyclic parallelogram, so ∠ABC = ∠ADC (opposite angles of a parallelogram are equal).
Also ∠ABC + ∠ADC = 180° (opposite angles of a cyclic quadrilateral are supplementary).
So 2∠ABC = 180° ⇒ ∠ABC = ∠ADC = 90°.
A parallelogram with one angle 90° is a rectangle.

PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =_____.MCQ

a) 18°
b) 41°
c) 67°
d) 23°

✓ Correct Answer: (b) 41°

∠QPS = ∠QPR + ∠SPR = 67° + 72° = 139°.
Since PQRS is cyclic, ∠QRS + ∠QPS = 180°.
∠QRS = 180° − 139° = 41°.

In the given figure, O is the center of the circle and angle OAB = 55°, then angle ACB is equal to:MCQ

Figure for Q4

a) 70°
b) 30°
c) 55°
d) 35°

✓ Correct Answer: (d) 35°

In △OAB, OA = OB (radii), so ∠OBA = ∠OAB = 55°.
By angle sum property: ∠AOB = 180° − 55° − 55° = 70°.
Angle at centre is double the angle at the circumference on the same arc, so ∠ACB = ∠AOB ÷ 2 = 70°2 = 35°.

Let C be the mid-point of an arc AB of a circle such that arc AB = 183°. If the region bounded by the arc ACB and line segment AB is denoted by S, then the centre O of the circle liesMCQ

a) in the interior of S
b) on the segment AB
c) in the exterior of S
d) on AB and bisect AB

✓ Correct Answer: (a) in the interior of S

Since arc AB = 183° is (just) more than a semicircle (180°), arc ACB is a major arc.
The region S bounded by the major arc ACB and chord AB contains the centre of the circle inside it.
So the centre O lies in the interior of S.

AD is a diameter of a circle and AB is a chord. If AD = 34 cm and AB = 30 cm, then the distance of AB from the center of circle isMCQ

a) 17 cm
b) 8 cm
c) 15 cm
d) 4 cm

✓ Correct Answer: (b) 8 cm

Radius OA = AD2 = 342 = 17 cm.
Drop OM ⊥ AB; the perpendicular from the centre bisects the chord, so AM = AB2 = 15 cm.
In right △OAM: OM² = OA² − AM² = 17² − 15² = 289 − 225 = 64.
OM = √64 = 8 cm.

PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =MCQ

a) 18°
b) 41°
c) 67°
d) 23°

✓ Correct Answer: (b) 41°

∠QPS = ∠QPR + ∠RPS = 67° + 72° = 139°.
PQRS is a cyclic quadrilateral, so ∠QPS + ∠QRS = 180°.
∠QRS = 180° − 139° = 41°.

In the given figure, O is the centre of the circle and ∠ACB = 30°. Then, ∠AOB = ?MCQ

Figure for Q8

a) 30°
b) 60°
c) 90°
d) 15°

✓ Correct Answer: (b) 60°

The angle subtended by an arc at the centre is double the angle it subtends at any point on the remaining part of the circumference.
∠AOB = 2∠ACB = 2 × 30° = 60°.

Given O is center of the circle with chord AB = 8 cm, OA = 5 cm and OD ⊥ AB. The length of OD is:MCQ

Figure for Q9

a) 2 cm
b) 3 cm
c) 4 cm
d) 5 cm

✓ Correct Answer: (a) 2 cm

Since OD ⊥ AB, OC (foot of perpendicular) bisects AB, so AC = AB2 = 4 cm.
In right △OAC: OA² = OC² + AC² ⇒ 5² = OC² + 4² ⇒ OC² = 9 ⇒ OC = 3 cm.
OD = OA = 5 cm (both radii), so CD = OD − OC = 5 − 3 = 2 cm.

Consider a quadrilateral with vertices at coordinates A(1,1), B(5,1), C(5,5), D(1,5). Can these points be concyclic?MCQ

a) No, because they form a square and squares are never cyclic.
b) Yes, because the quadrilateral is a square, and the sum of opposite angles is 180°.
c) Yes, but only if its diagonals are equal.
d) No, because only circles can have concyclic points.

✓ Correct Answer: (b) Yes, because the quadrilateral is a square, and the sum of opposite angles is 180°.

Plotting the points shows ABCD is a square, so every interior angle is 90°.
For any pair of opposite angles, the sum is 90° + 90° = 180°.
Since opposite angles are supplementary, the quadrilateral is cyclic — in fact every rectangle (including a square) is cyclic.

The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle, isMCQ

a) 150°
b) 120°
c) 75°
d) 60°

✓ Correct Answer: (a) 150°

Given AO = OB = AB, so △AOB is equilateral, giving ∠AOB = 60°.
Take point Q on the major arc: ∠AQB = ∠AOB2 = 30°.
Take point P on the minor arc; APBQ is a cyclic quadrilateral, so ∠APB + ∠AQB = 180°.
∠APB = 180° − 30° = 150°.

When are two isosceles triangles formed by chords and the center of a circle congruent to each other based on their base lengths?MCQ

a) They are congruent if and only if their radii are different.
b) They are congruent if their base lengths are equal.
c) They are always congruent, regardless of base length.
d) They are congruent only if their central angles are unequal.

✓ Correct Answer: (b) They are congruent if their base lengths are equal.

Each triangle has two sides equal to the (common) radius of the circle.
If the base lengths (the chords) are also equal, all three sides match, so by SSS congruence the triangles are congruent.

In the given figure, O is the center of the circle, chords AB, CD and EF are equal whereas chords BC, DE and FA are separately equal. The angle AOC is equal to:MCQ

Figure for Q13

a) 120°
b) 80°
c) 90°
d) 100°

✓ Correct Answer: (a) 120°

Let ∠AOB = ∠COD = ∠EOF = x and ∠BOC = ∠DOE = ∠AOF = y (equal chords subtend equal angles at the centre).
Sum of all six angles at O = 360°: 3x + 3y = 360° ⇒ x + y = 120°.
∠AOC = ∠AOB + ∠BOC = x + y = 120°.

The sum of the opposite angles of a cyclic quadrilateral is:MCQ

a) 150°
b) 90°
c) 360°
d) 180°

✓ Correct Answer: (d) 180°

This is the cyclic-quadrilateral theorem: opposite angles of a cyclic quadrilateral are always supplementary, i.e. their sum is 180°.

If two circles intersect in two points, then the line through their centres to the bisector of common chord isMCQ

a) Neither parallel nor perpendicular
b) Adjacent
c) perpendicular
d) parallel

✓ Correct Answer: (c) perpendicular

Let circles with centres C, D intersect at A, B, common chord AB, meeting the line CD at M.
In △ACD and △BCD: CA = CB, AD = BD (radii), CD = CD (common) ⇒ △ACD ≅ △BCD, so ∠ACM = ∠MCB.
Then in △ACM and △BCM, using CA = CB, ∠ACM = ∠MCB and CM common, △ACM ≅ △BCM (SAS), giving AM = MB and ∠CMA = ∠CMB.
Since ∠CMA + ∠CMB = 180° (AMB is a line) and they are equal, each is 90°. So the line of centres is the perpendicular bisector of the common chord.

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then the distance of AB from the centre of the circle is _______.MCQ

a) 8 cm
b) 15 cm
c) 4 cm
d) 17 cm

✓ Correct Answer: (a) 8 cm

Radius AO = AD2 = 17 cm; draw OP ⊥ AB, so AP = AB2 = 15 cm (perpendicular from centre bisects the chord).
In right △APO: OP² = 17² − 15² = 289 − 225 = 64 ⇒ OP = 8 cm.

In the given figure, O is center of the circle, AB ∥ DC and ∠ACD = 32°, ∠DAB is equal to:MCQ

Figure for Q17

a) 180°
b) 148°
c) 90°
d) 122°

✓ Correct Answer: (d) 122°

Join DA. Since DC is a diameter’s chord through the pattern shown, ∠DAC = 90° (angle in a semicircle).
AB ∥ DC, so alternate angles give ∠CAB = ∠ACD = 32°.
∠DAB = ∠DAC + ∠CAB = 90° + 32° = 122°.

A pizza with a diameter of 30 cm is cut into several slices. If a cut is made along the longest possible straight line across the pizza, how long is this cut?MCQ

a) 45 cm
b) 60 cm
c) 30 cm
d) 15 cm

✓ Correct Answer: (c) 30 cm

The longest possible straight line segment across a circle is its diameter.
Given diameter = 30 cm, so the length of the cut is 30 cm.

An equilateral triangle of side 12 cm is inscribed in a circle, then the radius of the circle isMCQ

a) 5√3 cm
b) 2√3 cm
c) 3√3 cm
d) 4√3 cm

✓ Correct Answer: (d) 4√3 cm

Let AD be a median of equilateral △ABC; AD ⊥ BC and BD = 122 = 6 cm.
AD = √(AB² − BD²) = √(144 − 36) = √108 = 6√3 cm.
The centroid G (which is also the circumcentre) divides AD in ratio 2 : 1, so radius = AG = 23AD = 23(6√3) = 4√3 cm.

In the given figure, O is center of the circle. Chord BC = chord CD and angle A = 80°. Angle BOC is:MCQ

Figure for Q20

a) 80°
b) 160°
c) 120°
d) 100°

✓ Correct Answer: (a) 80°

Since BC = CD, let ∠BOC = ∠COD = x, so ∠BOD = 2x.
Angle at centre is double the angle at the circumference: ∠BOD = 2∠BAD ⇒ 2x = 2 × 80° ⇒ x = 80°.
So ∠BOC = 80°.

Section B · Match & Short Answer (Q21–Q45)

2 Marks each

Match the following question:

(a) Any straight line segment that’s both endpoints falls on the boundary of the circle	?
(b) Chord which passes through the centre of the Circle	?
(c) Any smooth curve joining two points of circle	?
(d) Circles having same centre and different radii area	?

Options: (i) concentric circles (ii) arc (iii) Diameter (iv) Chord

(a) A line segment with both end-points on the circle is a Chord → (iv).
(b) A chord through the centre is the Diameter → (iii).
(c) A smooth curve joining two points of the circle is an Arc → (ii).
(d) Circles with the same centre and different radii are Concentric circles → (i).

✓ (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)

Match the following:

(a) Opposite sides of a quadrilateral circumscribing a circle subtend ______ angles at the centre of the circle.	?
(b) A circle may have ______ parallel tangents.	?
(c) The tangents drawn at the midpoint of an arc of a circle is ______ to the chord joining the end points of the arc.	?
(d) The tangents at the extremities of any chord of a circle make ______ angles with the chord.	?

Options: (i) Equal (ii) Supplementary (iii) Two (iv) Parallel

(a) Opposite sides of a quadrilateral circumscribing a circle subtend Supplementary angles at the centre → (ii).
(b) A circle can have exactly Two parallel tangents (at the ends of a diameter) → (iii).
(c) The tangent at the midpoint of an arc is Parallel to the chord joining the arc’s end points → (iv).
(d) Tangents at the extremities of a chord make Equal angles with the chord → (i).

✓ (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

Match the following:

(a) The fixed point in the circle	?
(b) A line segment joining any two points on the circle	?
(c) A chord of a circle passing through the centre of the circle	?
(d) The line segment joining the centre to any point on the circle	?

Options: (i) diameter (ii) centre of the circle (iii) radius (iv) chord

(a) The fixed point in the circle is the Centre of the circle → (ii).
(b) A line segment joining any two points on the circle is a Chord → (iv).
(c) A chord passing through the centre is the Diameter → (i).
(d) The segment joining the centre to any point on the circle is the Radius → (iii).

✓ (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

Match the following question:

(a) If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is	?
(b) Parallelogram inscribed in a circle	?
(c) Opposite angles of a cyclic quadrilateral are	?
(d) Any closed shape with all points connected at equidistance from centre	?

Options: (i) supplementary (ii) cyclic (iii) circle (iv) rectangle

(a) A quadrilateral whose opposite angles sum to 180° is Cyclic → (ii).
(b) A parallelogram inscribed in a circle must be a Rectangle → (iv).
(c) Opposite angles of a cyclic quadrilateral are Supplementary → (i).
(d) All points equidistant from a centre form a Circle → (iii).

✓ (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)

Match the following question:

(a) The region between a chord and either of its arcs	?
(b) Angles in the same segment of a circle are	?
(c) The angle subtended by the diameter in a semicircle is a	?
(d) Sum of opposite angles in cyclic quadrilateral	?

Options: (i) 180° (ii) Segment (iii) equal (iv) right angle

(a) The region between a chord and an arc is a Segment → (ii).
(b) Angles in the same segment of a circle are Equal → (iii).
(c) The angle subtended by a diameter is a Right angle → (iv).
(d) Sum of opposite angles of a cyclic quadrilateral is 180° → (i).

✓ (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)

A circle is touching the side BC of △ABC at P and touching AB and AC produced at Q and R respectively. Prove that AQ = ½(perimeter of △ABC).

The two tangents drawn to a circle from an external point are equal in length.
So AQ = AR, BP = BQ and CP = CR.
Perimeter of △ABC = AB + BC + AC = AB + BP + PC + AC.
Replace BP by BQ and PC by CR: = AB + BQ + CR + AC = (AB + BQ) + (CR + AC) = AQ + AR.
Since AQ = AR, perimeter = 2AQ, i.e. AQ = ½(perimeter of △ABC).

✓ AQ = ½(perimeter of △ABC) — proved.

If O is the centre of the circle, find the value of x in given figure:

Figure for Q27

The angles 120°, 90° and ∠BOC together make a full turn at O, so 90° + 120° + ∠BOC = 360° ⇒ ∠BOC = 150°.
Angle at the centre is double the angle at the circumference on the same arc: ∠BOC = 2∠BAC.
150° = 2x° ⇒ x° = 75°.

✓ x = 75

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.

Each interior angle of rectangle ABCD is 90°, so ∠ABC = 90°.
An angle of 90° on the circumference is subtended by a diameter, so diagonal AC must be a diameter.
Similarly, since ∠BCD = 90°, diagonal BD must also be a diameter.
All diameters of a circle pass through exactly one point — the centre.
Since both AC and BD are diameters, their intersection point is necessarily the centre of the circle.

✓ Proved: the diagonals' intersection point is the centre.

ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠ABD.

△ABC and △ADC are right-angled at B and D respectively, so ∠ABC = ∠ADC = 90°.
A circle drawn on AC as diameter must pass through both B and D, since each sees AC at a right angle (angle in a semicircle).
So A, B, C, D are concyclic, with B and D lying in the alternate segments of arc AC.
Chord CD subtends ∠CBD and ∠CAD in the same segment, so these would relate; more directly, chord BD subtends ∠BCD and ∠BAD in the same segment.
Since ∠CAD and ∠CBD are angles in the same segment standing on arc CD, ∠CAD = ∠CBD, i.e. ∠CAD = ∠ABD.

✓ ∠CAD = ∠ABD — proved.

If O is the centre of given circle, find the value of x in the given figure:

Figure for Q30

Given ∠ABD = 40°. Since ∠ACD and ∠ABD are angles in the same segment on chord AD, ∠ACD = ∠ABD = 40°.
In △PCD, by the angle sum property: ∠PCD + ∠CPD + ∠PDC = 180°.
40° + 110° + x = 180° ⇒ x = 30°.

✓ x = 30°

A line segment AB subtends angles ∠ACB = (3x − 12)° and ∠ADB = (2x + 18)° at two points C and D lying on the same side of AB. Find the value of x for which the points A, B, C, D are strictly concyclic.

For A, B, C, D to be concyclic, C and D (on the same side of AB) must see AB at equal angles — angles in the same segment are equal.
So ∠ACB = ∠ADB ⇒ 3x − 12 = 2x + 18.
3x − 2x = 18 + 12 ⇒ x = 30.

✓ x = 30

Find the value of x from the given figure below: Given that points A, B, C, D lie on a circle such that ∠ACB = 58° and ∠ADB = (2x + 10)°.

Angles in the same segment of a circle (subtended by the same arc AB) are equal.
So ∠ADB = ∠ACB ⇒ 2x + 10 = 58.
2x = 58 − 10 = 48 ⇒ x = 24.

✓ x = 24

In the given figure, sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If ∠CBF = 130° and ∠CDE = x°, find the value of x.

Figure for Q33

ABCD is a cyclic quadrilateral, and in a cyclic quadrilateral the exterior angle equals the interior opposite angle.
So ∠CBF = ∠CDA (exterior angle at B equals interior opposite angle at D).
Also ∠CDA = 180° − ∠CDE (since ADE is a straight line), so ∠CBF = 180° − x°.
130° = 180° − x° ⇒ x° = 50°.

✓ x = 50

If O is the center of the circle, find the value of x in the following figure (using the given information):

Figure for Q34

∠ADB = ∠ACB = x° since angles in the same segment (standing on arc AB) are equal.
In △AOD, OD = OA (radii), so ∠ODA = ∠OAD = 62° (angles opposite equal sides are equal).
Comparing, x = 62°.

✓ x = 62°

In the given figure, If O is the centre of the circle then find ∠AOB.

Figure for Q35

OA = OC (radii of the same semi-circle), so ∠OCA = ∠OAC = 20°.
OB = OC (radii), so ∠OCB = ∠OBC = 30°.
∠ACB = ∠OCA + ∠OCB = 20° + 30° = 50°.
Angle at the centre is double the angle at the circumference: ∠AOB = 2∠ACB = 2 × 50° = 100°.

✓ ∠AOB = 100°

Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.

Let ABCD be a quadrilateral whose internal angle bisectors AH, BF, CF and DH form quadrilateral EFGH.
∠FEH = ∠AEB = 180° − ∠EAB − ∠EBA = 180° − ½(∠A + ∠B).
Similarly ∠FGH = ∠CGD = 180° − ½(∠C + ∠D).
Adding: ∠FEH + ∠FGH = 360° − ½(∠A + ∠B + ∠C + ∠D) = 360° − ½(360°) = 180°.
Since a pair of opposite angles of quadrilateral EFGH sum to 180°, EFGH is cyclic.

✓ Quadrilateral EFGH is cyclic — proved.

In a triangle △ABC, a circle is drawn with the side BC as diameter. The circle intersects sides AB and AC at points P and Q respectively. Find the measure of ∠BPC.

BC is given as the diameter of the circle, and P lies on the circle.
The angle subtended by a diameter at any point on the circle is 90° (angle in a semicircle).
Therefore ∠BPC = 90°.

✓ ∠BPC = 90°

A chord AB of length 10 cm is at a distance of 12 cm from the centre of a circle. Find the radius of the circle.

Wait — using AM = 5 cm (half of chord AB = 10 cm) and OM = 12 cm as the given distance from the centre.
In right △OMA: OA² = OM² + AM² = 12² + 5² = 144 + 25 = 169.
OA = √169 = 13 cm.

✓ Radius = 13 cm

In a circle with centre O, chord EF = 14 cm and chord GH = 10 cm. If the distance of EF from O is d₁ and the distance of GH from O is d₂, determine which distance is greater and why.

Between two unequal chords of a circle, the longer chord is always closer to the centre, and the shorter chord is farther away.
Comparing lengths: EF = 14 cm and GH = 10 cm, so EF > GH.
Since EF is longer, it is closer to the centre, meaning d₁ < d₂.
Therefore d₂ (the distance of the shorter chord GH) is the greater distance.

✓ d₂ > d₁ (GH is farther from the centre).

In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P. If ∠DBC = 70° and ∠BAC = 30°, find ∠BCD.

∠BDC = ∠BAC = 30° (angles in the same segment on chord BC are equal).
In △BCD, by angle sum property: ∠BCD + ∠BDC + ∠DBC = 180°.
∠BCD + 30° + 70° = 180° ⇒ ∠BCD = 180° − 100° = 80°.

✓ ∠BCD = 80°

In the given circle with diameter AB, find the value of x.

Figure for Q41

∠ABD = ∠ACD = 30° (angles in the same segment, standing on chord AD, are equal).
Since AB is a diameter, ∠ADB = 90° (angle in a semicircle).
In △ABD, by angle sum property: ∠DAB + ∠ADB + ∠ABD = 180°.
x + 90° + 30° = 180° ⇒ x = 60°.

✓ x = 60°

The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centre.

Let PQ be the chord with centre O, radius 13 cm and PQ = 24 cm. Draw OM ⊥ PQ.
Since the perpendicular from the centre bisects the chord, PM = MQ = 242 = 12 cm.
In right △OMP: OP² = OM² + PM² ⇒ 13² = OM² + 12² ⇒ OM² = 169 − 144 = 25.
OM = 5 cm.

✓ Distance from centre = 5 cm

Two chords PQ and RS of a circle are parallel to each other and AB is the perpendicular bisector of PQ. Without using any construction, prove that AB bisects RS.

Since AB is the perpendicular bisector of chord PQ, AB passes through the centre O (the perpendicular bisector of any chord always passes through the centre).
Since PQ ∥ RS and AB ⊥ PQ, it follows that AB ⊥ RS as well.
AB passes through the centre O and is perpendicular to RS.
A line through the centre, perpendicular to a chord, always bisects that chord — so AB bisects RS.

✓ AB bisects RS — proved.

An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

Let AD be a median of the equilateral △ABC (side 9 cm); AD ⊥ BC and BD = 92 = 4.5 cm.
AD = √(AB² − BD²) = √(9² − 4.5²) = 9√32 cm.
The centroid (= circumcentre for an equilateral triangle) divides the median in ratio 2 : 1.
Radius = AG = 23AD = 239√32 = 3√3 cm.

✓ Radius = 3√3 cm

In the given figure, O is the centre of the circle and △ABC is equilateral. Find:

Figure for Q45

∠BDC
∠BEC.

(i) Since △ABC is equilateral, ∠A = ∠B = ∠C = 60°.
∠BDC = ∠BAC (angles in the same segment on chord BC) = 60°.
(ii) BDCE is a cyclic quadrilateral, so its opposite angles sum to 180°.
∠BDC + ∠BEC = 180° ⇒ 60° + ∠BEC = 180° ⇒ ∠BEC = 120°.

✓ ∠BDC = 60°, ∠BEC = 120°

Section C · Short Answer / Proofs (Q46–Q65)

3 Marks each

In the given figure, the two chords AC and BC are equal and the radius OC intersects AB at M. Then, find the ratio AM : BM.

Figure for Q46

Given AC = BC. In △OAC and △OBC: OA = OB (radii), AC = BC (given), OC = OC (common) ⇒ △OAC ≅ △OBC (SSS).
So ∠OCA = ∠OCB (CPCT), i.e. ∠MCA = ∠MCB.
Now in △AMC and △BMC: AC = BC (given), MC = MC (common), ∠ACM = ∠BCM (proved) ⇒ △AMC ≅ △BMC (SAS).
So AM = BM (CPCT).

✓ AM : BM = 1 : 1

In the given figure, ABCD is a cyclic quadrilateral, O is the centre of the circle. If ∠BOD = 160°, then find the measure of ∠BPD and ∠BCD.

Figure for Q47

Find ∠BPD
Find ∠BCD

Arc BCD subtends ∠BOD = 160° at the centre and ∠BAD at point A on the circumference.
∠BAD = ½∠BOD = 160°2 = 80°.
Since ABPD is a cyclic quadrilateral, ∠BAD + ∠BPD = 180° ⇒ ∠BPD = 180° − 80° = 100°.
∠BPD and ∠BCD are angles in the same segment, so ∠BCD = ∠BPD = 100°.

✓ ∠BPD = 100°, ∠BCD = 100°

AB and BC are two chords of a circle whose centre is O such that ∠ABO = ∠CBO. Prove that AB = CB.

Draw OL ⊥ BC and OM ⊥ BA.
In △OBL and △OBM: ∠OBL = ∠OBM (given), ∠OLB = ∠OMB = 90° (by construction), OB = OB (common) ⇒ △OBL ≅ △OBM (AAS).
So BL = BM (CPCT) ⇒ 2BL = 2BM ⇒ BC = BA.
[L and M are mid-points of BC and BA, since the perpendicular from the centre to a chord bisects it.]

✓ AB = CB — proved.

If two equal chords of a circle intersect within the circle, then prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Let AB and CD be equal chords (AB = CD) intersecting at X, with O the centre.
Draw OM ⊥ AB and ON ⊥ CD.
In △OMX and △ONX: OM = ON (equal chords are equidistant from the centre), ∠OMX = ∠ONX = 90° (construction), OX = OX (common) ⇒ △OMX ≅ △ONX (RHS).
So ∠OXM = ∠OXN (CPCT), i.e. ∠OXA = ∠OXD.

✓ OX makes equal angles with the chords — proved.

If circles are drawn taking two sides of a triangle as diameters, prove the point of intersection of these circles lie on the third side:

Figure for Q50

Let the two circles (on AP and AQ as diameters) intersect at A and B; join A and B.
Since AP is a diameter, ∠1 = ∠ABP = 90° (angle in a semicircle).
Since AQ is a diameter, ∠2 = ∠ABQ = 90° (angle in a semicircle).
∠1 + ∠2 = 90° + 90° = 180°, so ∠PBQ = 180°, meaning P, B, Q are collinear.
Thus point B, the second point of intersection of the two circles, lies on the third side PQ.

✓ The intersection point lies on the third side PQ — proved.

When two chords intersect, each of them 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.

Let AB and CD be two equal chords (AB = CD) intersecting at point P, with O the centre.
Draw OM ⊥ AB and ON ⊥ CD, and join OP.
In △OMP and △ONP: OM = ON (equal chords are equidistant from the centre), ∠OMP = ∠ONP = 90°, OP = OP (common hypotenuse) ⇒ △OMP ≅ △ONP (RHS).
So MP = NP (CPCT). Also since AB = CD, their halves are equal: AM = CN.
Adding: AM + MP = CN + NP ⇒ AP = CP. Subtracting from the full chords: AB − AP = CD − CP ⇒ PB = PD.

✓ AP = CP and PB = PD — the segments of equal chords are equal.

The radii of two concentric circles are 17 cm and 10 cm; a line PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR = 12 cm, calculate PQ.

Draw OM ⊥ QR (and hence ⊥ PS, since P, Q, R, S are collinear).
In the smaller circle, right △OQM: OQ² = OM² + QM² ⇒ 10² = OM² + 6² (since QM = QR2 = 6) ⇒ OM² = 100 − 36 = 64 ⇒ OM = 8 cm.
In the larger circle, right △OPM: OP² = OM² + PM² ⇒ 17² = 8² + PM² ⇒ PM² = 289 − 64 = 225 ⇒ PM = 15 cm.
PQ = PM − QM = 15 − 6 = 9 cm.

✓ PQ = 9 cm

In a triangle △ABC, the altitudes BD and CE are dropped to sides AC and AB respectively. If ∠AED = 65°, find the measure of ∠ACB by establishing concyclicity.

In quadrilateral BCDE, ∠BDC = 90° and ∠BEC = 90° since BD and CE are altitudes.
Since BC subtends equal right angles at D and E on the same side, points B, C, D, E are concyclic (form a cyclic quadrilateral BCDE).
By the exterior-angle property of a cyclic quadrilateral, an exterior angle equals the interior opposite angle.
∠AED is an exterior angle of cyclic quadrilateral BCDE at vertex E, so ∠ACB = ∠AED (i.e. ∠ACB = ∠AED, the interior opposite angle).
Given ∠AED = 65°, we get ∠ACB = 65°.

✓ ∠ACB = 65°

In a circle with centre O, a chord AB subtends an angle of 90° at the centre. If the radius of the circle is 8 cm, find the length of the chord AB leaving your answer in simplest radical form.

Let OA and OB be radii, so OA = OB = 8 cm, and ∠AOB = 90°.
△AOB is a right-angled isosceles triangle at O. By the Pythagoras theorem:
AB² = OA² + OB² = 8² + 8² = 64 + 64 = 128.
AB = √128 = √(64 × 2) = 8√2 cm.

✓ AB = 8√2 cm

If a pair of opposite sides of a cyclic quadrilateral are equal, then the diagonals are also equal.

Let ABCD be a cyclic quadrilateral with AB = DC.
∠1 = ∠2 (angles in the same segment, standing on equal arcs AB and DC, are equal) and similarly ∠3 = ∠4.
Adding: ∠1 + ∠3 = ∠2 + ∠4, i.e. ∠ADC = ∠BAD (angles subtended by the diagonals AC and BD respectively at the circle).
But these angles are subtended by the diagonals AC and BD in the same circle, so equal angles correspond to equal chords.
Therefore AC = BD.

✓ Diagonal AC = Diagonal BD — proved.

Suppose you are given a circle. Describe a method by which you can find the centre of this circle.

Take any two different chords AB and CD of the circle (without using a compass to draw the circle itself, since it is already given).
Draw the perpendicular bisector of chord AB.
Draw the perpendicular bisector of chord CD.
These two perpendicular bisectors intersect each other at a point O.
Since the perpendicular bisector of every chord passes through the centre, O (the common intersection point) is the centre of the given circle.

✓ O, the intersection of the two perpendicular bisectors, is the centre.

In the given figure, ∠OAB = 30° and ∠OCB = 57°. Find ∠BOC and ∠AOC.

Figure for Q57

Find ∠BOC
Find ∠AOC

In △OBC, OB = OC (radii), so ∠OCB = ∠OBC = 57°.
By angle sum property: ∠OCB + ∠OBC + ∠BOC = 180° ⇒ 57° + 57° + ∠BOC = 180° ⇒ ∠BOC = 66°.
In △AOB, OA = OB (radii), so ∠OBA = ∠OAB = 30°.
By angle sum property: ∠OAB + ∠OBA + ∠AOB = 180° ⇒ 30° + 30° + (∠AOC + ∠BOC) = 180°.
60° + ∠AOC + 66° = 180° ⇒ ∠AOC = 180° − 126° = 54°.

✓ ∠BOC = 66°, ∠AOC = 54°

Prove that a cyclic parallelogram is a rectangle.

Let ABCD be a cyclic parallelogram.
Since ABCD is a cyclic quadrilateral, ∠1 + ∠2 = 180° (opposite angles of a cyclic quadrilateral are supplementary) — (1)
Since ABCD is also a parallelogram, its opposite angles are equal: ∠1 = ∠2 — (2)
From (1) and (2): ∠1 = ∠2 = 90°.
A parallelogram with a 90° angle is a rectangle, so ∥gm ABCD is a rectangle.

✓ Cyclic parallelogram ABCD is a rectangle — proved.

AB is an arc of a circle having centre O. If ∠AOB = 60°, then prove that the chord AB is of radius length.

Let O be the centre and r the radius; OA = OB = r, and ∠AOB = 60° is given.
Since OA = OB, ∠OAB = ∠OBA (angles opposite equal sides).
By angle sum property in △OAB: ∠O + ∠A + ∠B = 180° ⇒ 60° + 2∠A = 180° ⇒ ∠A = ∠B = 60°.
So ∠O = ∠A = ∠B = 60°, meaning △OAB is equilateral.
Therefore AB = OA = OB = r.

✓ Chord AB = radius r — proved.

Prove that if chords of congruent circles subtend equal angles at their centre, then the chords are equal.

Let ∠AOB and ∠CO′D be the equal angles subtended by chords AB and CD of two congruent circles with centres O and O′ respectively.
In △OAB and △O′CD: OA = O′C (radii of congruent circles), OB = O′D (radii of congruent circles), ∠AOB = ∠CO′D (given).
By SAS, △OAB ≅ △O′CD.
So AB = CD (CPCT).

✓ AB = CD — proved.

In figure, ABCD is a cyclic quadrilateral. A circle passing through A and B meets AD and BC in the points E and F, respectively. Prove that EF ∥ DC.

Figure for Q61

ABFE is a cyclic quadrilateral (on the smaller circle), so ∠1 + ∠2 = 180° (opposite angles supplementary) — (1)
ABCD is also a cyclic quadrilateral, so ∠1 + ∠3 = 180° (opposite angles supplementary) — (2)
From (1) and (2): ∠2 = ∠3.
But ∠2 and ∠3 form a pair of equal corresponding angles for lines EF and DC cut by the transversal BC (or AD).
Since corresponding angles are equal, EF ∥ DC.

✓ EF ∥ DC — proved.

OC radius equal to chord CD and AB is diameter and AC and BD produced meet at P so prove that ∠CPD = 60°

Join BC. In △OCD: OC = OD (radii) and OC = CD (given), so OC = OD = CD, making △OCD equilateral.
Hence ∠COD = 60°.
Angle at the centre is double the angle at the circumference: ∠CBD = ½∠COD = 30°.
Since AB is a diameter, ∠ACB = 90° (angle in a semicircle). This is the exterior angle of △BCP at C: ∠ACB = ∠CBP + ∠CPB.
90° = 30° + ∠CPB ⇒ ∠CPB = 60°, i.e. ∠CPD = 60°.

✓ ∠CPD = 60° — proved.

ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12√5 cm and BC = 24 cm, find the radius of the circle.

Let O be the centre with radius r cm; AD is the median (also the altitude) from A, so BD = DC = 242 = 12 cm (perpendicular from the vertex of an isosceles triangle bisects the base, and D, O, A are collinear).
In right △ABD: AB² = BD² + AD² ⇒ (12√5)² = 12² + AD² ⇒ 720 = 144 + AD² ⇒ AD² = 576 ⇒ AD = 24 cm.
OD = AD − OA = (24 − r) cm.
In right △OBD: OB² = OD² + BD² ⇒ r² = (24 − r)² + 12² = 576 + r² − 48r + 144.
0 = 720 − 48r ⇒ 48r = 720 ⇒ r = 15 cm.

✓ Radius = 15 cm

ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°; calculate:

Figure for Q64

∠DAB
∠BDC.

(i) ∠DAB = ∠BED = 65° (angles subtended by the same chord DB on the circle, in the same segment, are equal).
(ii) Since AB is a diameter, ∠ADB = 90° (angle in a semicircle).
So ∠ABD = 90° − ∠DAB = 90° − 65° = 25°.
Since AB ∥ DC, ∠BDC = ∠ABD = 25° (alternate angles).

✓ ∠DAB = 65°, ∠BDC = 25°

In the given figure, O is the centre of the circle. If ∠AOB = 140° and ∠OAC = 50°, find:

Figure for Q65

∠ABC
∠BCO
∠OAB
∠BCA

(i) OA = OC (radii), so in △AOC, ∠OCA = ∠OAC = 50°. By angle sum: ∠AOC = 180° − 50° − 50° = 80°.
Angle at centre is double angle at circumference: ∠ABC = ½∠AOC = ½(80°) = 40°.
(ii) ∠BOC = ∠AOB − ∠AOC = 140° − 80° = 60°. Since OB = OC (radii), let ∠OBC = ∠OCB = x; then 2x + 60° = 180° ⇒ x = 60°, so ∠BCO = 60°.
(iii) In △AOB, OA = OB (radii), so ∠OBA = ∠OAB. By angle sum: 2∠OAB + 140° = 180° ⇒ ∠OAB = 20°.
(iv) ∠BCA = ∠BCO + ∠OCA = 60° + 50° = 110°.

✓ ∠ABC = 40°, ∠BCO = 60°, ∠OAB = 20°, ∠BCA = 110°

Section D · Case Study Based (Q66–Q70)

4 Marks each

Case Study: Mr. Vivekananda purchased a plot QRUT to build his house. He leaves space of two congruent semicircles for gardening and a rectangular area of breadth 3 cm for car parking.

Plot QRUT with garden semicircles and parking rectangle (Q66)

Find the total area of Garden. (1)
Find the area of rectangle left for car parking. (1)
Find the radius of semi-circle. (2)
OR — Find the area of a semi-circle. (2)

Radius of each semicircle: the two semicircles are stacked along PS, whose total length equals the 27 cm side.
Diameter of each semicircle = 272 = 13.5 cm ⇒ radius r = 13.52 = 6.75 cm.
(i) Area of one semicircle = ½πr² = ½ × 227 × 6.75² ≈ 71.60 cm².
Total garden area = 2 × 71.60 = 143.20 cm² (two congruent semicircles).
(ii) The parking rectangle is TPSU: length TP = 3 cm, breadth PS = 27 cm.
Area of parking rectangle = 3 × 27 = 81 cm².
(iii) As found above, radius of semicircle = 6.75 cm. OR Area of a semicircle = ½πr² = ½ × 227 × 6.75 × 6.75 ≈ 71.60 cm².

✓ Garden area ≈ 143.20 cm²; Parking area = 81 cm²; Radius = 6.75 cm (semicircle area ≈ 71.60 cm²)

Case Study: A sports stadium has a circular running track with centre O and radius 17 m. Four athletes A, B, C, D are positioned at four points on the track, forming a cyclic quadrilateral. The coach observes that ∠DAB = (2x + 15)° and ∠BCD = (3x − 10)°. Two chords EF and GH represent water pipe lines. EF has length 16 m and is at a distance of 6 m from the centre. GH is at a distance of 8 m from the centre. The coach also notes that arc AB subtends ∠AOB = 80° at the centre O.

Find the value of x and the measures of ∠DAB and ∠BCD. Also find ∠ABC and ∠CDA if ∠ABC = 95°. (1)
Verify that EF = 16 m is consistent with the given radius 17 m and distance 6 m. Then find the length of pipe GH using chord length = 2√(r² − d²). (1)
Arc AB subtends ∠AOB = 80° at centre O. Find the angle ∠ACB subtended by arc AB at any point C on the major arc. Four athletes A, B, C, D all lie on the same circle. What condition on ∠ACB and ∠ADB would confirm that all four are concyclic? (2)
OR — The coach claims that EF is longer than GH since EF is closer to the centre. Verify this by computing GH and comparing. Which theorem is being applied here? State the theorem. Also find which chord is farther from the centre and explain the relationship. (2)

(i) ABCD is cyclic, so opposite angles are supplementary: ∠DAB + ∠BCD = 180°.
(2x + 15) + (3x − 10) = 180 ⇒ 5x + 5 = 180 ⇒ x = 35.
∠DAB = 2(35) + 15 = 85°; ∠BCD = 3(35) − 10 = 95°. Check: 85° + 95° = 180° ✓
Given ∠ABC = 95°, and ∠ABC + ∠CDA = 180° (opposite angles), so ∠CDA = 180° − 95° = 85°.
(ii) Using chord length = 2√(r² − d²) with r = 17 m, d = 6 m: chord = 2√(289 − 36) = 2√253 ≈ 31.81 m.
This does not match the stated EF = 16 m — for a 16 m chord at radius 17 m, the correct distance from the centre works out to √(17² − 8²) = √225 = 15 m, not 6 m. (So the given EF-and-distance pairing in the case study is inconsistent — a useful check to flag.)
For pipe GH, using the stated distance d = 8 m: GH = 2√(r² − d²) = 2√(289 − 64) = 2√225 = 2 × 15 = 30 m.
(iii) Angle at the centre is double the angle at the circumference: ∠ACB = ½∠AOB = ½(80°) = 40°.
For A, B, C, D to be concyclic, points C and D (on the same side of AB) must subtend equal angles at AB: the condition is ∠ACB = ∠ADB (converse of the “angles in the same segment” theorem).
OR: Theorem being applied: “Of two unequal chords of a circle, the one nearer the centre is longer (equivalently, equal chords are equidistant from the centre).” Comparing the two given distances (6 m and 8 m), the chord at distance 6 m is nearer the centre, so — by this theorem — it is the longer chord; the chord at distance 8 m (GH = 30 m) is farther from the centre and shorter, consistent with the general relationship even though the specific 16 m label for EF does not check out exactly.

✓ x = 35; ∠DAB = 85°, ∠BCD = 95°, ∠CDA = 85°; GH = 30 m; concyclic condition: ∠ACB = ∠ADB

Case Study: Given below is the map giving the position of four housing societies in a township connected by a circular road A. Society 2 and 3 are connected by straight road B, society 4 and 2 are connected by straight road C, and society 4 and 3 are connected by road D. Point P denotes the position of a park, equidistant from all four societies. Rubina claims that it is not possible to construct another circular road connecting all four societies.

Township map with four societies and roads (Q68)

Which of the following options justifies Rubina’s claim? a. Equal chords of congruent circles subtend equal angles at the centre. b. The perpendicular from the centre of a circle to a chord bisects the chord. c. There is a unique circle passing through three non-collinear points. d. Points equidistant from a given point will lie on a circle.
What is the position of the park P with respect to road A? a. Chord b. Centre c. Sector d. Segment
The length of Road B is equal to the length of Road D. Which option can be true for the roads in the township? a. Road B bisects Road D. b. Road B and Road C make an acute angle. c. Road B, Road C and Road D are of equal length. d. Road B and Road D subtend equal angles at society 1.
Alex says the angle made by road B on road D is a right angle. Jai says “angles in the same segment of a circle are equal”; Angad says “the angle in a semicircle is a right angle”. Who has given the correct justification?

(i) Since three non-collinear points determine one and only one circle, once societies 1, 2, 3 (any three of the four) fix a unique circle passing through them, no second, different circular road through all four points can exist. So option (c) justifies Rubina’s claim.
(ii) P is equidistant from all four societies, which all lie on circular road A — so P is the point equidistant from every point of the circle, i.e. the Centre. Answer: (b).
(iii) Equal chords of a circle subtend equal angles at the centre; road B and road D are equal chords, so they subtend equal angles at the centre — and by the inscribed-angle relationship this corresponds to equal angles at society 1 (a point on the circle) too. Answer: (d).
(iv) The correct justification is Angad’s: “the angle in a semicircle is a right angle” applies here because the angle in question is subtended by a diameter of the circle. Jai’s statement about equal angles in the same segment is a true theorem in general but does not, by itself, establish that the angle is 90°.

✓ (i) c, (ii) b, (iii) d, (iv) Angad is correct.

Case Study: Sanjay and his mother visited a mall. He observes that three shops are situated at P, Q, R as shown in the figure. Distance between shop P and Q is 8 m and between shop P and R is 6 m. Considering O as the centre of the circle (QR is a diameter).

Circle with shops P, Q, R, S and centre O (Q69)

Find the Measure of ∠QPR. (1)
Find the radius of the circle. (1)
Find the Measure of ∠QSR. (2)
OR — Find the area of △PQR. (2)

(i) QR is a diameter and P lies on the circle, so the angle in a semicircle is 90°: ∠QPR = 90°.
(ii) Since ∠QPR = 90°, by Pythagoras: QR² = PQ² + PR² = 8² + 6² = 64 + 36 = 100 ⇒ QR = 10 m.
Radius = QR2 = 102 = 5 m.
(iii) ∠QSR and ∠QPR are angles in the same segment (both stand on chord QR), so ∠QSR = ∠QPR = 90°.
OR: Area of △PQR = ½ × PQ × PR = ½ × 8 × 6 = 24 m².

✓ ∠QPR = 90°; radius = 5 m; ∠QSR = 90° (or Area of △PQR = 24 m²)

Case Study: Govind has his home located at A and his college located at B. Govind drives his motorbike three days in a week and rides his bicycle in the remaining 3 days, to go to his college and back home. AOB is a sector of a circle with centre O, central angle 60° and radius 4.2 km. Path AOB (via O) is the route for driving by motorbike and path ACB (the arc) is for bicycle only.

Sector AOB — motorbike route via O, bicycle route along arc ACB (Q70)

Find the total distance travelled by Govind through the motorbike in a week to go to college. (1)
Find the total distance travelled by Govind through the bicycle in a week to go to college. (1)
Find the area of sector AOB. (2)
OR — If the cost of fuel for the motorbike is ₹20 per km, then find the total cost of fuel used in a week in going to college. (2)

(i) Motorbike route AOB (one way) = OA + OB = 4.2 + 4.2 = 8.4 km. Round trip per day = 2 × 8.4 = 16.8 km.
Over 3 motorbike days: total = 16.8 × 3 = 50.4 km.
(ii) Arc length ACB (one way) = θ360° × 2πr = 60°360° × 2 × 227 × 4.2 = ⅙ × 26.4 = 4.4 km.
Round trip per day = 2 × 4.4 = 8.8 km. Over 3 bicycle days: total = 8.8 × 3 = 26.4 km.
(iii) Area of sector AOB = θ360° × πr² = 60°360° × 227 × 4.2 × 4.2 = ⅙ × 55.44 = 9.24 km².
OR: Weekly fuel cost = total motorbike distance × rate = 50.4 × ₹20 = ₹1008.

✓ Motorbike (week) = 50.4 km; Bicycle (week) = 26.4 km; Sector area = 9.24 km² (or fuel cost = ₹1008)

Section E · Long Answer / Proofs (Q71–Q75)

5 Marks each

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Figure for Q71

Given: △ABC is isosceles with AB = AC. A circle is drawn with AB as diameter, intersecting the base BC at D.
Construction: Join AD.
Since AB is a diameter and D lies on the circle, ∠ADB = 90° (angle in a semicircle) — call this ∠1.
Since BDC is a straight line, ∠1 + ∠ADC = 180° (linear pair), so ∠ADC = 90° — call this ∠2.
Now in right △ABD and right △ACD: hypotenuse AB = AC (given), AD = AD (common) ⇒ △ABD ≅ △ACD (RHS criterion).
So BD = DC (CPCT), which means D is the mid-point of BC.

✓ The circle on AB as diameter bisects the base BC at D — proved.

OD is perpendicular on chord AB of the circle with centre O. If BC is a diameter, then prove that CA = 2OD.

Figure for Q72

Given OD ⊥ AB. The perpendicular from the centre to a chord bisects the chord, so D is the mid-point of AB, i.e. AD = BD.
O is the mid-point of BC, since BC is a diameter of the circle (centre O lies on every diameter, exactly at its midpoint).
In △ABC, D is the mid-point of AB and O is the mid-point of BC.
By the Mid-point Theorem, OD (joining the mid-points of two sides) is parallel to the third side AC, and OD = ½AC.
So AC = 2 × OD, i.e. CA = 2OD.

✓ CA = 2OD — proved.

In the given figure, O is centre of the circle. If ∠BAD = 75° and BC = CD, then find:

Figure for Q73

∠BOD
∠BCD
∠BOC
∠OBD

(i) Angle at the centre is double the angle at the circumference on the same arc: ∠BOD = 2∠BAD = 2 × 75° = 150°.
(ii) ABCD is a cyclic quadrilateral, so ∠BAD + ∠BCD = 180° ⇒ ∠BCD = 180° − 75° = 105°.
(iii) Since BC = CD (given), equal chords subtend equal angles at the centre: ∠BOC = ∠COD.
∠BOC + ∠COD = ∠BOD = 150° ⇒ 2∠BOC = 150° ⇒ ∠BOC = 75°.
(iv) In △OBD, OB = OD (radii), so ∠OBD = ∠ODB. By angle sum property: ∠BOD + ∠OBD + ∠ODB = 180°.
150° + 2∠OBD = 180° ⇒ 2∠OBD = 30° ⇒ ∠OBD = 15°.

✓ ∠BOD = 150°, ∠BCD = 105°, ∠BOC = 75°, ∠OBD = 15°

In the given figure, a square ABCD is inscribed in a circle with center O. Find: (i) ∠BOC (ii) ∠OCB (iii) ∠COD (iv) ∠BOD. Is BD a diameter of the circle?

Figure for Q74

∠BOC
∠OCB
∠COD
∠BOD, and is BD a diameter?

Since ABCD is a square, its four vertices divide the circumscribing circle into four equal arcs, so each of ∠AOB, ∠BOC, ∠COD, ∠DOA (the angles between consecutive radii) is 360°4 = 90°.
(i) ∠BOC = 90°.
(ii) In △OCB, OB = OC (radii), so it is isosceles with ∠OBC = ∠OCB. By angle sum: ∠OBC + ∠OCB + ∠BOC = 180° ⇒ 2∠OCB + 90° = 180° ⇒ ∠OCB = 45°.
(iii) By the same symmetric reasoning as (i) (square’s diagonals and radii are evenly spaced), ∠COD = 90°.
(iv) ∠BOD = ∠BOC + ∠COD = 90° + 90° = 180°, so B, O, D are collinear.
Since ∠BOD = 180° and O is the centre, BD is a straight line passing through the centre — so BD is indeed a diameter of the circle.

✓ ∠BOC = 90°, ∠OCB = 45°, ∠COD = 90°, ∠BOD = 180°; Yes, BD is a diameter.

M and N are the mid-points of two equal chords AB and CD respectively of a circle with center O. Prove that: (i) ∠BMN = ∠DNM (ii) ∠AMN = ∠CNM.

Figure for Q75

Prove ∠BMN = ∠DNM
Prove ∠AMN = ∠CNM

Construction: Drop OM ⊥ AB and ON ⊥ CD.
Since the perpendicular from the centre bisects the chord, OM bisects AB and ON bisects CD — consistent with M, N being the given mid-points.
Since AB = CD, their halves are equal: BM = ½AB = ½CD = DN — (1)
By Pythagoras in △OMB and △OND: OM² = OB² − BM² = OD² − DN² (using OB = OD, radii, and (1)) = ON².
So OM = ON, which makes △OMN isosceles, giving ∠OMN = ∠ONM — (2)
(i) ∠OMB = ∠OND = 90° each (both are the constructed right angles). Subtracting equation (2) from each: ∠OMB − ∠OMN = ∠OND − ∠ONM, i.e. ∠BMN = ∠DNM.
(ii) Similarly ∠OMA = ∠ONC = 90° each. Adding equation (2) to each: ∠OMA + ∠OMN = ∠ONC + ∠ONM, i.e. ∠AMN = ∠CNM.

✓ ∠BMN = ∠DNM and ∠AMN = ∠CNM — both proved.

📋 Details

Class	Class IX (CBSE / NCERT)
Subject	Maths
Chapter	Chapter 5: I'm Up and Down, and Round and Round
Resource Type	Worksheet
Last Updated	04 September 2026
Session	2026-27 (Latest NCERT Syllabus)
Downloads	0+
Prepared by	Sumeet Sahu, Unique Study Point, Indore
Cost	Free

Circles Class 9 Worksheet (Ganita Manjari Ch 5) with Answers

📌 How to use this Worksheet

"I'm Up and Down, and Round and Round"

Class 09 · Maths (Ganita Manjari) · Practice Worksheet with Solutions

Section A · Multiple Choice / Fill in the Blanks (Q1–Q20)

Section B · Match & Short Answer (Q21–Q45)

Section C · Short Answer / Proofs (Q46–Q65)

Section D · Case Study Based (Q66–Q70)

Section E · Long Answer / Proofs (Q71–Q75)

📋 Details

Algebraic Identities Class 9 PPT Notes PDF – Ganita Manjari

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