Circles

Let AB be a chord of a circle with center O. OCAB, then

AB = 16 cm, and OA = 10 cm.

OCAB

Therefore,

OC bisects AB at C

AC = (1/2) AB

⇒ AC = (1/2) 16

⇒ AC = 8 cm

In triangle OAC,

OA² = OC² + AC²

⇒ 10² = OC² + 8²

⇒ 100 = OC² + 64

⇒ OC²= 36

⇒ OC= 6

Let distance OC = 3 cm

Radius = OA = 5 cm

Draw OCAB

In triangle OCA,

OA² = OC² + AC²

⇒ 5² = 3² + AC²

⇒ AC²= 16

⇒ AC = 4 cm __________________ (i)

Now,

AB = 2 AC

⇒ AB = 8 cm [From equation (i)]

Hence, length of a chord = 8 cm.

Let distance OC = 8 cm

Chord AB = 30 cm

Draw OCAB

Therefore,

OC bisects AB at C

AC = (1/2) AB

⇒ AC = (1/2) 30

⇒ AC = 15 cm

In triangle OCA,

OA² = OC² + AC²

⇒ OA² = 8² + 15²

⇒ OA²= 64 + 225

⇒ OA²= 289

⇒ OA = 17 cm

Hence, radius of the circle = 17 cm.

(i)

Let radius OB = OD = 5 cm

Chord AB = 8 cm

Chord CD = 6 cm

BP = (1/2) AB

⇒ BP = (1/2) 8 = 4 cm

DQ = (1/2) CD

⇒ DQ = (1/2) 6 = 3 cm

In triangle OPB,

OP² = OB² - BP²

⇒ OP² = 5² - 4²

⇒ OP²= 25 - 16

⇒ OP²= 9

⇒ OP = 3 cm

In triangle OQD,

OQ² = OD² - DQ²

⇒ OQ² = 5² - 3²

⇒ OQ²= 25 - 9

⇒ OQ²= 16

⇒ OQ = 4 cm

Now,

PQ = OQ – OP = 4 – 3 = 1

Hence, distance between chords = 1 cm.

(ii)

Let radius OA = OC = 5 cm

Chord AB = 8 cm

Chord CD = 6 cm

AP = (1/2) AB

⇒ AP = (1/2) 8 = 4 cm

CQ = (1/2) CD

⇒ CQ = (1/2) 6 = 3 cm

In triangle OAP,

OP² = OA² - AP²

⇒ OP² = 5² - 4²

⇒ OP²= 25 - 16

⇒ OP²= 9

⇒ OP = 3 cm

In triangle OQD,

OQ² = OC² - CQ²

⇒ OQ² = 5² - 3²

⇒ OQ²= 25 - 9

⇒ OQ²= 16

⇒ OQ = 4 cm

Now,

PQ = OP + OQ = 3 + 4 = 7

Hence, distance between chords = 7 cm.

Let radius OA = OC = 17 cm

Chord AB = 30 cm and CD = 16 cm

Draw OL and OM

Therefore,

AP = (1/2) AB

⇒ AP = (1/2) 30 = 15 cm

CQ = (1/2) CD

⇒ CQ = (1/2) 16 = 8 cm

In triangle OAP,

OP² = OA² - AP²

⇒ OP² = 17² - 15²

⇒ OP²= 289 - 225

⇒ OP²= 64

⇒ OP = 8 cm

In triangle OQD,

OQ² = OC² - CQ²

⇒ OQ² = 17² - 8²

⇒ OQ²= 289 - 64

⇒ OQ²= 225

⇒ OQ = 15 cm

Now,

PQ = OP + OQ = 8 + 15 = 23

Hence, distance between chords = 23 cm.

Let radius OA = OC = OD = r

Chord AB = 12 cm

OE = OC – CE

⇒ OE = r – 3

AE = (1/2) AB

⇒ AE = (1/2) 12 = 6 cm

In triangle AOE,

OA² = AE² + OE²

⇒ r² = 6² + (r - 3)²

⇒ r² = 36 + r² + 9 - 6r

⇒ 6r = 45

⇒ r = 7.5 cm

Hence, radius of circle = 7.5 cm.

Let radius OA = OB = OD = r

DE = 8 cm

OE = OB – BE

⇒ OE = r – 4

In triangle ODE,

OD² = DE² + OE²

⇒ r² = 8² + (r - 4)²

⇒ r² = 64 + r² + 16 - 8r

⇒ 8r = 80

⇒ r = 10 cm

Hence, radius of circle = 10 cm.

Given ODAB

In triangle ABC,

D is the mid-point of AB

∴ AD = DB

O is the mid-point of BC

∴ OC = OB

We say, AC||OD

(1/2) AC = OD [Mid-point theorem in triangle ABC]

⇒ AC = 2 × OD Proved.

Proof

In ΔOEP and ΔOFP,

∠OEP = ∠OFP [equal to 90°]

OP = OP [common]

∠OPE = ∠OPF [OP bisects ∠BPD]

Therefore,

ΔOEP = ΔOFP [By angle-side-angle]

∴ OE = OF

AB = CD [Chords are equidistant from the center]

Hence, AB = CD Proved.

∠PFD = ∠PEB [equal to 90°]

∴PFCD and OFCD

We know that the perpendicular from the center of a circle to chord, bisect the chord.

Therefore,

CF = FD Proved.

Let two different circles intersect at three distinct points A, B and C.

Then, these points are already non-collinear.

A unique circle can be drawn to pass through these points. This is a contradiction.

Hence, two different circles cannot intersect each other at more than two points.

Let,

Radius OA = 10 cm and O’A = 8 cm

Chord AB = 12 cm

Now,

AD = (1/2) AB

⇒ AD = (1/2) 12 = 6 cm

In triangle OAD,

OD² = OA² - AD²

⇒ OD² = 10² - 6²

⇒ OD²= 100 - 36

⇒ OD²= 64

⇒ OD = 8 cm

In triangle O’AD,

O’D² = O’A² - AD²

⇒ O’D² = 8² - 6²

⇒ O’D²= 64 - 36

⇒ O’D²= 28

⇒ O’D = 2 √7 cm

Now,

OO’ = OD + O’D =

Hence, distance between their centers =

Join PQ,

PQ is the common chord of both the circles.

Thus,

arc PCQ = arc PDQ

∴ ∠QAP = ∠QBP

∴ QA = QB Proved.

Let AB and CD are two chords of a circle with center O.

Diameter POQ bisect s them at L and M.

Then,

OLAB and OMCD

∴ ∠ALM = ∠LMD

∴ AB||CD [Alternate angles]

Join AP.

Let PQ intersect AB at L,

Then, AB = 5 – 3 = 2 cm

PQ is the perpendicular bisector of AB,

Then,

AL = (1/2) AB

⇒ AL = (1/2) 2 = 1 cm

In triangle APL,

PL² = PA² - AL²

⇒ PL² = 5² - 1²

⇒ PL²= 25 - 1

⇒ PL²= 24

⇒ PL = 2 √6 cm

Now,

PQ = 2 PL

⇒ PQ = 2 × 2 √6

⇒ PQ = 4 √6 cm

Given, OB = OC

Then, ∠BOC = ∠BCO = y°

External ∠OBA = ∠BOC + ∠BCO = (2y)°

Now,

OA = OB

Then, ∠OAB = ∠OBA = (2y)°

External ∠AOD = ∠OAC + ∠ACO

= ∠OAB + ∠BCO = (3y)°

∴ x° = (3y)°[Given ]

Given AB = AC

∴ (1/2)AB = (1/2)AC

OPAB and OQAC

∴ MB = NC

⇒ ∠PMB = ∠QNC [90°]

Equal chords are equidistant from the center.

⇒ OM = ON

OP = OQ

⇒ OP – OM = OQ – ON

⇒ PM = QN

∴ ΔABC ≅ ΔABC [By side-angle-side criterion of congruence]

∴ PB = QC Proved.

Draw, OPAB and OQCD

In triangle OBP and triangle OQC,

∠OPB = ∠OQC [Angle = 90°]

∠OBP = ∠OCD [Alternate angle]

OB = OC [Radius]

By side-angle-side criterion of congruence

ΔOBP ≅ ΔOQC

∴ OP = OQ

The chords equidistant from the center are equal.

∴ AB = CD Proved.

Let ABC be an equilateral triangle of side 9 cm.

And AD be one of its medians.

Then,

ADBC

BD = (1/2) BC

⇒ BD = (1/2) 9 = 4.5 cm

In triangle ADB,

AD² = AB² - BD²

⇒ AD² = 9² – (9/2)²

⇒ AD² = 81 – (81/4)

⇒ AD = (9√3)/2

In an equilateral triangle the centroid and circumcenter coincide and AO: OD = 2: 1

∴ radius AO = (2/3) AD

= (2/3) (9√3)/2

= 3√3 cm

Hence, radius of circle = 3√3 cm.

In triangle OAB and triangle OAC,

AB = AC [Given]

OB = CO [Radius]

OA = OA [Common]

By side-side-side criterion of congruence

ΔOAB ≅ ΔOAC

∴ ∠OAB = ∠OAC Proved.

In triangle OPX and triangle ORY,

OX = OY [Radius]

∠OPX = ∠ORY [Common]

OP = OR [Sides of square]

By side-angle-side criterion of congruence,

ΔOPX ≅ ΔORY

∴ PX = RY

⇒ PQ – PX = QR – RY [PQ = QR]

⇒ QX = QY Proved.

_{(i) Join OB.}

∠OAB = ∠OBA = 40°[Because OB = OA]

∠OCB = ∠OBC = 30°[Because OB = OC]

∠ABC = ∠OBA + ∠OBC

⇒ ∠ABC = 40°+ 30°

⇒ ∠ABC = 70°

∠AOC = 2 × ∠ABC

⇒ ∠AOC = 2 × ∠ABC

⇒ ∠AOC = 2 × 70°

⇒ ∠AOC = 140°

(ii)

∠BOC = 360° - (∠AOB + ∠AOC) [Sum of all angles at a point = 360°]

⇒ ∠BOC = 360° - (90° + 110°)

⇒ ∠BOC = 360° - 200°

⇒ ∠BOC = 160°

We know that ∠BOC = 2 × ∠BAC

⇒ ∠BAC = (1/2) × ∠BOC

⇒ ∠BAC = (1/2) × 160°

⇒ ∠BAC = 80°

(i) ∠AOC + ∠AOB = 180°[Because BC is a straight line]

⇒ ∠AOC + 70°= 180°

⇒ ∠AOC + 70°= 180°

⇒ ∠AOC = 110°

OA = OC [Radius]

∴ ∠OAC = ∠OCA ________________ (i)

In triangle AOC,

∠OAC + ∠OCA + ∠AOC = 180°[Sum of angles of triangle]

⇒ 2 ∠OCA + 110° = 180°[From equation (i)]

⇒ 2 ∠OCA = 70°

⇒ 2 ∠OCA = 70°

⇒ ∠OCA = 35°

(ii)

∠AOC + ∠AOB = 180°[Because BC is a straight line]

⇒ ∠AOC + 70°= 180°

⇒ ∠AOC + 70°= 180°

⇒ ∠AOC = 110°

OA = OC [Radius]

∴ ∠OAC = ∠OCA ________________ (i)

In triangle AOC,

∠OAC + ∠OCA + ∠AOC = 180°[Sum of angles of triangle]

⇒ 2 ∠OAC + 110° = 180°[From equation (i)]

⇒ 2 ∠OAC = 70°

⇒ 2 ∠OAC = 70°

⇒ ∠OAC = 35°

∠BPC + ∠APB = 180°[Because APC is a straight line]

⇒ ∠BPC + 110° = 180°

⇒ ∠BPC = 70°

In triangle BPC,

∠BPC + ∠PBC + ∠PCB = 180°[Sum of angles of triangle]

⇒ 70°+ 25° + ∠PCB = 180°

⇒ ∠PCB = 85°

∴ ∠ADB = ∠PCB = 85°[Angles in the same segment of a circle]

In triangle ABD,

∠ABD + ∠BAD + ∠ADB = 180°[Sum of angles of triangle]

⇒ 35°+ 90° + ∠ADB = 180°

⇒ ∠ADB = 55°

∴ ∠ACB = ∠ADB = 55°[Angles in the same segment of a circle]

∠AOB = 2 × ∠ACB

⇒ ∠AOB = 2 × 50°

⇒ ∠AOB = 100°

OA =OB [Radius of the circle]

∴ ∠OAB = ∠OBA _________________ (i)

In triangle AOB,

∠OAB + ∠OBA + ∠AOB = 180°[Sum of angles of triangle]

⇒ 2 ∠OAB + 100° = 180°[From equation (i)]

⇒ 2 ∠OAB = 80°

⇒ ∠OAB = 40°

(i)

∠ABD and ∠ACD are in the segment AD.

∴ ∠ACD = ∠ABD [Angles in the same segment of a circle]

∠ACD = 54°

(ii) ∠BAD = 43°

∠BAD and ∠BCD are in the segment BD.

∴ ∠BAD = ∠BCD [Angles in the same segment of a circle]

∠BAD = 43°

(iii)

In triangle ABD,

∠ABD + ∠BAD + ∠BAD = 180°[Sum of angles of triangle]

⇒ 54° + 43° + ∠BAD = 180°

⇒ 97° + ∠BAD = 180°

⇒ ∠BAD = 83°

∠CAD and ∠CBD are in the segment BD.

∴ ∠CAD = ∠CBD [Angles in the same segment of a circle]

∠CAD = 60°

In triangle ACD,

∠CAD + ∠ADC + ∠ACD = 180°[Sum of angles of triangle]

⇒ 60° + 90° + ∠ACD = 180°

⇒ 150° + ∠ACD = 180°

⇒ ∠ACD = 30°

∴ ∠CDE = ∠ACD = 30°[Alternate angles]

Join OC and OD.

∠ABC = ∠BCD = 25°[Alternate angles]

The angle subtended by an arc of a circle at the center is double the angle subtended by the arc at any point on the circumference.

∴ ∠BOD = 2 × ∠BCD

⇒ ∠BOD = 2 × 25°

⇒ ∠BOD = 50°

Similarly,

∠AOC = 2 × ∠ABC

⇒ ∠AOC = 2 × 25°

⇒ ∠AOC = 50°

Now,

∠AOB = 180° [AOB is a straight line]

⇒ ∠AOC + ∠COD + ∠BOD = 180°

⇒ 50° + ∠COD + 50° = 180°

⇒ 100° + ∠COD = 180°

⇒ ∠COD = 80°

∴∠CED = (1/2) ∠COD

⇒ ∠CED = (1/2) 80°

⇒ ∠CED = 40°

(i)

In triangle CDE,

∠CDE + ∠CED + ∠DCE = 180°[Sum of angles of triangle]

⇒ 40° + 90° + ∠DCE = 180°

⇒ 130° + ∠DCE = 180°

⇒ ∠DCE = 50°

(ii)

∠AOC + ∠BOC = 180°[Because AOB is a straight line]

⇒ 80° + ∠BOC = 180°

⇒ ∠BOC = 100°

In triangle BOC,

∠OCB + ∠BOC + ∠OBC = 180°[Sum of angles of triangle]

⇒ 50° + 100° + ∠OBC = 180°[∠DCE = 50°]

⇒ 150° + ∠OBC = 180°

⇒ ∠OBC = 30°

∴ ∠ABC = ∠OBC = 30°

∠DCB = (1/2) ∠AOB [∠DCB = ∠ACB]

⇒ ∠DCB = (1/2) 40°

⇒ ∠DCB = 20°

In triangle BCD,

∠BDC + ∠DCB + ∠DBC = 180°[Sum of angles of triangle]

⇒ 100° + 20° + ∠OBC = 180°

⇒ 120° + ∠DBC = 180°

⇒ ∠DBC = 60°

∴ ∠OBC = ∠DBC = 60°

Join OB,

∴ OA = OB [Radius]

∴ ∠OAB = ∠OBA = 25°

In triangle AOB,

∠AOB + ∠OAB + ∠OBA = 180°[Sum of angles of triangle]

⇒ ∠AOB + 25° + 25° = 180°

⇒ ∠AOB + 50° = 180°

⇒ ∠AOB = 130°

Now,

∠ACB = (1/2) ∠AOB

⇒ ∠ACB = (1/2) 130°

⇒ ∠ACB = 65°

In triangle BEC,

∠EBC + ∠ECB + ∠BEC = 180°[Sum of angles of triangle]

⇒ ∠EBC + 65° + 90° = 180°

⇒ ∠EBC + 155° = 180°

⇒ ∠EBC = 25°

(i)

OB = OC [Radius]

∴ ∠OBC = ∠OCB = 55°

In triangle OCB,

∠OBC + ∠OCB + ∠BOC = 180°[Sum of angles of triangle]

⇒ 55° + 55° + ∠BOC = 180°

⇒ 110° + ∠BOC = 180°

⇒ ∠BOC = 70°

(ii)

OA = OB [Radius]

∴ ∠OBA = ∠OAB = 20°

In triangle AOB,

∠OBA + ∠OAB + ∠AOB = 180°[Sum of angles of triangle]

⇒ 20° + 20° + ∠AOB = 180°

⇒ 40° + ∠AOB = 180°

⇒ ∠AOB = 140°

∴ ∠AOC = ∠AOB - ∠BOC

⇒ ∠AOC = 140° - 70°

⇒ ∠AOC = 70°

∠BOC = 2 × ∠BAC

⇒ ∠BOC = 2 × 30°

⇒ ∠BOC = 60°______________ (i)

OB = OC

∴ ∠OBC = ∠OCB ________________ (ii)

In triangle AOB,

∠OBC + ∠OCB + ∠BOC = 180°[Sum of angles of triangle]

⇒ 2 ∠OCB + 60° = 180°

⇒ 2 ∠OCB = 120°

⇒ ∠OCB = 60°

∴ ∠OBC = 60°[From equation (ii)]

From equation (i) and (ii),

∠OBC = ∠OCB = ∠BOC = 60°

∴ BOC is an equilateral triangle.

∴ OB = OC = BC

Hence, BC is the radius of the circumcircle.

In triangle PQR,

∠QPR + ∠PQR + ∠PRQ = 180°[Sum of angles of triangle]

⇒ ∠QPR + 65° + 90° = 180°

⇒ ∠QPR + 155° = 180°

⇒ ∠QPR = 25°______________ (i)

In triangle PMQ,

∠QPM + ∠PMQ + ∠PQM = 180°[Sum of angles of triangle]

⇒ ∠QPM + 90° + 50° = 180°

⇒ ∠QPM + 140° = 180°

⇒ ∠QPM = 40°

Now,

∠PRS = ∠QPR = 25°[Alternate angles]

(i)

∠BAC = ∠BDC = 40°[Angles in the same segment]

In triangle BCD,

∠BCD + ∠DBC + ∠BDC = 180°[Sum of angles of triangle]

⇒ ∠BCD + 60° + 40° = 180°

⇒ ∠BCD + 100° = 180°

⇒ ∠BCD = 80°

(ii)

∠CAD = ∠CBD [Angles in the same segment]

⇒ ∠CAD = 40°

In cyclic quadrilateral PQRS,

∠PSR + ∠PQR = 180°[Opposite angles]

⇒ 150° + ∠PQR = 180°

⇒ ∠PQR = 30°

In triangle PQR,

∠RPQ + ∠PQR + ∠PRQ = 180°[Sum of angles of triangle]

⇒ ∠RPQ + 30° + 90° = 180°

⇒ ∠RPQ + 120° = 180°

⇒ ∠RPQ = 60°

(i)

∠BAD + ∠BCD = 180°[Opposite angles of a cyclic quadrilateral arc supplementary]

⇒ 100° + ∠BCD = 180°

⇒ ∠BCD = 80°

(ii)

∠BAD + ∠ADC = 180°[Interior angles of same side]

⇒ 100° + ∠ADC = 180°

⇒ ∠ADC = 80°

(iii)

∠BCD + ∠ABC = 180°[Interior angles of same side]

⇒ 80° + ∠ABC = 180°

⇒ ∠ABC = 100°

Reflex ∠AOC = 360° - ∠AOC

= 360° - 130°

= 230°

∴∠ABC = (1/2) ∠AOC

⇒ ∠ABC = (1/2) 230°

⇒ ∠ABC = 115°

Now,

∠ABC + ∠PBC = 180°[Because ABP is a straight line]

⇒ 115° + ∠PBC = 180°

⇒ ∠PBC = 65°

ABCD is cyclic quadrilateral.

∴ ∠ABC + ∠ADC = 180°

⇒ 92° + ∠ADC = 180°

⇒ ∠ADC = 88°

AE || CD

∴∠EAD = ∠ADC = 88°

Now,

∠BCD = 180° - ∠DAB

⇒ ∠BCD = ∠DAF = ∠EAD + ∠EAF

⇒ ∠BCD = 88° + 20°

⇒ ∠BCD = 108°

BD = CD

∴∠CBD = ∠BCD = 30°

In triangle BCD,

∠BDC + ∠BCD + ∠CBD = 180°[Sum of angles of triangle]

⇒ ∠BDC + 30° + 30° = 180°

⇒ ∠BDC + 60° = 180°

⇒ ∠BDC = 120°

Now,

∠BDC + ∠BAC = 180°[ABCD is a cyclic quadrilateral]

⇒ 120° + ∠BAC = 180°

⇒ ∠BAC = 60°

∠ADC = (1/2) ∠AOC

⇒ ∠ADC = (1/2) 100°

⇒ ∠ADC = 50°

Now,

∠ADC + ∠ABC = 180°[ABCD is a cyclic quadrilateral]

⇒ 50° + ∠ABC = 180°

⇒ ∠ABC = 130°

(i)

ABC is equilateral triangle.

∴ ∠ABC = ∠ACB = ∠BAC = 60°____________________ (i)

∠BDC = ∠BAC = 60°[Angles in the same segment of a circle are equal]

(ii)

ABCD is a cyclic quadrilateral

∴ ∠BAC + ∠BEC = 180°

⇒ 60° + ∠BEC = 180°

⇒ ∠BEC = 120°

ABCD is a cyclic quadrilateral

∴ ∠BCD + ∠BAD = 180°[Opposite angle of a cyclic quadrilateral are supplementary]

⇒ 100° + ∠BAD = 180°

⇒ ∠BAD = 80°

In triangle ABD,

∠ADB + ∠ABD + ∠BAD = 180°[Sum of angles of triangle]

⇒ ∠ADB + 50° + 80° = 180°

⇒ ∠ADB + 130° = 180°

⇒ ∠ADB = 50°

Reflex ∠BOD = (360° – ∠BOD)

⇒ Reflex ∠BOD = (360° – 150°)

⇒ Reflex ∠BOD = 210°

Now,

X = (1/2) (Reflex ∠BOD)

⇒ X = (1/2) 210°

⇒ X = 105°

X + Y = 180°

⇒ 105° + Y = 180°

⇒ Y = 75°

OA = OB [Radius]

∴ ∠OAB = ∠OBC = 50°

In triangle AOB,

∠AOB + ∠OAB + ∠OBC = 180°[Sum of angles of triangle]

⇒ ∠AOB + 50° + 50° = 180°

⇒ ∠AOB + 100° = 180°

⇒ ∠AOB = 80°

∴ x = 180°– ∠AOB [AOD is a straight line]

⇒ x = 180°- 80°

⇒ x = 100°

∴ X + Y = 180°[Opposite angle of a cyclic quadrilateral are supplementary]

⇒ 100° + Y = 180°

⇒ Y = 80°

∠ABC + ∠CBF = 180°[Because ABF is a straight line]

⇒ ∠ABC + 130° = 180°

⇒ ∠ABC = 50°

∴ x = ∠ABC = 50°[Exterior angle = interior opposite angle]

(i)

ABCD is a cyclic quadrilateral.

∴ ∠BAD + ∠BCD = 180°

⇒ ∠BAD + 120° = 180°

⇒ ∠BAD = 60°

(ii)

∠BDA = 90°[Angle in a semi-circle]

In triangle ABD,

∠ABD + ∠BDA + ∠BAD = 180°[Sum of angles of triangle]

⇒ ∠ABD + 90° + 60° = 180°

⇒ ∠ABD + 150° = 180°

⇒ ∠ABD = 30°

(iii)

OD = OA [Radius]

∴ ∠OAD = ∠ODA = ∠BAD = 180°

∴∠ODB = 90° - ∠ODA

⇒ ∠ODB = 90° - 60°

⇒ ∠ODB = 30°

(iv)

∠ADC = ∠ADB + ∠CDB

⇒ ∠ADC = 90° + 30°

⇒ ∠ADC = 120°

In triangle AOD,

∠AOD + ∠OAD + ∠ODA = 180°[Sum of angles of triangle]

⇒ ∠AOD + 60° + 60° = 180°

⇒ ∠AOD + 120° = 180°

⇒ ∠AOD = 60°

∴ Triangle AOD is an equilateral triangle.

Two chords AB and CD of a circle intersect each other at P outside the circle.

∴ AP × BP = CP × PD

⇒ (AB + BP) × BP = (CD + PD) × PD

⇒ (6 + 2) × 2 = (CD + 2.5) × 2.5

⇒ 16 = 2.5 CD + 6.25

⇒ 2.5 CD = 9.75

⇒ CD = 3.9 cm

(i)

∠BOD + ∠AOD = 180°[AOB is a straight line]

⇒ ∠BOD + 140° = 180°

⇒ ∠BOD = 40°

OB = OD

∴ ∠OBD = ∠ODB

In triangle AOD,

∠BOD + ∠OBD + ∠ODB = 180°[Sum of angles of triangle]

⇒ 40°+ 2 ∠OBD = 180°

⇒ 2 ∠OBD = 140°

⇒ ∠OBD = 70°

∴ ∠OBD = ∠ODB = 70°

ABDC is a cyclic quadrilateral.

∴ ∠CAB + ∠BDC = 180°

⇒ ∠CAB + ∠ODB + ∠ODC = 180°

⇒ 50°+ 70°+ ∠ODC = 180°

⇒ ∠ODC = 60°

Now,

∠EDB = 180° – ∠BDC [Because CDE is a straight line]

⇒ ∠EDB = 180° – (∠ODB + ∠ODC)

⇒ ∠EDB = 180° – (70°+ 60°)

⇒ ∠EDB = 180° – 130°

⇒ ∠EDB = 50°

(ii)

∠BOD + ∠AOD = 180°[AOB is a straight line]

⇒ ∠BOD + 140° = 180°

⇒ ∠BOD = 40°

OB = OD

∴ ∠OBD = ∠ODB

In triangle AOD,

∠BOD + ∠OBD + ∠ODB = 180°[Sum of angles of triangle]

⇒ 40°+ 2 ∠OBD = 180°

⇒ 2 ∠OBD = 140°

⇒ ∠OBD = 70°

∴ ∠OBD = ∠ODB = 70°

Now,

∠EBD + ∠OBD = 180°[Because OBE is a straight line]

⇒ ∠EBD + 70° = 180°

⇒ ∠EBD = 110°

In ΔEBC and ΔEDA,

∠EBC = ∠CDA

⇒ ∠EBC = ∠CDA ________________ (i)

∠ECB = ∠BAD

⇒ ∠ECB = ∠EAD ________________ (ii)

∠BEC = ∠DEA ________________ (iii)

From equation (i), (ii) and (iii),

ΔEBC ≅ ΔEDA Proved.

Given AB = AC

∴ ∠ACB = ∠ABC

Ext. ∠ADE = ∠ACB = ∠ABC

∴ ∠ADE = ∠ABC

∴ DE || BC Proved.

Given, ABC is an isosceles triangle in which AB = AC. D and E are midpoints of AB and AC respectively.

∴ DE || BC

⇒ ∠ADE = ∠ABC ______________ (i)

AB = AC

⇒ ∠ABC = ∠ACB ______________ (ii)

Now,

∠ADE + ∠EDB = 180°[Because ADB is a straight line]

⇒ ∠ACB + ∠EDB = 180°

The opposite angles are supplementary.

∴ D, B, C, E are concyclic.

Let, ABCD be a cyclic quadrilateral and O be the center of

the circle passing through A, B, C, and D.

Then,

Each of AB, BC, CD and DA being a chord of the

circle, its right bisector must pass through O.

Therefore,

The right bisectors of AB, BC, CD and DA pass through and are concurrent.

Let diagonals BD and AC of the rhombus ABCD intersect at O.

We know that the diagonals of a rhombus bisect each other at right angles.

∴ ∠BOC = 90°

∴∠BOC lies in a circle.

The circle drawn with BC as diameter will pass through O.

Let O be the point of intersection of the diagonals BD and AC of rectangle ABCD.

Since, the diagonals of a rectangle are equal and bisect each other.

∴ OA = OB = OC = OD

Hence, O is the center of the circle through A, B, C, D.

Let A, B, C, D be the given points.

With B as center and radius equal to AC draw an arc.

With C as center and AB as radius draw another arc.

Which cuts the previous arc at D,

Then, D is the required point BD and CD.

In ΔABC and ΔDCB,

AB = DC

AC = DB

BC = CB

∴ ΔEBC ≅ ΔEDA

⇒ ∠BAC = ∠CDB

Thus, BC subtends equal angles, ∠BAC and ∠CDB on the same side of it.

Therefore, points A, B, C, D are concyclic.

Given, ∠B - ∠D = 60°____________________ (i)

ABCD is a cyclic quadrilateral,

∴ ∠B + ∠D = 180°____________________ (ii)

From equation (i) and (ii),

2 ∠B = 240°

⇒ ∠B = 120°________________ (iii)

From equation (ii),

∠B + ∠D = 180°

⇒ 120° + ∠D = 180°[From equation (iii)]

⇒ ∠D = 60°

Hence, the smaller of the two angle ∠D = 60°.

In ΔADE and ΔBCF,

AD = BC

∠AED = ∠BFC

∠ADE = ∠BCF [∠ADC - 90° = ∠BCD - 90°]

∴ ΔADE ≅ ΔBCF

The Cross ponding parts of the congruent triangles are equal.

∴ ∠A = ∠B

Now,

∠A + ∠B + ∠C + ∠D = 360°

⇒ 2 ∠B + 2 ∠D = 360°

⇒ ∠B + ∠D = 180°

∴ ABCD is a cyclic quadrilateral.

∠DCF = ∠DAB

⇒ x = 75° [Exterior angle is equal to the interior opposite angle.]

Now,

∠DCF + ∠DEF = 180° [Opposite angles of a cyclic quadrilateral]

⇒ x + y = 180°

⇒ 75° + y = 180°

⇒ y = 105°

Given: Let ABCD be a cyclic quadrilateral, diagonals AC and BD intersect at O at right angles.

∠OCN = ∠OBM [Angles in the same segment] ___________ (i)

∠OBM + ∠BOM = 90° [Because ∠OLB = 90°] ______________ (ii)

∠BOM + ∠CON = 90°[LOM is a straight line and ∠BOC = 90°] ______________ (iii)

From equation (ii) and (iii),

∠OBM + ∠BOM = ∠BOM + ∠CON

⇒ ∠OBM = ∠CON

Thus, ∠OCN = ∠OBM and ∠OBM = ∠CON

⇒ ∠OCN = ∠CON

∴ON = CN ____________________ (iv)

Similarly, ON = ND ___________________ (v)

From equation (iv) and (v),

CN = ND Proved.

If one side of a cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle.

Chord AB of a circle is produced to E.

∴ ext. ∠BDE = ∠BAC = ∠EAC ___________________ (i)

Chord CD of a circle is produced to E.

∴ ext. ∠DBE = ∠ACD = ∠ACE ____________________ (ii)

In ΔEDB and ΔEAC,

∠BDE = ∠CAE [From equation (i)]

∠DBE = ∠ACE [From equation (ii)]

∠E = ∠E [Common angle]

∴ ΔEDB ∼ ΔEAC Proved.

Given: AB and CD are two parallel chords of a circle.

If one side of a cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle.

∴ ext. ∠DCE = ∠B and ext. ∠EDC = ∠A

A || B

∴ ∠EDC = ∠B and ∠DCE = ∠A

∴∠A = ∠B

Hence, ΔAEB is isosceles.

(i)

∠BDA = 90°= ∠EDB [Semi circle angle]

In triangle EBD,

∠DBE + ∠EDB + ∠BED = 180°

⇒ ∠DBE + 90° + 25° = 180°

⇒ ∠DBE + 115° = 180°

⇒ ∠DBE = 65°

Now,

∠DBC + ∠DBE = 180°[CBE is a straight line]

⇒ ∠DBC + 65° = 180°

⇒ ∠DBC = 115°

(ii)

∠DCB = ∠BAD [Angle in the same segment]

∴∠DCB = 35°

(iii) ∠BDC = 30°

In triangle BCD,

∠BDC + ∠DCB + ∠DBC = 180°

⇒ ∠BDC + 35° + 115° = 180°

⇒ ∠BDC + 150° = 180°

⇒ ∠BDC = 30°

Given radius(AO) = 13cm

Length of the chord (AB) = 10cm

Draw a perpendicular bisector from center to the chord and name it OC.

AC = BC = 5cm

Now in ∆ AOC,

Using Pythagoras theorem

AO² = AC² + OC²

13² = 5² + OC²

OC² = 13² – 5²

OC² = 169 – 25

OC² = 144

OC = 12cm

The distance of the chord from the centre is 12cm.

Given radius(AO) = 17cm

Length of the chord (AB) = x

distance of the chord from the centre is 8cm.

Draw a perpendicular bisector from center to the chord and name it OC.

AC = BC

Now in ∆ AOC

Using Pythagoras theorem

AO² = AC² + OC²

17² = AC² + 8²

AC² = 17² – 8²

AC² = 289 – 64

AC² = 225

AC = 15cm

BC = 15cm

The length of the chord is AC + BC = 15 + 15 = 30 cm.

Given: BOC is the diameter of the circle

AB = AC

Here, BAC forms a semicircle.

We know that angle in a semicircle is always 90

BAC = 90

Here ABC = ACB (since angles opposite equal sides are equal in a triangle)

We know that sum of all the angles in the triangle is 180

That is

ABC + ACB + BAC = 180

⇒ 2 × ABC + BAC = 180

⇒ 2 × ABC + 90 = 180

⇒ 2 × ABC = 180 – 90

⇒ 2 ×ABC = 90

⇒ ABC = 45

Given:

We know that

2 × ACB = AOB (The angle subtended by an arc at the center is twice the angle subtended by the same arc on any point on the remaining part of the circle).

2 × 30 = AOB

AOB = 60.

AOB = 60

In Δ AOB OA = OB( radius)

OAB = OBA (Angles opposite to equal sides are equal)

OBA = 40

By angle sum property

OAB + OBA + AOB = 180°

AOB = 180° – OAB – OBA

AOB = 180° – 40° – 40° = 100°

We know that

2 ×ACB = AOB (The angle subtended by an arc at the center is twice the angle subtended by the same arc on any point on the remaining part of the circle).

2 ×ACB = 100°

ACB =

ACB = 50

Given: AB 34 cm and CD = 30 cm

Here OL is the perpendicular bisector to CD

∴ CL = LD = 15 cm

Construction: Join OD(radius)

OD = 17cm

Now in Δ ODL

By Pythagoras theorem

OD² = OL² + LD²

17² = OL² + 15²

OL² = 17² + 15²

OL² = 289—225

OL² = 64

OL = 8

∴The distance of from is = OL = 8cm

Given:

AB = CD

We know that angles subtended from equal chords at center are equal.

∴ ∠ AOB = ∠ COD

∴ ∠ COD = 80°

Given: AB = 12cm, CE = 3cm

AB = AE + EB

AE = EB (OC is perpendicular bisector to AB)

∴ AE = 6 cm

Let CD = 2x (diameter)

AO = OC = x (radius)

In Δ AOE

AO² = AE² + OE²

x² = 6² + (OC – EC)²

x² = 6² + (x – 3)²

x² = 6² + x² + 3² – 2(x)(3)

x² = 36+ x² + 9 – 6x

6x = 36+ 9 + x² – x²

6x = 45

x = = 7.5

∴ Radius = x = 7.5 cm

Given: CE = ED = 8 cm and EB = 4 cm

Construction: Join OC (OC is radius)

Let AB = 2x (diameter)

OB = OC = x (radius)

In Δ COE

CO² = CE² + OE²

x² = 8² + (OB – EB)²

x² = 8² + (x – 4)²

x² = 8² + x² + 4² – 2(x)(4)

x² = 64+ x² + 16 – 8x

8x = 64+ 16 + x² – x²

8x = 80

x = = 10

∴ Radius = x = 10 cm

Given: AB||CD and AB = 10cm

Construction: Drop perpendiculars OE and OF on to AB and CD respectively.

Now,

Consider ΔBOE and ΔCOF

Here,

OB = OC (radius)

∠OEB = ∠OFC (right angle)

∠COF = ∠BOE (vertically opposite angles)

∴ By AAS congruency ΔBOE ΔCOF

∴ OE = OF (by congruent parts of congruent triangles)

Chords equidistant from center are equal in length

That is CD = AB = 10cm

∴ CD = 10cm

Given: BC = OB and ∠ACD = 25°

Here in Δ OBC

∠BOC = ∠BCO (angles opposite to equal sides are equal)

∴ ∠BOC = 25°

By angle sum property

∠BOC + ∠BCO + ∠OBC = 180°

25° + 25° + ∠OBC = 180°

50° + ∠OBC = 180°

∠OBC = 180° – 50°

∴ ∠OBC = 130°

Here

∠ABC = ∠ABO + ∠OBC = 180°

∠ABO + 130° = 180°

∠ABO = 180° – 130°

∴ ∠ABO = 50°

Now, in ΔAOB

OB = OA (radius)

∠ABO = ∠BAO = 50° (angles opposite to equal sides are equal)

By angle sum property

∠ABO + ∠BAO + ∠AOB = 180°

50° + 50° + ∠AOB = 180°

∠AOB = 180° – (50° + 50°) = 180° – 100° = 80°

∴ ∠AOB = 80°

Here

∠DOC = ∠AOD + ∠AOB + ∠BOC = 180°

∠AOD + 80° + 25° = 180°

∠AOD + 105° = 180°

∠AOD = 180° – 105°

∠AOD = 75°

∴ ∠AOD = 75°

Given: and OD = 6cm

Here OB is radius

Let OB = x cm

In ΔBOD, By Pythagoras theorem

OB² = BD² + OD²

x² = BD² + 6²

x² = BD² + 36

BD² = x² – 36

Now consider Δ ABC

Here BC = 2x

By Pythagoras theorem

BC² = AB² + AC²

(2x)² = 4(x² – 36)+ AC²

4x² = 4x² –144 + AC²

AC² = 144

AC = 12 cm

∴ AC = 12 cm

Given: Equilateral triangle of side 9 cm is inscribed in a circle.

Construction: Join OA, OB, OC and drop a perpendicular bisector from center O to BC.

Here,

Area (ΔABC) = 3× area (ΔOBC)

Area (ΔABC) = a² = × 9² =

Now,

Area (ΔOBC) = × AC × OD = × 9 × OD

We know that,

Area (ΔABC) = 3× area (ΔOBC)

= × 9 × OD

OD =

Now, in ΔODC

By Pythagoras theorem

OC² = OD² + DC²

OC² = ² + ²

OC² = + = = 27

OC =

∴ Radius = OC =

Angle in a semicircle measures 90°

Angles in the same segment of a circle are always equal.



Proof:

As we know angle subtended by an arc is double the angle subtended at any other point.

So,

∠POQ = 2∠PAQ .... (1)


∠POQ= 2∠PBQ .... (2)

From (1) and (2),

∠PAQ = ∠PBQ

Hence proved

Given: Two triangles ΔABC and ΔBCD, ∠BAC = 60° and ∠DBC = 50°

We know that ∠BAC = ∠BDC = 60° ( angles in the same segment drawn from same chord are

equal).

Now consider ΔBCD

By angle sum property

∠DBC + ∠BDC + ∠BCD = 180°

50° + 60° + ∠BCD = 180°

∠BCD = 180° – 50° – 60°

∠BCD = 70°

∴ ∠BCD = 70°

Given:

Here,

∠BAC = 90° (angle in the semicircle)

Now, in ΔABC

By angle sum property

∠BCA + ∠BAC + ∠ABC = 180°

30° + 90° + ∠ABC = 180°

∠ABC = 180° – 30° – 90°

∠ABC = 60°

Here,

∠ABC = ∠ADC (angles in the same segment)

∴ ∠CDA = 60°

Given: ∠OAC = 50°

Consider ΔAOC

∠OAC = ∠OCA = 50° ( OA = OC = radius, angles opposite to equal sides are equal)

Now, by angle sum property

∠OAC + ∠OCA + ∠AOC = 180°

50° + 50° + ∠AOC = 180°

∠AOC = 180° – 50° – 50°

∠AOC = 80°

Now angle ∠BOD = ∠AOC = 80° (vertically opposite angles)

Now, consider ΔBOD

Here,

OB = OD (radius)

∠OBD = ∠ODB (angles opposite to equal angles are equal)

Let ∠ODB = x

By angle sum property

∠ODB + ∠OBD + ∠BOD = 180°

x + x + 80° = 180°

2x = 180° – 80°

2x = 100°

x = 50°

∴ ∠ODB = 50°

Given: and

Consider ΔOAB

Here,

OA = OB (radius)

∠OBA = ∠OAB = 20° (angles opposite to equal sides are equal)

By angle sum property

∠AOB + ∠OBA + ∠OAB = 180°

∠AOB + 20° + 20° = 180°

∠AOB = 180° – 20° – 20°

∠AOB = 140°

Similarly, in ΔAOC

OA = OC (radius)

∠OCA = ∠OAC = 30° (angles opposite to equal sides are equal)

By angle sum property

∠AOC + ∠OCA + ∠OAC = 180°

∠AOC + 30° + 30° = 180°

∠AOC = 180° – 30° – 30°

∠AOC = 120°

Here,

∠CAB = ∠OAB + ∠OAC = 50°

Here,

2CAB = BOC (The angle subtended by an arc at the center is twice the angle subtended by the same arc on any point on the remaining part of the circle).

2CAB = BOC

2 × 50° = BOC

BOC = 100°.

BOC = 100

Given: ∠AOB = 100° and ∠AOC = 90°,

Consider ΔOAB

Here,

OA = OB (radius)

Let ∠OBA = ∠OAB = x (angles opposite to equal sides are equal)

By angle sum property

∠AOB + ∠OBA + ∠OAB = 180°

100° + x + x = 180°

2x = 180° – 100°

2x = 80°

x = 40°

Similarly, in ΔAOC

OA = OC (radius)

Let ∠OCA = ∠OAC = y (angles opposite to equal sides are equal)

By angle sum property

∠AOC + ∠OCA + ∠OAC = 180°

90° + y + y = 180°

2y = 180° – 90°

2y = 90°

y = 45°

Here,

∠BAC = ∠OAB + ∠OAC = x + y = 40° + 45° = 85°

∴ ∠BAC = 85°

Given: and

In ΔOAB

Here,

OA = OB (radius)

Let ∠OBA = ∠OAB = x (angles opposite to equal sides are equal)

By angle sum property

∠AOB + ∠OBA + ∠OAB = 180°

50° + x + x = 180°

2x = 180° – 50°

2x = 130°

x = 60°

∴ ∠OAB = 60°

Given: ∠AOC = 120°

Construction: Join OD

We know that,

∠AOC = 2 × ∠ADC

120° = 2 ∠ADC

∠ADC = 60°

Here,

∠ADB = 90° (angle in a semicircle)

∠ADB = ∠ADC + ∠CDB = 90°

∠ADC + ∠CDB = 90°

60° + ∠CDB = 90°

∠CDB = 90° – 60°

∠CDB = 30°

∴ ∠BDC = 30°

Given: ∠OAB = 50°

Construction: Join AC

Here,

In ΔAOB

OA = OB (radius)

∠OAB = ∠OBA (angles opposite to equal sides are equal)

∴ ∠OBA = 50°

∠OBA = ∠CDA (angles in the same segment)

∴ ∠CDA = 50°

Given: and ∠BCD = 80°

Here,

∠CAB = ∠CDB = 40° ( angles in the same segment drawn from same chord are equal).

Now, in ΔBCD

By angle sum property

∠BCD + ∠CDB + ∠CBD = 180°

80° + 40° + ∠CBD = 180°

∠CBD = 180° – 40° – 80°

∠CBD = 60°

∴ ∠CBD = 60°

Given: ∠AEB = 110° and ∠CBE = 30°

∠AEC = ∠AEB + ∠BEC = 180°

∠AEB + ∠BEC = 180°

110° + ∠BEC = 180°

∠BEC = 180° – 110°

∠BEC = 70°

In ΔBEC

By angle sum property

∠CBE + ∠BEC + ∠ECB = 180°

30° + 70° + ∠ECB = 180°

∠ECB = 180° – 30° – 70°

∠ECB = 80°

Here,

∠ECB = ∠ADB (angles in the same segment)

∴ ∠ECB = ∠ADB = 80°

∴ ∠ADB = 80°

Given: ∠OAB = 20° and ∠OCB = 50°

Here,

In ΔAOB

OA = OB (radius)

∠OAB = ∠OBA (angles opposite to equal sides are equal)

∴ ∠ OBA = 20°

Now, by angle sum property

∠AOB + ∠OBA + ∠OAB = 180°

∠AOB + 20° + 20° = 180°

∠AOB = 180° – 20° – 20°

∠AOB = 140°

Now, Consider Δ BOC

OC = OB (radius)

∠OCB = ∠OBC (angles opposite to equal sides are equal)

∴ ∠ OBA = 50°

Now, by angle sum property

∠COB + ∠OBC + ∠OCB = 180°

∠COB + 50° + 50° = 180°

∠COB = 180° – 50° – 50°

∠COB = 80°

Here,

∠AOB = ∠AOC + ∠COB

140° = ∠AOC + 80°

∠AOC = 140° – 80°

∠AOC = 60°

∴ ∠AOC = 60°

Given: ABCD is cyclic quadrilateral and ∠ADC = 120°

Here,

∠ADC + ∠ABC = 180° (opposite angles in cyclic quadrilateral are supplementary)

120° + ∠ABC = 180°

∠ABC = 180° – 120°

∠ABC = 60°

Here,

∠ACB = 90° (angle in semicircle)

Now, consider ΔABC

By angle sum property

∠BAC + ∠ABC + ∠ACB = 180°

∠BAC + 60° + 90° = 180°

∠BAC = 180° – 60° – 90°

∠BAC = 30°

Given: ABCD is a cyclic quadrilateral, AB||DC and ∠BAD = 100°

Here,

∠BAD + ∠BCD = 180° (opposite angles in cyclic quadrilateral are supplementary)

100° + ∠BCD = 180°

∠BCD = 180° – 100°

∠BCD = 80°

Here, AB||DC and BC is the transversal

∠ABC + ∠BCD = 180° (interior angles along the transversal are supplementary)

∠ABC + 80° = 180°

∠ABC = 180° – 80° = 100°

∴ ∠ABC = 100°

Given: ∠AOC = 130°

Here,

(Exterior ∠AOC) = 360° – (interior ∠AOC)

(Exterior ∠AOC) = 360° – 130°

(Exterior ∠AOC) = 230°

We know that,

(Exterior ∠AOC) = 2 × ∠ABC

230° = 2 × ∠ABC

∠ABC = = 115°

∴ ∠ABC = 115°

Given: CD||AB and ∠BAD = 30°

Consider ΔABD

∠ADB = 90° (angle in semicircle)

Now, by angle sum property

∠ABD + ∠BAD + ∠ADB = 180°

∠ABD + 30° + 90° = 180°

∠ABD = 180° – 30° – 90°

∠ABD = 60°

Here,

∠ABD + ∠ACD = 180° (opposite angles in cyclic quadrilateral are supplementary)

60° + ∠ACD = 180°

∠BCD = 180° – 60°

∠BCD = 120°

Here, CD||AB and AC is the transversal

∠CAB + ∠ACD = 180° (interior angles along the transversal are supplementary)

∠CAB + 120° = 180°

∠ABC = 180° – 120° = 60°

∠ABC = 60°

∠ABC = ∠CAD + ∠DAB

60° = ∠CAD + 30°

∠CAD = 60° – 30° = 30°

∴ ∠CAD = 30°

Given: ∠AOC = 100°

Here,

(Exterior ∠AOC) = 360° – (interior ∠AOC)

(Exterior ∠AOC) = 360° – 100°

(Exterior ∠AOC) = 260°

We know that,

(Exterior ∠AOC) = 2 × ∠ADC

260° = 2 × ∠ABC

∠ABC = = 130°

∴ ∠ABC = 130°

Here,

∠ABD = ∠ABC + ∠CBD

180° = 130° + ∠CBD

∠CBD = 180° – 130° = 50°

∴ ∠CBD = 50°

Given: ∠OAB = 50°

Consider ΔAOB

Here,

OA = OB (radius)

∠OAB = ∠OBA = 50° (In a triangle, angles opposite to equal sides are equal)

By angle sum property

∠AOB + ∠OAB + ∠OBA = 180°

∠AOB + 50° + 50° = 180°

∠AOB = 180° – 50° – 50°

∠AOB = 80°

Here,

∠AOD = ∠AOB + ∠BOD

180° = 80° + ∠BOD

∠BOD = 180° – 80° = 100°

∴ ∠BOD = 100°

Given: CB = CD and ∠CBD = 35°

Consider ΔBCD

Here,

CB = CD (given)

∠CBD = ∠CDB = 35° (In a triangle, angles opposite to equal sides are equal)

By angle sum property

∠BCD + ∠CBD + ∠CDB = 180°

∠BCD + 35° + 35° = 180°

∠BCD = 180° – 35° – 35° = 110°

We know that,

In a cyclic quadrilateral opposite angles are supplementary

∴ ∠BCD + ∠BAD = 180°

110° + ∠BAD = 180°

∠BAD = 180° – 110° = 70°

∴ ∠BAD = 70°

Given: ΔABS is equilateral

In ΔABC

∠BAC = 60° (All angles in equilateral triangle are equal to 60°)

We know that,

In a cyclic quadrilateral opposite angles are supplementary

∴ ∠BAC + ∠BDC = 180°

60° + ∠BDC = 180°

∠BDC = 180° – 60° = 120°

∴ ∠BDC = 120°

Given: ∠CBE = 100°

Here,

∠ABE = ∠ABC + ∠CBE

180° = ∠ABC + 100°

∠ABC = 180° – 100° = 80°

We know that,

In a cyclic quadrilateral opposite angles are supplementary

∴ ∠ABC + ∠ADC = 180°

80° + ∠ADC = 180°

∠ADC = 180° – 80° = 100°

Here,

∠ADF = ∠ADC + ∠CDF

180° = 100° + ∠CDF

∠CDF = 180° – 100° = 80°

∴ ∠CDF = 80°

Given: ∠AOB = 140°

Here,

(Exterior ∠AOB) = 360° – (interior ∠AOB)

(Exterior ∠AOB) = 360° – 140°

(Exterior ∠AOB) = 220°

We know that,

(Exterior ∠AOB) = 2 × ∠ACB

220° = 2 × ∠ACB

∠ACB = = 110°

∴ ∠ACB = 110°

Given: ∠AOB = 130°

Here,

(Exterior ∠AOB) = 360° – (interior ∠AOB)

(Exterior ∠AOB) = 360° – 130°

(Exterior ∠AOB) = 230°

We know that,

(Exterior ∠AOB) = 2 × ∠ACB

230° = 2 × ∠ACB

∠ACB = = 115°

∴ ∠ACB = 115°

Given: ABCD, ABEF are two cyclic quadrilaterals and ∠BCD = 110°

In Quadrilateral ABCD

We know that,

In a cyclic quadrilateral opposite angles are supplementary

∴ ∠BCD + ∠BAD = 180°

110° + ∠BAD = 180°

∠BAD = 180° – 110° = 70°

Similarly in Quadrilateral ABEF

∴ ∠BAD + ∠BEF = 180°

70° + ∠BEF = 180°

∠BEF = 180° – 70° = 110°

∴ ∠BEF = 110°

Given: is a cyclic quadrilateral, CF||AB, and

Here, CF|| AB

Hence BC is transversal

∴ ∠ABC = ∠BCF = 85° (Alternate interior angles)

Here,

∠DCB + ∠BCF + ∠ECF = ∠DCE

∠DCB + 85° + 20° = 180°

∠DCB = 180° – 85° – 20° = 75°

We know that,

In a cyclic quadrilateral opposite angles are supplementary

∴ ∠DCB + ∠BAD = 180°

75° + ∠BAD = 180°

∠BAD = 180° – 75° = 105°

∴ ∠BAD = 105°

Given: AB = 11cm, BE = 3cm and DE = 3.5cm

Construction: Join AC

Here,

AE: CE = DE: BE

AE×BE = DE×CE

(AB + BE)×BE = DE×(CD + DE)

(11 + 3)×3 = 3.5×(CD + 3.5)

14×3 = 3.5×(CD + 3.5)

3.5×(CD + 3.5) = 42

(CD + 3.5) = = 12

CD = 12—3.5 = 8.5

∴ CD = 8.5

Given: two circles having radii 6 cm and 3 cm

Construction: join AP

Consider ΔABP

Here,

AP² = AB² + BP²

5² = 4² + 3²

25 = 16 + 9

25 = 25

∴ ΔABP is right angled triangle

PQ = 2×BP

PQ = 2×3 = 6cm

∴PQ = 6cm

Given: and

Construction: join CD

We know that,

∠AOB = 2 × ∠ACB

90° = 2 × ∠ACB

∠ACB = = 45°

Similarly,

∠COA = 2 × ∠CBA

∠COA = 2 × 30

∠COA = 60°

Here,

∠COD + ∠COA = ∠AOD

∠COD + 60° = 180°

∠COD = 180° – 60° = 120°

Again

∠COD = 2 × ∠CAO

∠CAO = = 60°

∴∠CAO = 60°

Here, Clearly I and III are correct.

Let us check for II statement

Construction: Let B and C be the centers of two circles having radii 10cm and 17 cm respectively and let AD be the common chord cutting BC at E.

Here,

AE = ED = 8cm

Now, in ΔABE

BE² = AB² – AE²

BE² = (10)² – (8)²

BE² = 100– 64 = 36

BE = 6cm

Now, in ΔAEC

EC² = AC² – AE²

EC² = (17)² – (8)²

EC² = 289– 64 = 225

EC = 25cm

Here,

BC = BE + EC = 6 + 15 = 21cm

But, it is given BC = 23cm

∴ Statement II is false

Here,

ABCD is said to be cyclic quadrilateral

If either of any point is satisfied

i)Points and lie on a circle.

ii)∠ B + ∠C = 180°

Here,

right – angled at B

If both the conditions satisfy

i)is a cyclic quadrilateral

ii)

Assertion (A):

Construction: Draw a Δ ABC in which AB = AC, Let O be the midpoint of AB and with O as centre and OA as radius draw a circle, meeting BC at D

Now, In Δ ABD

∠ ADB = 90° (angle in semicircle)

Also, ∠ ADB + ∠ ADC = 180°

90° + ∠ ADC = 180°

∠ ADC = 180° – 90°

∠ ADC = 90°

Consider Δ ADB and Δ ADC

Here,

AB = AC (given)

AD = AD (common)

∠ ADB = ∠ ADC ( 90° )

∴ By SAS congruency, Δ ADB Δ ADC

So, BD = DC(C.P.C.T)

Thus, the given circle bisects the base. So, Assertion (A) is true

Reason (R) :

Let ∠ BAC be an angle in a semicircle with centre O and diameter BOC

Now, the angle subtended by arc BOC at the centre is ∠ BOC = 2× 90°

∠ BOC = 2× ∠ BAC = 2× 90°

So, ∠ BAC = 90° (right angle)

So, reason (R) is true

Clearly, reason (R) gives assertion (A)

Hence, correct choice is A

Assertion (A) :

Let O be the centre of the circle and AB be the chord

Construction: Draw, L is the midpoint of AB

Here,

OA = 10cm

AL = AB = 8cm

In ΔOAL,

OL² = OA ² – AL²

OL² = (10)² – (8) ²

OL² = 100 – 64

OL = = 6cm

Thus, Assertion (A) is true.

Clearly, reason (R) given Assertion (A).

Hence, the correct choice is A.

Clearly, reason (R) is true.

Assertion (A) :

OA = 13cm

OL = 12cm

In ΔOAL,

AL² = OA² – OL²

AL² = (13)² – (12)²

AL² = 169 – 144

OL = = 5cm

Now, AB = 2 × AL = 2 × 5 = 10cm

Thus, Assertion (A) is true

∴ Reason (R) and Assertion (A) are both true but reason (R) does not gives Assertion (A).

Hence, correct choice is B

Assertion (A) :

Here, in ΔABC

By angle sum property

∠ABC + ∠BCA + ∠CAB = 180°

70° + 30° + ∠CAB = 180°

∠CAB = 180° – 70° – 30° = 80°

∠CAB = ∠BDC = 80° (angles in same segment)

But given that ∠BDC = 70°

∴ Assertion(A) is wrong.

Reason (R) :

∠ ADC = ∠ AOC = × 130° = 65°

∠ ABC + ∠ ADC = 180°

∠ ABC + 65° = 180°

∠ABC = 180° – 65° = 115°

Reason (R) is true

Assertion (A) :

∠ABC + ∠BCA + ∠BAC = 180°

70° + 30° + ∠BAC = 180°

∠BAC = 180° – 70° – 30°

∠ BAC = 80°

∴ ∠BDC = ∠BAC = 80° (angles in the same segment)

This is false.

Thus, Assertion (A) is false and Reason (R) is true.

Hence, correct choice is D

Clearly, Assertion (A) is false and Reason (R) is true.

Hence, correct choice is D

Clearly, Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).

(i) T

(ii) T

(iii) T (The region inside the circle, region outside the circle and region on the circle).

(iv) T ( because point P lies inside the circle)

(v) F (A circle can have infinite number of chords)

(a) Angle in a semicircle measures – 90° (r)

(b) In the given figure, O is the centre of a circle. If ∠AOB = 120°, then ∠ACB = ?

∠AOB = ∠ACB

∠ACB = × 120° = 60°

∠ACB = 60° (s)

(c) In the given figure, O is the centre of a circle. If and then

Here, OP = OR = OQ (radius)

In ΔPOR

∠OPR = ∠ORP (angles opposite to equal sides are equal)

By angle sum property

∠POR + ∠OPR + ∠ORP = 180°

90° + 2×∠OPR = 180°

2×∠OPR = 180°—90°

2×∠OPR = 90°

∠OPR = 45°

Similarly in Δ POQ

∠OPQ = ∠OQP (angles opposite to equal sides are equal)

By angle sum property

∠POQ + ∠OPQ + ∠OQP = 180°

110° + 2×∠OQP = 180°

2×∠OQP = 180°–110°

2×∠OQP = 70°

∠OQP = 35°

∠QPR = ∠QPO + ∠OPR = 45° + 35° = 80°

∴ ∠QPR = 80° (q)

(d) In cyclic quadrilateral it is given that and is a diameter of the circle

through and Then,

Here,

∠ADC + ∠ABC = 180° (opposite angles in cyclic quadrilateral are supplymentary)

130° + ∠ABC = 180°

∠ABC = 180°–130° = 50°

In ΔABC

By angle sum property

∠BAC + ∠ABC + ∠ACB = 180°

∠BAC + 50° + 90° = 180°

∠BAC = 180°–50°–90° = 40°

∴∠BAC = 40° (p)

∴ Answers are: (a) – (r), (b) – (s), (c) – (q), (d) – (p)

(i) Two circles having the same centre and different radii are called concentric cricles.

(ii) Diameter is the longest chord of a circle.

(iii) A continuous piece of a circle is called the arc of the circle.

(iv) An arc of a circle is called a semicircle if the ends of the arc are the ends of a diameter.

(v) A segment of a circle is the region between an arc and a chord of the circle.

(vi) A line segment joining the centre to any point on the circle is called its radius.

Given: and

Here,

∠ACB = ∠ADB = 40° (angles in same segment)

∠BEC = ∠AED = 105° (vertically opposite angles)

In ΔAED

By angle sum property

∠ADE + ∠AED + ∠EAD = 180°

40° + 105° + ∠EAD = 180°

∠EAD = 180° – 40° – 105° = 35°

∴ ∠EAD = 35°

Given: and

We know that,

∠AOB = 2× ∠ACB

∠AOB = ∠ACB

×90° = ∠ACB

∠ACB = 45°

Now, consider ΔABC

By angle sum property

∠ACB + ∠ABC + ∠CAB = 180°

45° + 30° + ∠CAB = 180°

∠CAB = 180° — 45° — 30° = 105°

Consider ΔAOB

Here,

OA = OB (radius)

Let OA = OB = x

By angle sum property

∠AOB + ∠OAB + ∠OBA = 180°

90° + x + x = 180°

2x = 180° – 90° = 90°

x = 45°

Now,

∠CAB = ∠BAO + ∠CAO = 105°

∠CAO = 105° – 45° = 60°

∴ ∠CAO = 60°

Given: ∠OAB = 40°

Consider ΔAOB

Here,

OA = OB (radius)

∠OBA = ∠OAB = 40° (angles opposite to equal sides are equal)

By angle sum property

∠OBA + ∠OAB + ∠AOB = 180°

40° + 40° + ∠AOB = 180°

∠AOB = 180° — 40° — 40° = 100°

We know that,

∠AOB = 2× ∠ACB

∠AOB = ∠ACB

×100° = ∠ACB

∠ACB = 50°

∴∠ACB = 50°

Given: ∠DAB = 60° and ∠ABD = 50°

In ΔABD

By angle sum property

∠DAB + ∠ABD + ∠ADB = 180°

60° + 50° + ∠ADB = 180°

110° + ∠ADB = 180°

∠ADB = 180° – 110° = 70°

Here,

∠ADB = ∠ACB = 70° (angles in same segment)

∴∠ACB = 70°

Given:

Consider ΔAOB

Here,

OA = OB (radius)

∠OBA = ∠OAB = 60° (angles opposite to equal sides are equal)

By angle sum property

∠OBA + ∠OAB + ∠AOB = 180°

60° + 60° + ∠AOB = 180°

∠AOB = 180° — 60° — 60° = 60°

Here,

∠BOC = ∠BOA + ∠AOC = 180°

60° + ∠AOC = 180°

∠AOC = 180° – 60° = 120°

We know that,

∠AOC = 2× ∠ADC

∠AOC = ∠ADC

×120° = ∠ADC

∠ADC = 60°

∴∠ADC = 60°

Given radius(AO) = 15cm

Length of the chord (AB) = x

distance of the chord from the centre is 9cm.

Draw a perpendicular bisector from center to the chord and name it OC.

AC = BC

Now in ∆ AOC

Using Pythagoras theorem

AO² = AC² + OC²

15² = AC² + 9²

AC² = 15² – 9²

AC² = 225 – 81

AC² = 144

AC = 12cm

BC = 12cm

The length of the chord is AC + BC = 12 + 12 = 24 cm.

Given: AB = CD

Construction: Drop perpendiculars OX and OY on to AB and CD respectively and join OA and OD.

Here, OX AB (perpendicular from center to chord divides it into two equal halves)

AX = BX = – – (1)

OY CD (perpendicular from center to chords divides it into equal halves

CY = DY = – – (2)

Now, given that

AB = CD

∴ =

AX = DY (from –1 and –2 ) – – (3)

In ΔAOX and ΔDOY

∠OXA = ∠OYD (right angle)

OA = OD (radius)

AX = DY (from –3 )

∴ BY RHS congruency

ΔAOXΔDOY

OX = OY (by C.P.C.T)

Hence proved.

We know that,

∠POQ = 2∠PAQ

= ∠PAQ

= ∠PAQ

90° = ∠PAQ

∠PAQ = 90°

Hence proved

We know that,

A chord nearer to the center is longer than the chord which is far from the center

∴ Diameter is the longest chord in the circle (because it passes through the center and other chords are far from the center)

Given: and

Consider ΔOAB

Here,

OA = OB (radius)

∠OBA = ∠OAB = 30° (angles opposite to equal sides are equal)

Similarly, in ΔAOC

OA = OC (radius)

∠OCA = ∠OAC = 40° (angles opposite to equal sides are equal)

Here,

∠CAB = ∠OAB + ∠OAC = 30° + 40° = 70°

Here,

2×CAB = BOC (The angle subtended by an arc at the center is twice the angle subtended by the same arc on any point on the remaining part of the circle).

2×CAB = BOC

2×70° = BOC

BOC = 140°.

BOC = 140

Given: arc

Here,

2×AXB = BYC

∴ 2×∠AOB = ∠BOC

∠AOB = ∠BOC –1

Here,

∠AOC = ∠AOB + ∠BOC = 180°

∠BOC + ∠BOC = 180° (from –1)

∠BOC = 180°

∠BOC = ×180° = 120°

∴

Given: ∠ABC = 45°

We know that ,

∠ AOC = 2×∠ABC

∠ AOC = 2×45 = 90°

∴ ∠AOC = 90°

Therefore

Hence proved.

Given: ∠ADC = 130° , BC = BE

We know that,

(exterior ∠AFC ) = (2×∠ADC)

(exterior ∠AFC ) = (2×130)

(exterior ∠AFC ) = 260

∠AFC = 360° – (exterior ∠AFC) = 360°–260° = 100°

∠AFB = ∠AFC + ∠CFB = 180°

∠AFC + ∠CFB = 180°

100° + ∠CFB = 180°

∠CFB = 180° – 100° = 80°

In quadrilateral ABCD

∠ADC + ∠ABC = 180° (opposite angles in cyclic quadrilateral are supplementary)

130° + ∠ABC = 180°

∠ABC = 180° – 130° = 50°

In ΔBCF

By angle sum property

∠CBF + ∠CFB + ∠BCF = 180°

50° + 80° + ∠BCF = 180°

∠BCF = 180° – 50° – 80° = 50°

Now,

∠CFE = ∠CFB + ∠BFE = 180°

∠CFB + ∠BFE = 180°

80° + ∠BFE = 180°

∠BFE = 180° – 80° = 100°

Here,

In ΔBCE

BC = BE (given)

∠BCE = ∠BEC = 50° (angles opposite to equal sides are equal)

By angle sum property

∠BCE + ∠BEC + ∠CBE = 180°

50° + 50° + ∠CBE = 180°

∠CBE = 180° – 50° – 50° = 100°

∴

Given: ∠ACB = 40°

We know that ,

∠ AOB = 2×∠ACB

∠ AOB = 2×40 = 80°

∴ ∠AOB = 80°

In ΔAOB

OA = OB (radius)

∠OAB = ∠OBA (angles opposite to equal sides are equal)

Let ∠OAB = ∠OBA = x

By angle sum property

∠AOB + ∠OAB + ∠OBA = 180°

80 + x + x = 180°

80 + 2x = 180°

2x = 180° – 80° = 100°

x = = 50°

∴

Given: and

Here,

In ΔAOB

OA = OB (radius)

∠OAB = ∠OBA (angles opposite to equal sides are equal)

∴ ∠ OBA = 30°

Now, by angle sum property

∠AOB + ∠OBA + ∠OAB = 180°

∠AOB + 30° + 30° = 180°

∠AOB = 180° – 30° – 30°

∠AOB = 120°

Now, Consider Δ BOC

OC = OB (radius)

∠OCB = ∠OBC (angles opposite to equal sides are equal)

∴ ∠ OBA = 55°

Now, by angle sum property

∠BOC + ∠OBC + ∠OCB = 180°

∠BOC + 55° + 55° = 180°

∠BOC = 180° – 55° – 55° = 70°

∴ ∠BOC = 70°

Here,

∠AOB = ∠AOC + ∠BOC

120° = ∠AOC + 70°

∠AOC = 120° – 70°

∠AOC = 50°

∴ ∠AOC = 50°

∴

Given: and

In ΔOBD

OB = OD = DB

∴ Δ OBD is equilateral

∴ ∠ODB = ∠DBO = ∠BOD = 60°

Consider ΔDEB and ∠BEC

Here,

BE = BE (common)

∠CEB = ∠DEB (right angle)

CE = DE ( OE is perpendicular bisector)

∴ By SAS congruency

ΔDEB ∠BEC

∴∠DEB = ∠EBC (C.P.C.T)

∴∠EBC = 60°

Now, in ΔABC

∠EBC = 60°

∠ACB = 90° (angle in semicircle)

By angle sum property

∠EBC + ∠ACB + ∠CAB = 180°

60° + 90° + ∠CAB = 180°

∠CAB = 180° – 60° – 90° = 30°

∴∠CAB = 30°

Here,

In cyclic Quadrilateral ABFE

∠ABF + ∠AEF = 180° (opposite angles in cyclic quadrilateral are supplementary) –1

In cyclic Quadrilateral ABCD

∠ABC + ∠ADC = 180° (opposite angles in cyclic quadrilateral are supplementary) –2

From –1 and –2

∠ABF + ∠AEF = ∠ABC + ∠ADC

∠AEF = ∠ADC (∠ABF = ∠ABC)

Since these are corresponding angles

We can say that EF || DC

∴ EF||DC

Hence proved.

Construction: Join AE

Consider cyclic quadrilateral ACDEA

Here,

∠ACD + ∠DEA = 180° (opposite angles in cyclic quadrilateral are supplementary)

Also,

∠AEB = 90° (angle in semicircle)

∴∠ACD + ∠DEA + ∠AEB = 180° + 90°

∠ACD + ∠BED = 270° (∠DEA + ∠AEB = ∠BED)

∴∠ACD + ∠BED = 270°

Hence proved.

Given:

In ΔEOC

By angle sum property

∠EOC + ∠OEC + ∠OCE = 180°

∠EOC + 90° + 30° = 180°

∠EOC = 180° – 90° – 30° = 60°

∠EOC = 60°

Here,

∠EOD = ∠EOC + ∠COD = 90°

∠EOC + ∠COD = 90°

60° + ∠COD = 90°

∠COD = 90° – 60° = 30°

Now,

∠ AOC = ∠AOD + ∠COD = 90° + 30° = 120°

We know that ,

∠ COD = 2×∠CBD

∠ COD = ∠CBD

∠CBD = × 120° = 60°

Consider ΔABE

By angle sum property

∠AEB + ∠ABE + ∠BAE = 180°

90° + 60° + ∠BAE = 180°

∠BAE = 180° – 90° – 60° = 30°

∴ x = 30°

We know that ,

∠ AOC = 2×∠ABC

∠ AOC = ∠ABC

∠ABC = × 30° = 15°

∴y = 15°

∴

Given: chords PQ and RQ are equidistant from center.

Here consider ΔPQS and ΔRQS

Here,

QS = QS (common)

∠QPS = ∠QRS (right angle)

PQ = QS (chords equidistant from center are equal in length)

∴ By RHS congruency ΔPQS ΔRQS

∴ ∠RQS = ∠SQP and ∠RSQ = ∠QSP (by C.P.C.T)

Therefore we can say that diameter passing through Q bisects and

Given: Three non collinear points P, Q and R

Construction: Join PQ and QR.

Draw perpendicular bisectors AB of PQ and CD of QR. Let the perpendicular bisectors intersect at the point O.

Now join OP, OQ and OR.

A circle is obtained passing through the points P, Q and R.

Proof:

We know that,

Every point on the perpendicular bisector of a line segment is equidistant from its ends

points.

Thus, OP = OQ (Since, O lies on the perpendicular bisector of PQ)

and OQ = OR. (Since, O lies on the perpendicular bisector of QR)

So, OP = OQ = OR.

Let OP = OQ = OR = r.

Now, draw a circle C(O, r) with O as centre and r as radius.

Then, circle C(O, r) passes through the points P, Q and R.

Next, we prove this circle is the only circle passing through the points P, Q and R.

If possible, suppose there is a another circle C(O′, t) which passes through the points P, Q, R.

Then, O′ will lie on the perpendicular bisectors AB and CD.

But O was the intersection point of the perpendicular bisectors AB and CD.

So, O ′ must coincide with the point O. (Since, two lines cannot intersect at more than one point)

As, O′P = t and OP = r; and O ′ coincides with O, we get t = r .

Therefore, C(O, r) and C(O, t) are congruent.

Thus, there is one and only one circle passing through three the given non – collinear points.

Construction: Join OX and OY

In Δ OPX and ΔORY,

OX = OY (radii of the same circle)

OP = OR (sides of the square)

∴ΔOPX Δ ORY (RHS rule)

∴ PX = RY (CPCT)—1

OPQR is a square

∴ PQ = RQ

∴ PX + QX = RY + QY

QX = QY (from –1)

Hence proved

Given: AB = AC

Construction: join OA, OB and OC

Proof:

Consider ΔAOB and ΔAOC

Here,

OC = OB (radius)

OA = OA (common)

AB = AC (given)

∴ By SSS congruency

ΔAOB ΔAOC

∴ ∠OAC = ∠OAB (by C.P.C.T)

Hence, we can say that OA is the bisector of ∠BAC, that is O lies on the bisector of ∠BAC.