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Given, PQ is a chord of a circle with center O. Also, ∠QPT = 60°.

let x be the point on the tangent PT.

Now, ∠QPT + ∠QPX = 180°  [Linear pair]

⇒ ∠QPX = 180° - ∠QPT = 180° - 60° = 120°

Again, ∠QPX = ∠PRQ  [Alternate segment theorem]

⇒ ∠PRQ = 120°

PQ is a chord in a circle with center O and MN is a tangent drawn at point R on the circle PQ is parallel to MN

To Prove : R bisects the arc PRQ i.e. arc PR = arc PQ

Proof :

∠1 = ∠2 [Alternate Interior angles]

∠1 = ∠3 [angle between tangent and chord is equal to angle made by chord in alternate segment]

So, we have

∠2 = ∠3

QR = PR [Sides opposite to equal angles are equal]

As the equal chords cuts equal arcs in a circle.

Arc PR = arc RQ

R bisects the arc PRQ .

Hence Proved

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. Distance between AB and CD is 6 cm. 

To Prove: Radius of the circle, OP = ? 

Construction: Join OP and OQ, OB and OD. 

Proof: Chord AB || Chord CD. 

AB = 5 cm, and CD =11 cm. OP⊥AB 

∴ BP = AP = \(\frac{5}{2}\) = 2.5 cm. 

OQ⊥CD 

∴ CQ = QD = \(\frac{11}{2}\) = 5.5 cm. 

PQ = 6 cm. (Data) 

Let OQ = 2 cm then, OP = (6 – x) cm. 

In ∆BPO, ∠P = 90° 

As per Pythagoras theorem, 

OB2 = BP² + PO² = (2.5)² + (6 – x)² 

= 6.25 + 36 – 12x + x² 

OB² = x²– 12x + 42.25 …………….. (i) 

In ∆OQD, ∠Q= 90° 

∴ OD² = OQ² + QD² = (x)² + (5.5)² 

OD² = x² + 30.25 ……………….. (ii) 

OB = OD (∵ radii of same circle) 

From (i) and (ii). 

x² – 12x + 42.25 

= x² + 30.25 -12x 

= 30.25 – 42.25 -12x 

= -12 

12x = 12 

∴ x = \(\frac{12}{12}\)

∴ x = 1 cm. 

From (ii), OD² = x² + 30.25 

= (1)2 + 30.25 

= 1 + 30.25 

∴ OD² = 31.25 

OD = √(31.25)

∴ OD = 5.59 cm. 

∴ Radius of circle OP = OD = 5.59 cm.

It is given that AB and CD of a circle intersect each other at P outside the circle. If AB = 6cm, BP = 2cm and PD = 2.5cm

So we get

AP × BP = CP × DP

From the figure we know that CP = CD + DP

By substituting the values

8 × 2 = (CD + 2.5) × 2.5 cm

Consider x = CD

So we get

8 × 2 = (x + 2.5) × 2.5

On further calculation

16 = 2.5x + 6.25

It can be written as

2.5x = 16 – 6.25

By subtraction

2.5x = 9.75

By division

x = 9.75/2.5

So we get

x = 3.9cm

Therefore, CD = 3.9cm.

Proof: 

arc ABC is intercepted by the inscribed angle ∠ADC. 

∴ ∠ADC m(arcABC) (i) [Inscribed angle theorem] 

Similarly, ∠ABC is an inscribed angle. It intercepts arc ADC. 

∴ ABC = 1/2 m(arc ADC) (ii) [Inscribed angle theorem] 

∴ ∠ADC + ∠ABC 

= 1/2 m(arcABC) + 1/2 m(arc ADC) [Adding (i) and (ii)] 

∴ ∠D + ∠B = 1/2 m(areABC) + m(arc ADC)] 

∴ ∠B + ∠D = 1/2 × 360° [arc ABC and arc ADC constitute a complete circle] = 180°

∴ ∠B + ∠D = 180° 

Similarly we can prove, 

∠A + ∠C = 180°

Given: C is the centre of circle. ∠PQR and ∠PSR are inscribed in same arc PTR. 

To prove: ∠PQR ≅ ∠PSR

Proof:

i. arc PTR is intercepted by ∠PQR. 

ii. arc PTR is intercepted by ∠PSR. 

iii. ∠PQR = 1/2 m(arc PTR), and (i) [inscribed angle theorem] 

∠PSR = 1/2 m(arcPTR) (ii) [Inscribed angle theorem] 

∴ ∠PQR ≅ ∠PSR [From (i) and (ii)]

If ∠PQR = 100°, then 

∠OPR = ? 

∠PQR = 100°. 

∴ ∠POR = ? 

∠POR = 2 × ∠PQR = 2 × 100° 

∠POR = 200° 

∴ ∠POR – Reflex angle 

∠POR = 360° 

∴ ∠POR = 360 – 200 

∴ ∠POR = 160° 

In ∆POR, 

OP = OR radii. 

∴ ∠OPR = ∠ORB 

∴ ∠OPR + ∠ORP + ∠POR = 180° 

∠OPR + ∠OPR + 160 = 180 

2∠OPR+ 160 = 180 

2∠OPR = 180 – 160 

2∠OPR = 20 

∠OPR = \(\frac{20}{2}\)

∴ ∠OPR= 10°.

(i) Angle subtended in the 

circumference, ∠BAD = ? 

(ii) Angle subtended in the circumference, 

∠BCD = ? 

In this figure BD is chord, 

OB is radius, it is equal to OD. 

∴ OB = OD = BD 

∴ ∆OBD is equilateral triangle. 

∴ each angle is equal to 60°. 

∴ angle subtended at the centre 

∠BOD = 60°. 

(i) Angle subtended in the circumference 

∠BAD= \(\frac{1}{2}\)× angle subtended at centre ∠BOD 

= \(\frac{1}{2}\) × 60° 

∴ ∠BAD = 30°. 

(ii) The sum of either pair of opposite angles of a cyclic quadrilateral is 180°. 

∴ In cyclic quadrilateral ABCD, 

∠BAD + ∠ACD = 180 

30 + ∠ACD = 180 (∵ ∠BAD = 30°) 

∴ ∠ADC = 180 – 30 

∴ ∠ACD = 150°.

∠AOB + ∠BOC = 60° + 30° = 90° 

∴ ∠AOC = 90° 

(Angle subtended at the centre) 

∠ADC = ? 

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. 

∴ Angle subtended at the centre 

∠AOC= 2 × ∠ADC 

90° = 2 × ∠ADC 90 

∴ ∠ADC = \(\frac{90}{2}\)= 45° 

∴ ∠ADC = 45°.

Two circles are drawn taking PQ and PR of a triangle as diameter. 

Let these intersect at P and S. 

To Prove: ∠ACP = ∠QCD 

Proof: ∠ABP = ∠QBD ………….. (i) 

(vertically opposite angles) 

∠ABP = ∠ACP ……….. (ii) 

(angles in the same segment) 

Similarly, ∠QCD = ∠QBD …………. (iii)(angles in the same segment) 

From (i), (ii), and (iii), 

∠ACP = ∠QCD.

PA = PB------(i)

As tangent drawn from external points to a circle are equal in length

PL + AL = PN + BN------(ii)

PLA and PNB are the two tengent which are equal

AL = ML and BN = MN--------(iii)

From (ii) and (iii)

PL = ML

Given : AB, BC and AC are tangents to the circle at E, D and F. 

BD = 30 cm and DC = 7 cm and ∠BAC = 90° 

Recall that tangents drawn from an exterior point to a circle are equal in length 

Hence BE = BD = 30 cm 

Also FC = DC = 7 cm 

Let AE = AF = x → (1) 

Then AB = BE + AE = (30 + x) 

AC = AF + FC = (7 + x) 

BC = BD + DC = 30 + 7 = 37 cm 

Consider right Δ ABC, by Pythagoras theorem we have 

BC2 = AB² + AC² 

⇒ (37)² = (30 + x)² + (7 + x⁾² 

⇒ 1369 = 900 + 60x + x² + 49 + 14x + x² 

⇒ 2x² + 74x + 949 – 1369 = 0 

⇒ 2x²+ 74x – 420 = 0 

⇒ x² + 37x – 210 = 0 

⇒ x² + 42x – 5x – 210 = 0 

⇒ x (x + 42) – 5 (x + 42) = 0 

⇒ (x – 5) (x + 42) = 0 

⇒ (x – 5) = 0 or (x + 42) = 0 

⇒ x = 5 or x = – 42 

⇒ x = 5 [Since x cannot be negative] 

∴ AF = 5 cm [From (1)] 

Therefore AB = 30 +x = 30 + 5 = 35 cm 

AC = 7 + x = 7 + 5 = 12 cm 

Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle. 

Join point O, F; points O, D and points O, E. 

From the figure, 

Area of (ΔABC) = Area (ΔAOB) + Area (ΔBOC) + Area (ΔAOC) 

∴ r = 5 

Thus the radius of the circle is 5 cm

Proof: 

i. ∆QRS is an equilateral triangle, [Given] 

∴ seg RS ≅ seg QS ≅ seg QR [Sides of an equilateral triangle] 

∴ arc RS ≅ arc QS ≅ arc QR [Corresponding arcs of congruents chords of a circle are congruent] 

ii. Let m(arc RS) = m(arc QS)= m(arc QR) = x

m(arc RS) + m(arc QS) + m(arc QR) = 360° [Measure of a circle is 360° , arc addition property] 

∴ x + x + x = 360° 

∴ 3x = 360° 

∴ x = 360°/3 = 120° 

∴ m(arc RS) = m(arc QS) = m(arc QR) = 120° (i) 

Now, m(arc QRS) = m(arc QR) + m(arc RS) [Arc addition property]

= 120° + 120° [From (i)] 

∴ m(arc QRS) = 240°

∠OAB = ∠OCD (Given)

OA = OC (radius).

∠P = ∠Q = 90°

∴ (OP ⊥ AB, OQ ⊥ CD ) ∴ ΔOAP ≅ ΔOCQ;

∴ OP = OQ, Therefore AB = CD

[Equal chords of a circle are equidistant from die centre]

Given that, 

∠OBD = 50° 

Since, 

AB and CD are the diameters of the circles and O is the centre of the circle 

Therefore, 

∠PBC = 90° 

(Angle in the semi-circle) 

∠OBD + ∠DBC = 90° 

50° + ∠DBC = 90° 

∠DBC = 40° 

By degree measure theorem, 

∠AOC = 2∠ABC 

∠AOC = 2 x 40° 

= 80°

Correct answer is

(D) 8.8 or 2.2 cm

Two circles can touch each other internally or externally.

∴ Distance between centres = 5.5 + 3.3 or 5.5 – 3.3 

= 8.8 or 2.2

Option : (C)

Let O and O' be the centre of two circles 

OA and O'A = Radius of the circles 

AB be the common chord of both the circles 

OM perpendicular to AB 

And, 

O'M perpendicular to AB 

ΔAOO' is an equilateral triangle. 

AM = Altitude of AOO' 

Height of ΔAOO' = \(\frac{\sqrt 3}{2}\)r 

AB = 2 AM

= 2\(\frac{\sqrt 3}{2}\)r

= \(\sqrt{3}\)r

(C) The angles B and D is 60⁰ .

 The correct answer is  (B) 4√3cm.

Correct option is  A) AB = DC

Answer : (c) ∠z 

∠QSR = ∠QTR = \(\frac{1}{2}\) ×∠QOR = \(\frac{z}{2}\) 

(Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)

∠PSM = ∠PTM = 180º - \(\frac{z}{2}\) (Straight line) 

Also, ∠SMT = ∠QMR = y (vert. opp. ∠s) 

∴ In quad. PSMT, ∠SMT + ∠PTM + ∠TPS + ∠PSM = 360° 

⇒ y + 180° – \(\frac{z}{2}\) + x + 180° – \(\frac{z}{2}\) = 360° 

⇒ x + y = z.

(i) ∠COB = 2 ∠CAB = 2x° (angle at the centre = 2 × angle at the remaining part of the circumference) 

(ii) ∠OCD = ∠COB = 2x° (alternate ∠s, DC || AB) 

OD = OC (radii of the same circle) 

⇒ ∠OCD = ∠ODC 

⇒ ∠ODC = 2x° 

∴ In ∆DOC, ∠DOC = 180° – (2x° + 2x°) 

= 180° – 4x°  (∠sum prop. of a ∆) 

(iii) ∠DAC = \(\frac{1}{2}\)∠DOC = \(\frac{1}{2}\) (180 − 4x)°  

(angle made by arc DC at the centre = Twice the angle at the remaining part of the circumference) 

= (90 – 2x)° 

(iv) In ∆ADC, ∠ACD = ∠CAB = x° (alt ∠s; DC || AB) 

∴ ∠ADC = 180° – (x° + 90° – 2x°) 

= (90 + x)°.         (∠sum prop. of a ∆)

Answer is C. 8 cm

Given: 

OA = 5 cm 

OQ = 3 cm 

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency. 

Property 2: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal. 

By property 1, ∆OAQ is right-angled at ∠OQA (i.e., ∠OQA = 90°). 

By Pythagoras theorem in ∆OAQ, 

OA² = QA² + OQ²

⇒ QA² = OA² – OQ² 

⇒ QA² = 5² – 3² 

⇒ QA² = 25² – 9² 

⇒ QA² = 16

 ⇒ QA = √16 

⇒ QA = 4 cm 

By property 2,

BQ = BP (tangent from B) 

And, 

AQ = BQ = 4 cm [∵ Q is midpoint of AB] 

PB = PC = 4 cm [∵ P is midpoint of BC] 

Now, 

BC = BP + PC 

⇒ BC = 4 cm + 4 cm 

⇒ BC = 8 cm 

Hence, BC = 8 cm

Answer is C. 3

Given: 

AO’ = r 

O’P = r 

PO = r 

AO = AO’ + O’P + PO 

⇒ AO = r + r + r 

⇒ AO = 3r 

Now,

\(\frac{AO}{AO'}=\frac{3r}{r}\) = 3

Hence, AO: AO’ = 3

Correct option is: (A) 100°

We know that angle made by an arc at centre is double than the angle made by that arc at any point on the remaining circumference of circle.

\(\because\) \(\angle\) ACB is angle made by arc AB at point of circumference of circle.

&  \(\angle\) AOB (or x) is angle made by arc AB at centre of the circle.

\(\therefore\) x = 2 \(\angle\) ACB = 2 x 50° = 100°.

Correct option is: (A) 100°

Answer is C. 120°

Given: 

∠RPQ = 60° 

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency. 

Property 2: Sum of all angles of a triangle = 180°. 

By property 1, ∆OPR is right-angled at ∠OPR (i.e., ∠OPR = 90°). 

OP = OQ [∵ radius of circle] 

∴ ∠OPQ = ∠OQP = 30° 

Now by property 2, 

∠OPQ + ∠OQP + ∠POQ = 180° 

⇒ 30° + 30° + ∠POQ = 180° 

⇒ 60° + ∠POQ = 180° 

⇒ ∠POQ = 180° - 60° 

⇒ ∠POQ = 120° 

Hence, ∠POQ = 120°

Answer is A. 9 cm

Given: 

QP = 4.5 cm 

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal. 

By the above property, 

PQ = PT = PR = 4.5 cm (tangent from P) 

Now, 

QR = PQ + PR 

QR = PQ + PQ [∵PQ = PR] 

QR = 2PQ 

QR = 2 × 4.5 cm

QR = 9 cm 

Hence, QR = 9 cm

Answer is A. PQ

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal. 

By the above property,

PD = PA (tangent from P) 

QB = QA (tangent from Q) 

RC = RB (tangent from R) 

SC = SD (tangent from S) 

Now, 

PD + QB = PA + QA 

⇒ PD + QB = PQ [∵PQ = PA + QA] 

Hence, PD + QB = PQ

Correct option is: (D) PS

Lengths of tangents from a fixed outer point to circle are equal.

\(\therefore\) PB = PA (Tangents drawn from outer point P)

and SD = AS (Tangents drawn from outer point  S)

\(\therefore\) PB + SD = PA + AS = PS

Correct option is: (D) PS

In the given figure,

PA and PB are the tangents drawn from P, to the outer circle and inner circle respectively.

PA = 10 cm

OA and OB are the radii and

PA and PB are two tangents to the circles respectively

So,

OA ⊥ PA and OB ⊥ PB

In right ∆OAP,

By Pythagoras Theorem:

OP² = OA² + PA²

= (6)² + (10)²

OP² = 136 …(1)

From right ∆OBP,

OP² = OB² + PB²

136 = (4)² + PB²

136 = 16 + PB²

[Using equation (1)]

We have PB² = 136 – 16 = 120

Or PB = √120 cm = 2√30 cm = 2 x 5.47 = 10.9

Answer: Length of PB is 10.9 cm

∆ABC is an isosceles triangle where AB = AC and circumscribed a circle.

The circle touches its sides BC, CA and AB at P, Q and R respectively.

To Prove : P bisects the base BC, i.e. BP = PC

Now form figure,

BR and BP are tangents to the circle.

So, BR = BP …..(1)

AR and AQ are tangents to the circle.

AR = AQ But AB = AC

AB – AR = AC – AQ

BR = CQ …..(2)

Similarly, CP and CQ are tangents to the circle.

CP = CQ ……(3)

From (1), (2) and (3)

BP = PC

Hence Proved.

Solution: 12 = a + 2d
106 = a + 49d
So, 106-12 = 47d
Or, 94 = 47d
Or, d = 2
Hence, a = 8
And, n₂₉ = 8 + 28x2 = 64

Answer : (a) \(\frac{3}{4}\) 

∠CED = 120° (\(\because\) CEDA is a cyclic quad.)

⇒ ∠BED = 60° 

\(\therefore\) In ∆ EDB, ∠EDB = 90°

\(\therefore\)  \(\frac{BD}{BE}\) = cos 30° ⇒ \(\frac{6}{BE}\) = \(\frac{\sqrt{3}}{2}\)  ⇒ BE = 4\(\sqrt{3}\) cm.

BC = BE + CE =  4\(\sqrt{3}\) + 5\(\sqrt{3}\) =  9\(\sqrt{3}\)  cm

\(\because\) AB and CB are secants of the given circle, 

BD × BA = BE × EC 

⇒ 6 × BA = 4\(\sqrt{3}\)   × 9\(\sqrt{3}\) 

⇒ BA = 18 cm.

\(\because\) ∠ACB = 90°, ∆ ABC is a rt. ∠d ∆ 

⇒ AC2 = \(\sqrt{AB^2 -BC^2}\) = \(\sqrt{{18}^2 -(9\sqrt{3})^2}\) 

= \(\sqrt{324 -243}\) = \(\sqrt{81}\) = 9 cm. 

\(\therefore\) AD = AB – BD = 12 cm. 

\(\therefore\) AC : AD = 9 : 12  = 3 : 4 

= \(\frac{3}{4}\)

1. B) diameter

2. C) AC = AD

If four points are concyclic, then the line joining any two points subtend equal angles at the other two points which are on the same side of that line.