# Tangents and Intersecting Chords Chapter-18 Concise Maths Solutions

## Tangent Properties of Circles Concise Mathematics Solutions

**Tangents and Intersecting Chords Chapter-18** Concise Maths Solutions. Tangents and Chords properties of Circles for Concise Maths Solutions.** Solutions** of Exercise – 18 (A), Exercise – 18 (B), Exercise – 18 (C) for** Concise **Selina Maths of ICSE Board Class 10th. **Concise Maths Solutions Tangents and Intersecting Chords Chapter-18** for ICSE Maths Class 10 is available here.

## Tangents and Intersecting Chords Chapter-18 Concise Maths Solutions

All **Solutions **of **Concise** Selina **Maths** of **Tangents and Intersecting Chords** Chapter-18 has been solved according instruction given by council. This is the **Solutions **of **Tangents and Intersecting Chords** Chapter-18 for ICSE Class 10th. ICSE **Maths** text book of **Concise** is In series of famous ICSE writer in maths publications. **Concise** is most famous among students

### Concise Maths Solutions Chapter-18 Tangents and Intersecting Chords

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**Exercise – 18 (A), Exercise – 18 (B), Exercise – 18 (C) **

Concise Solutions Chapter-18 Tangents and Intersecting Chords

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**EXERCISE -18 (A)**, Tangents and Intersecting Chords Concise Solutions for ICSE Maths Class 10

**Question 1**

**The radius of a circle is 8 cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10 cm from its centre.**

**Answer 1**

OP = 10 cm,

radius OT = 8 cm

∵ OT ⊥ PT

∴ In right ΔOTP,

OP^{2 }= OT^{2}+PT^{2}

⇒ (10)^{2} =(8)^{2}+PT^{2
}and ⇒ 100 = 64+PT^{2
}⇒ PT^{2} = 100-64 = 36

∴ PT = = 6 cm

**Question 2.**

**In the given figure O is the centre of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.
**

**Answer 2**

∠OBA = 90° (Radius through the point of contact is perpendicular to the tangent)

and ⇒ OB^{2} = OA^{2} – AB^{2} ⇒ r^{2} = (r + 7.5)^{2} – 15^{2}

so ⇒ r^{2} = r^{2} + 56.25 + 15r – 225 168.75

hence ⇒ 15r= 168.75

⇒ r = ⇒ r=11.25

Hence, radius of the circle = 11.25 cm

**Question 3.**

**Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.**

**Answer 3**

**Given:** Two circles with centre O and O’ touches at P externally. Q is a point on the common tangent through P.

QA and QB are tangents from Q to the circles respectively.

**To Prove:** QA=QB.

**Proof:** From Q, QA and QP are the tangents to the circle with centre O

∴ QA=QP ….(i)

Similarly, QP and QB are the tangents to the circle with centre O’

∴ QP=QB ….(ii)

From (i) and (ii)

QA=QB Q.ED.

**Question 4.**

**Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent, are equal in length.**

**Answer 4**

**Given:** Two circles with centre O and O’ touch each other internally at P. Q is a point on the common tangent through P. QP an QB are tangents from Q to the circles respectively.

**Question 5.**

**Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.**

**Answer 5**

**Given:** Two concentric circles with radius 5 cm and 3 cm with centre O. PQ is the chord of the outer circle which touches the inner circle at L. Join OL and OP.

OL=3 cm, OP = 5 cm

**Question 6.**

**Three circles touch each other externally. A triangle is formed when the centres of these circles are j oined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.**

**Answer 6**

Three circles touches each other externally

Δ ABC is formed by joining the centres A, B and C of the circles.

AB = 6 cm, AC = 8 cm and BC = 9 cm

Let radii of the circles having centres A, B and C be r_{1}, r_{2}, r_{3} respectively

**Question 7.**

**If the sides of a quadrilateral ABCD touch a circle, prove that: AB + CD = BC + AD.**

**Answer 7**

**Given:** A circle touches the sides AB, BC, CD and DA of quad. ABCD at P,Q,R and S respectively.

**To Prove:** AB + CD = BC + AD

**Proof:** Since AP and AS are the tangents to the circle from external point A.

∴ AP = AS ….(i)

Similarly, we can prove that,

BP=BQ ….(ii)

CR=CQ …(iii)

DR=DS ….(iv)

Adding, we get:

AP + BP + CR + DR = AS + BQ + CQ + DS

and AP + BP + CR + DR = AS + DS + BQ + CQ

AB + CD = AD + BC

Hence AB + CD = BC + AD. Q.E.D.

**Question 8.**

**If the sides of a parallelogram touch a circle (refer figure of Q/7) prove that the parallelogram is a rhombus.
**

**Answer 8**

**Given :** The sides AB, BC, CD and DA of ||gm ABCD touches the circle at P, Q, R and S respectively.

**To Prove :** ABCD is a rhombus.

**Proof :** From A, AP and AS are the tangents to the circle.

∴ AP = AS ….(i)

**Question QUESTION 9.**

**From the given figure, prove that:**

**AP + BQ + CR = BP + CQ + AR.**

**Answer 9**

**Given:** In the figure, sides of Δ ABC touch a circle at P, Q, R.

**To Prove:
**(i) AP + BQ + CR = BP + CQ + AR

(ii) AP + BQ + CR = Perimeter of Δ ABC.

**Proof :**

∵ From B, BQ and BP are the tangents to the circle.

∴ BQ = BP ….(i)

Similarly we can prove that

AP= AR ……… (ii)

and CR = CQ …..(iii)

Adding we get

AP + BQ + CR = BP + CQ + AR ….(iv)

Adding AP + BQ + CR both sides,

2(AP + BQ + CR) = AP + PQ + CQ + QB + AR+CR.

⇒ 2 (AP + BQ + CR) = AB + BC + CA

∴ AP + BQ + CR = (AB + BC + CA)

= Perimeter of Δ ABC. Q.E.D.

**Question 10.**

**In the figure of Q.9 if AB = AC then prove that BQ = CQ.
**

**Answer 10**

**Given:** A circle touches the sides AB, BC, CA of Δ ABC at P, Q and R respectively, and AB = AC

**Question 11.**

**Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centres if **

**(i) they touch each other externally,**

**(ii) they touch each other internally.**

**Answer 11**

Radius of bigger circle = 6.3 cm

and raduis of smaller circle = 3.6 cm.

(i) Two circles touch each other at P externally. 0 and O’ are the centres of the circles.

Join OP and OP’

OP = 3.6 cm, O’P = 6.3 cm.

Adding we get

OO’ = OP + O’P = 3.6 + 6.3 = 9.9 cm

(ii) If the circles touch each other internally at P.

OP = 3.6 cm and O’P = 6.3 cm.

∴ OO’ = O’P – OP

= 6.3 – 3.6 = 2.7 cm

**Question 12.**

**From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that:**

**(i) ∠AOP = ∠BOP,**

**(ii) OP is the ⊥ bisector of chord AB.
**

**Answer 12**

**Given: ** A circle with centre 0. A point P out side the circle. From P, PA and PB are the tangents to the circle, OP and AB are joined.

**To prove:
**(i) ∠AOP = ∠BOP

(ii) OP is the perpendicular bisector of chord AB.

Proof : In ∆ AOP and ∆ BOP,

AP = BP (Tangents from P to the circle.)

OP = OP (Common)

OA = OB (Radii of the same circle)

∴ ∆ AOP s ∆ BOP (SSS postulate)

∴∠AOP = ∠BOP (C.P.C.T.)

Now in ∆ OAM and ∆ OBM,

OA = OB (Radii of the same circle)

OM = OM (Common)

∠AOM = ∠BOM (Proved ∠AOP = ∠BOP)

∴ ∆ OAM = ∆ OBM (S.A.S. Postulate)

∴ AM = MB (C.P.C.T.)

and ∠OMA = ∠OMB (C.P.C.T.)

But ∠OMA + ∠OMB = 180° (Linear pair)

∴ ∠OMA = ∠OMB = 90°

Hence OM or OP is the perpendicular bisector of AB. Q.E.D.

**Question 13.**

**In the given figure, two circles touch each other externally at point P, AB is the direct common tangent of these circles. Prove that:**

**(i) tangent at point P bisects AB.**

**(ii) Angle APB = 90°**

**Answer 13**

Given : Two circles with centre O and O’ touch each other at P externally. AB is the direct common tangent touching the circles at A and B respectively.

AP, BP are joined. TPT’ is the common tangent to the circles.

**To Prove :** (i) TPT’ bisects AB (ii) ∠APB = 90°

**Proof :**

∵ TA and TP arc the tangents to the circle

∴ TA = TP …(i)

Similarly TP. = TB ….(ii)

From (i) and (ii)

TA = TB

∴ TPT’ is the bisector of AB.

Now in ∆ ATP

TA = TP

∴ ∠TAP = ∠TPA

Similarly in A BTP.

∠TBP = ∠TPB

Adding we get.

∠TAP + ∠TBP = ∠APB

But ∠TAP + ∠TBP + ∠APB = 180°

∴ ∠APB = ∠TAP + ∠TBP = 90°. Q.E.D.

**Question 14.**

**Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:**

**∠PAQ = 2∠OPQ**

**Answer 14**

**Given:** A circle with centre O. two tangents PA and QA are drawn from a point A out side the circle OP, OQ. OA and PQ are joined.

**Question 15.**

**Two parallel tangents of a circle meet a third tangent at points P and Q. Prove that PQ subtends a right angle at the centre.**

**Answer 15**

**Given:** A circle with centre O, AP and BQ are two parallel tangents. A third tangent PQ intersect them at P and Q. PO and QO are joined

**Question 16.**

**ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.**

**Calculate the value of x, the radius of the inscribed circle.
**

**Answer 16**

In ∆ ABC, ∠B = 90°

OL ⊥ AB, OM ⊥ BC and

ON ⊥ AC.

**Question 17.**

**In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate :**

**(i) ∠ QOR**

**(ii) ∠ QPR given that ∠ A = 60°.
**

**Answer 17**

**Question 18.**

**In the following figure, PQ and PR are tangents to the circle, with centre O. If ∠ QPR = 60°, calculate:**

**(i) ∠ QOR**

**(ii) ∠ OQR**

**(iii) ∠ QSR.**

**Answer 18**

**Question 19.**

**In the given figure, AB is the diameter of the circle, with centre O, and AT is the tangent. Calculate the numerical value of x.**

**Answer 19**

**Question 20.**

**In quadrilateral ABCD; angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle. |1990]
**

**Answer 20**

BQ and BR arc the tangenls from B to the circle.

∴ BR = BQ = 27 cm.

∴ RC = 38-27 = 11 cm.

Since CR and CS are the tangents from C to the circle

∴ CS = CR= 11 cm.

∴ DS = 25 – 11 = 14 cm.

DS and DP are the tangents to the circle

∴ DS = DP

∴ ∠ PDS = 90° (given)

and OP ⊥ AD, OS ⊥ DC

∴ Radius = DS = 14 cm

**Question 21.**

**In the given, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.**

**Given ∠ SPR = x° and ∠ QRP = y°;**

**Prove that**

**(i) ∠ ORS = y°**

**(ii) Write an expression connecting x and y. [1992]**

**Answer 21**

∠ QRP = ∠ OSR = y (Angles in the alternate segment)

But OS = OR (radii of the same circle)

and ∴ ∠ ORS = ∠ OSR = y°

so ∴ OQ = OR (radii of the same circle)

hence ∴ ∠ OQR = ∠ ORQ = 90° – y° ….(i) (OR ⊥ PT)

But in ∆ PQR,

Ext. ∠ OQR = x° + y° ….(ii)

from (i) and (ii)

x° + y° = 90° – y°

⇒ x° + 2y° = 90°

**Question 22.**

**PT is a tangent to the circle at T. If ∠ ABC = 70° and ∠ ACB = 50°; calculate :**

**Answer 22**

**Question 23.**

**In the given figure. O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC 80°; find the angle BAC. [1996]**

**Answer 23**

Join OC

∴ PA and PC are the tangents

∴ OA ⊥ PA and OC ⊥ PC

In quad APCO,

∴ ∠ APC + ∠ AOC = 180°

⇒ 80° + ∠ AOC = 180°

∴ ∠ AOC = 180° – 80° = 100°

∠ BOC = 360° – (∠ AOB + ∠ AOC)

= 360° – (140° + 100°)

= 360° – 240° = 120°

Now arc BC subtends ∠ BOC at the centre and ∠BAC at the remaining part of the circle.

∴ ∠ BAC = ∠BOC= x 120° = 60°

**Question 24.**

In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.

**Answer 24**

∠CAB = ∠BAQ = 30°……(AB is angle bisector of ∠CAQ)

∠CAQ = 2∠BAQ = 60°……(AB is angle bisector of ∠CAQ)

∠CAQ + ∠PAC = 180°……(angles in linear pair)

∴∠PAC = 120°

∠PAC = 2∠CAD……(AD is angle bisector of ∠PAC)

∠CAD = 60°

Now,

∠CAD + ∠CAB = 60 + 30 = 90°

∠DAB = 90°

Thus, BD subtends 90° on the circle

So, BD is the diameter of circle

**EXERCISE – 18 (B),** Tangents and Intersecting Chords Concise Solutions ICSE Maths

**Question 1.**

**(i) In the given figure 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.**

**(ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4 cm. Find CD.**

**(iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.**

#### Answer 1

**Question 2.**

**In the figure given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find:**

**(i) AB.**

**(ii) the length of tangent PT. (2014)**

#### Answer 2

PT is tangent and PDC is secant out to the circle

∴ PT² = PC x PD

PT² = (5 + 7.8) x 5 = 12.8 x 5

PT² = 64 ⇒ PT = 8 cm

In ΔOTP

PT² + OT² = OP²

8²+x² = (x + 4)²

⇒ 64 +x² = x² + 16 + 8x

64- 16 = 8x

⇒ 48 = 8x

x = = 6 cm

∴ Radius = 6 cm

AB = 2 x 6 = 12 cm

**Question 3.**

**In the following figure, ****PQ is the tangent to the circle at A, DB is the diameter and O is the centre of the circle. If ∠ ADB = 30° and ∠ CBD = 60°, calculate**

**(i) ∠ QAB**

**(ii) ∠ PAD**

**(iii) ∠ CDB.**

#### Answer 3

(i) PAQ is a tangent and AB is the chord, ∠ QAB = ∠ ADB (Angles in the alternate segment)

= 30°

(ii) OA = OD (Radii of the same circle)

∴ ∠ OAD = ∠ ODA = 30°

But OA ⊥ PQ

∴ ∠ PAD = ∠ OAP – ∠OAD = 90° – 30° = 60°

(iii) BD is diameter

∴ ∠ BCD = 90° (Angle in semi circle)

Now in ∆ BCD,

∠ CDB + ∠ CBD + ∠ BCD = 180°

⇒ ∠ CDB + 60° + 90° = 180°

⇒ ∠ CDB = 180°- (60° + 90°) = 180° – 150° = 30°

**Question 4.**

**If PQ is a tangent to the circle at R; calculate:**

**(i) ∠ PRS
(ii) ∠ ROT.**

**Given O is the centre of the circle and angle TRQ = 30°.**

#### Answer 4

PQ is tangent and OR is the radius

∴ OR ⊥ PQ

∴ ∠ORT = 90° =

∠TRQ = 90° – 30° = 60°

But in ∆ OTR, OT = OR (Radii of the circle)

∴ ∠ OTR = 60° or ∠STR = 60°

But ∠PRS = ∠STR (Angles in the alternate segment) = 60°

In ∆ ORT, ∠ OTR = 60°, ∠ TOR = 60°

∴ ∠ROT= 180°-(60°+ 60°)= 180°-120° = 60°

**Question 5.**

**AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersect AB produced in D. Show that BC = BD.**

#### Answer 5

Given: In a circle, O is the centre,

AB is the diameter, a chord AC such that ∠ BAC = 30°

and a tangent from C, meets AB in D on producing. BC is joined.

**To Prove:** BC = BD

**Construction:** Join OC

**Proof :** ∠ BCD = ∠ BAC =30°(Angle in alternate segment)

Arc BC subtends ∠ DOC at the centre of the circle and ∠ BAC at the remaining part of the circle.

∴ ∠ BOC = 2 ∠ BAC = 2 x 30° = 60°

Now in ∆ OCD,

∠ BOC or ∠ DOC = 60° (Proved)

∠ OCD = 90° (∵ OC ⊥ CD)

∴ ∠ DOC + ∠ ODC = 90°

⇒ 60° + ∠ ODC = 90°

∴ ∠ ODC = 90°- 60° = 30°

Now in ∆BCD,

∵∠ ODC or ∠ BDC = ∠ BCD (Each = 30°)

∴ BC = BD Q.E.D.

**Question 6.**

**Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that ∆PQR is isosceles.**

#### Answer 6

In a circumcircle of ∆PQR, a tangent TPS is drawn through P which is parallel to QR

**To prove :** ∆PQR is an isosceles triangle.

**Proof:**

∵ TS QR

∠TPQ = ∠PQR (Alternate angles) ….(i)

∵ TS is tangent and PQ is the chord of the circle

∴ ∠TPQ = ∠RP (Angles in the alternate segment) ….(ii)

From (i) and (ii),

∠PQR = ∠QRP

∴ PQ = PR (Opposite sides of equal angles)

∴ ∆PQR is an isosceles triangle

Hence proved

**Question 7.**

**Two circles with centres O and O’ are drawn to intersect each other at points A and B. Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O’ at A. Prove that OA bisects angle BAC.**

#### Answer 7

**Given:** Two circles with centre O and O’ intersect each other at A and B, O lies on the circumference, of the other circle. CD is a tangent at A to the second circle. AB, OA are joined.

**To Prove:** OA bisects ∠ BAC.

**Construction:** Join OB, O’A, O’B and OO’

**Proof:** CD is the tangent and AO is the chord

∠ OAC = ∠ OBA …(i)

(Angles in alt. segment)

In ∆ OAB, OA = OB (Radii of the same circle)

∴ ∠ OAB = ∠ OBA ….(ii)

From (i) and (ii),

∠ OAC = ∠ OAB

∴ OA is the bisector of ∠ BAC Q.E.D.

**Question 8.**

**Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠ CPA = ∠DPB.**

#### Answer 8

**Given:**Two circles touch each other internally at P. A chord AB of the bigger circle intersects the smaller circle at C and D. AP, BP, CP and DP are joined.

**To Prove:** ∠ CPA = ∠ DPB

**Construction:** Draw a tangent TS at P to the circles given.

**Proof:**

∵ TPS is the tangent, PD is the chord.

∴ ∠ PAB = ∠ BPS …(i) ( Angles in alt. segment)

Similarly we can prove that

∠ PCD = ∠ DPS …(ii)

Subtracting (i) from (ii), we gel

∠ PCD – ∠ PAB = ∠ DPS – ∠ BPS

But in ∆ PAC,

Ext. ∠ PCD = ∠ PAB + ∠ CPA

∴ ∠ PAB + ∠ CPA – ∠ PAB = ∠ DPS – ∠ BPS

∠ CPA = ∠ DPB Q.E.D.

**Question 9.**

**In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.**

#### Answer 9

**Given:** ABCD is a cyclic quadrilateral and diagonal AC bisects ∠ BCD. AT A. a tangent TAS is drawn. BD is joined.

**To Prove:** TS || BD.

**Proof:** ∠ ADB = ∠ ACB …….(i) (Angles in the same segment)

Similarly ∠ ABD = ∠ ACD ……..(ii)

But ∠ ACB = ∠ ACD (AC is the bisector of ∠ BCD)

∴ ∠ ADB = ∠ ABD |From (i) and (ii)]
TAS is a tangent and AB is chord

∴ ∠ BAS = ∠ ADB (Angles in all segment)

But ∠ ADB = ∠ ABD (Proved)

∴ ∠ BAS = ∠ ABD

But these are alternate angles.

∴ TS || BD. Q.E.D.

**Question 10.**

**In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠ BCG = 108° and O is the centre of the circle,**

**Find:**

**(i) angle BCT**

**(ii) angle DOC**

#### Answer 10

Join OC, OD and AC.

(i) ∠ BCG + ∠ BCD = 180° (Linear pair)

⇒ 108° + ∠ BCD = 180°

(∵∠ BCG = 108° given)

∴∠ BCD = 180° – 108° = 72°

BC = CD (given)

∴ ∠ DCP = ∠ BCT

But ∠ BCT + ∠ BCD + ∠ DCP = 180°

∴∠ BCT + ∠ BCT + 72° = 180°

(∵∠ DCP = ∠ BCT)

2 ∠ BCT = 180° – 72° = 108°

∴∠ BCT = = 54°

(ii) PCT is the tangent and CA is chord

∴ ∠ CAD = ∠ BCT = 54°

But arc DC subtends ∠ DOC at the centre and

∠ CAD at the remaining part of the circle

∴ ∠ DOC = 2 ∠ CAD = 2 x 54° = 108°.

**Question 11.**

**Two circles intersect each other at points A and B. A straight line PAQ cuts the circles at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T arc concyclic.**

#### Answer 11

**Given:** Two circles intersect each other at point A and B. PAQ is a line which intersects circles at P, A and Q. At P and Q, tangents are drawn to the circles which meet at T.

**To Prove:** P, B, Q, T are concyclic.

**Construction:** Join AB, BP and BQ.

**Proof:** TP is the tangent and PA. a chord

∴ ∠ TPA = ∠ ABP

(angles in alt. segment)

Similarly we can prove that

∠ TQA = ∠ ABQ …,(ii)

Adding (i) and (ii), we get

∠ TPA + ∠ TQA = ∠ ABP + ∠ ABQ

But in ∆ PTQ,

∠ TPA + ∠ TQA + ∠ PTQ = 180°

and ⇒ ∠ TPA + ∠ TQA = 180° – ∠ PTQ

so ⇒ ∠ PBQ = 180°- ∠ PTQ

hence ⇒ ∠ PBQ + ∠PTQ = 180°

But there are the opposite angles of the quadrilateral

∴ Quad. PBQT is a cyclic

Hence P, B. Q and T are concyclic Q.E.D.

**Question 12.**

**In the figure; PA is a tangent to the circle. PBC is secant and AD bisects angle BAC.**

**Show that triangle PAD is an isosceles triangle. Also shaw that:**

#### Answer 12

**Given:** In a circle PA is the tangent, PBC is the secant and AD is the bisector of ∠BAC which meets the secant at D.

**To Prove:**

(i) ∆ PAD is an isosceles triangle.

(ii) ∠CAD = [(∠PBA – ∠PAB)]
**Proof:**

(i) PA is the tangent and AB is chord.

∠PAB = ∠C ….(i)

(Angles in the alt. segment)

AD is the bisector is ∠BAC

∴ ∠1 = ∠2 ….(ii)

In ∆ ADC,

Ext. ∠ADP = ∠C + ∠1

= ∠PAB + ∠2 = ∠PAD

∴ ∆ PAD is an isosceles triangle.

(ii) In A ABC,

Ext. ∠PBA = ∠C + ∠BAC

∴∠BAC = ∠PBA – ∠C

and ⇒ ∠1 + ∠2 = ∠PBA – ∠PAB [from (i)]
so ⇒ 2 ∠1 = ∠PBA – ∠PAB

therefore ⇒ ∠1 = – [∠PBA – ∠PAB]
⇒ ∠CAD = – [∠PBA – ∠PAB] Q.E.D.

**Question 13.**

**Two circles intersect each other at points A and B. Their common tangent touches the circles v at points P and Q as shown in the figure . Show that the angles PAQ and PBQ arc supplementary. [2000]**

Answer 13

**Given:** Two circles intersect each other at A and

B. A common tangent touches the circles at P and

Q. PA. PB, QA and QB are joined.

**To Prove:** ∠ PAQ + ∠ PBQ = 180°

or ∠ PAQ and ∠ PBQ are supplementary.

**Construction:** Join AB.

**Proof:** PQ is the tangent and AB is the chord

∴ ∠ QPA = ∠ PBA (alternate segment) ,…(i)

Similarly we can prove that

∠ PQA = ∠ QBA ,…(ii)

Adding (i) and (ii), we get

∠ QPA + ∠ PQA = ∠ PBA + ∠ QBA

But ∠ QPA + ∠ PQA = 180° – ∠ PAQ ,…(iii) (In ∆ PAQ)

and ∠ PBA + ∠ QBA = ∠ PBQ ,…(iv)

from (iii) and (iv)

∠ PBQ = 180° – ∠ PAQ

⇒ ∠ PBQ + ∠ PAQ = 180°

= ∠ PAQ + ∠ PBQ = 180°

Hence ∠ PAQ and ∠ PBQ arc supplementary Q.E.D.

**Question 14.**

**In the figure, chords AE and BC intersect each other at point D.**

**(i) If ∠ CDE = 90°.**

**AB = 5 cm, BD = 4 cm and CD 9 cm;**

**Find DE.**

**(ii) If AD = BD, show that AE = BC.**

#### Answer 14

**Question 15.**

**Circles with centres P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles arc congruent; show that CE = BD.**

#### Answer 15

**Given:** Two circles with centre P and Q intersect each other at A and B. CBD is a line segment and EBM is tangent to the circle with centre Q, at B. Radii of the cirlces are equal.

**To Prove:** CE = BD

**Construction:** Join AB and AD.

**Proof:** EBM is the tangent and BD is the chord

∴ ∠ DBM = ∠ BAD (Anglesi in alt. segment)

But ∠ DBM = ∠ CBE (Vertically opposite angles)

∴ ∠ BAD = ∠ CBE

∵ In the same circle or congruent circles, if angles are equal, then chords opposite to them are also equal.

∴ CE = BD Q.E.D.

**Question 16.**

**In the following figure O is the centre of ti.e circle and AB is a tangent to it at point B. ∠BDC = 65°, Find ∠BAO**

#### Answer 16

**Given:** ∠BDC = 65° and AB is tangent to circle with centre O.

∵ OB is radius

⇒ OB ⊥ AB

In ∆BDC

∠DBC + ∠BDC + ∠B CD = 180°

90° + 65° + ∠BCD =180°

⇒ ∠BCD = 25°

∵ OE = OC = radius

⇒ ∠OEC = ∠OCE

⇒ ∠OEC = 25°

Also, ∠BOE = ∠OEC + ∠OCE

[Exterior angle = sum of opposite interior angles in a ∆]
⇒ ∠BOE = 25°+ 25°

⇒∠BOE = 50°

⇒ ∠BOA = 50°

In ∆AOB

∠AOB + ∠BAO + ∠OBA = 180°

50° + ∠BAO + 90° = 180°

⇒ ∠BAO = 40°

**EXERCISE – 18 (C)**, Concise Maths Solutions Tangents and Intersecting Chords

**Question 1.**

**Prove that, of any two chords of a circle, the greater chord is nearer to the centre.**

**Answer 1**

Given: In circle with centre O and radius r.

OM ⊥ AB and ON ⊥ CD and AB > CD

To Prove: OM < ON

Construction: Join OA, OC

**Question 2.**

**OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.**

**(i) If the radius of the circle is 10 cm, find the area of the rhombus.**

**(ii) If the area of the rhombus is 32√3 cm² find the radius of the circle.**

**Answer 2**

**Question 3.**

**Two circles with centres A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.**

**Answer 3**

Two circles with centres A and B touch each other at C internally.

PQ is the perpendicular bisector of AB meeting the bigger circles at P and Q. Join AP.

Radius AC = 5 cm.

and radius BC = 3 cm.

AB = AC – BC = 5-3 = 2 cm.

**Question 4.**

**Two chords AB and AC of a circle are equal. Prove that the centre of the circle, lies on the bisector of angle BAC.**

**Answer 4**

Given: A circle in which two chords AC and AB are equal in length. AL is the bisector of ∠ BAC.

To Prove: O lies on the bisector of ∠ BAC

Proof: In ∆ ADC and ∆ ADB,

AD = AD (Common)

AB = AC (Given)

∠ BAD = ∠ CAD (Given)

∴ ∆ ADC = ∆ ADB (SAS postulate)

∴ BD = DC (C.P.C.T.)

and ∠ ADB = ∠ ADC (C.P.C.T.)

But ∠ ADB + ∠ ADC = 180° (Linear pair)

∴ ∠ ADB = ∠ ADC = 90°

∴ AD is the perpendicular bisector of chord BC.

∵ The perpendicular bisector of a chord passes through the centre of the circle.

∴ AD is the bisector of ∠ BAC passes through the centre O of the circle. Q.E.D.

**Question 5.**

**The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle ?**

**Answer 5**

AB is the diameter and AC is the chord

∴ AB = 20 cm and AC = 12 cm

Draw OL ⊥ AC

∵ OL ⊥ AC and hence it bisects AC, O is the center of the circle.

**Question 6.**

**ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.**

**Answer 6**

ABCD is a cyclic quad, in which AD || BC

∠ ADC = 110°, ∠ BAC = 50°

∠B + ∠D= 180° (Sum of opposite angles of a cyclic quad.)

⇒∠B + 110°= 180°

∴ ∠ B or ∠ ABC = 180° – 110° – 70″

Now. in ∆ ABC,

∠ BAC + ∠ ABC – ∠ ACB = 180°

⇒ 50° + 70° + ∠ ACB – 180°

⇒ 120° – ∠ ACB = 180°

∴ ∠ ACB = 180° – 120° = 60″

OL ⊥ AC and hence it bisects AC, O is the center of the circle.

OL ⊥ AC and hence it bisects AC, O is the center of the circle.

∵ AD || BC (Given)

∴ ∠ DAC = ∠ ACB (Alternate angles)

= 60°

Now, in ∆ ADC,

∠ DAC + ∠ ADC + ∠ DCA -= 180°

60°+ 110° + ∠ DCA = 180°

170° + ∠ DCA – 180°

∴ ∠ DCA = 180″ – 170° = 10°

**Question 7.**

**In the given figure, C and D arc points on the semi-circle described on AB as diameter. Given angle BAD = 70° and angle DBC = 30°. calculate angle BDC.**

**Answer 7**

∴ ABCD is a cyclic quad.

∠ BAD ∠ BCD 180 (Sum of opposite angles)

⇒ 70° + ∠ BCD = 180°

⇒ ∠ BCD = 180°- 70° = 110°

Now in ∆ BCD,

∠ BCD + ∠ DBC + ∠ BDC = 180°

⇒ 30°+ 110° + ∠ BDC = 180°

⇒ 140°+ ∠ BDC = 180°

∴ ∠ BDC = 180°- 140° = 40°

**Question 8.**

**In cyclic quadrilateral ABCD, ∠ A = 3 ∠ C and ∠ D = 5 ∠ B. Find the measure of each angle of the quadrilateral.**

**Answer 8**

**Question 9.**

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

**Answer 9**

**Given:** ∆ABC in which AB=AC and AB as diameter, a circle is drawn which intersects BC at D.

**To Prove:** BD = DC

**Construction:** Join AD

**Proof:** AB is the diameter

∴ ∠ ADB = 90° (Angle in a semi-circle)

But ∠ ADB + ∠ ADC =180° (A linear pair)

∴ ∠ ADC = 90°

Now in right angled ∆s ABD and ACD,

Hypotenuse AB = AC (given)

Side AD = AD (Common)

∴ ∆ ABD ≅ ∆ ACD (RHS postulate)

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

Hence the circle bisects base BC at D. Q.ED.

**Question 10.**

**Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at the points D, E and F respectively. Prove that.**

**angle EDF = 90° – ∠A.**

**Answer 10**

**Given:** A ∆ ABC whose bisectors of angles A, B and C intersect the circumcircle at D, E and F respectively. ED, EF and DF are joined.

∠ EDF = 90° – ∠A

Construction: Join BF, FA, AE and EC.

Proof: ∠ EBF = ∠ ECF = ∠ EDF ……..(i)

(Angles in the same segment)

**Question 11.**

**In the flgure; AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠ C = 20°, find angle AOD.**

**Answer 11**

Join OB

In ∆ OBC, BC = OD = OB (radii of the circle)

∴ ∠ BOC = ∠ BCO = 20°

and Ext. ∠ ABO = ∠ BCO + ∠ BOC

= 20°+ 20° = 40° ….(i)

In ∆ OAB, OA = OB (radii of the circle)

∴ ∠ OAB = ∠ OBA = 40° [from (i)]
∠ AOB = 180°- ∠ OAB – ∠ OBA

= 180°-40°-40°= 100°

∵ DOC is a line

∴ ∠ AOD + ∠ AOB + ∠ BOC = 180°

⇒ ∠ AOD + 100° + 20° = 180°

⇒ ∠ AOD + 120° = 180°

∴ ∠ AOD = 180° – 120° = 60°

**Question 12.**

**Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.**

**Answer 12**

**Given:** In ∆ ABC, ∠ B = 90° which is inscribed in a circle and O is the incentre of the incircle of ∆ABC.

D and d are the diameters of circumcircle and incircle of ∆ ABC.

**To Prove:** AB +BC + CA = 2D + d.

**Construction:** Join OL, OM and ON.

**Proof:** In circumcircle of ∆ ABC,

∠ B = 90° (given)

∴ AB is the diameter of circumcircle i.e. AB = D.

Let radius of incircle = r

∴ OL = OM = ON = r

Now from B, BL, BM are the tangents to the incircle

∴ BL = OM = r

Similarly we can prove that:

AM = AN and CL = CN = R (radius)

(Tagents from the point outside the circle)

Now AB + BC + CA = AM + BM + BL + CL + CA

and = AN + r + r + CN + CA

so = AN + CN + 2r + CA

hence = AC + AC + 2r = 2 AC + 2r = 2D + d Q.E.D.

**Question 13.**

**P is the mid point of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.**

**Answer 13**

**Given:** A circle with centre O, AB is an arc whose mid point is P and AB is chord. TPS is the tangent at P.

**To Prove:** TPS ||AB.

**Construction:** Join AP and BP.

**Prove:** TPS is tangent and PA is chord of the circle

∠ APT = ∠ PBA (angles in the alternate segment)

But ∠ PBA = ∠ PAB (∵ PA = BP)

∴ ∠ APT = ∠ PAB

But these are alternate angles

∴ TPS || AB Q.E.D

**Question 14.**

**In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent Prove that the line NM produced bisects AB at P.**

**Prove that the line NM produced bisects AB at P.**

**Answer 14**

**Given:** Two circles intersect each other at M and N. AB is their common tangent, chord MN intersect the tangent at P.

**To Prove:** P is mid point of AB.

**Proof:** From P, AP is the tangent and PMN is the secant of first circle.

∴ AP^{2} = PM x PN ….(i)

Again from P, PB is the tangent and PMN is the secant of the second circle.

PB2 = PM x PN ….(ii)

from (i) and (ii)

AP^{2} = PB^{2} ⇒ AP = PB

∴ P is the mid point of AB. Q.E.D.

**Question 15.**

**In the given figure, ABCD is a cyclic- quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠ DCQ = 40° and ∠ ABD = 60°, find :**

**(i) ∠ DBC**

**(ii) ∠ BCP**

**(iii) ∠ ADB**

**Answer 15**

(i) PQ is the tangent and CD is the chord

∴ ∠ DCQ = ∠ DBC (angles in the alternate segment)

∴ ∠ DBC = 40° (∵ ∠ DCQ = 40°)

(ii) ∠ DCQ + ∠ DCB + ∠ BCP = 180°

⇒ 40° + 900 + ∠ BCP = 180°(∵ ∠ DCB = 90°)

⇒ 130°+ ∠ BCP = 180°

∵ ∠ BCP =180° -130° = 50°

(iii) In Δ ABD, ∠ BAD = 90° (Angle in a semi circle) and ∠ ABD = 60°

∴ ∠ ADB = 180°- (60° + 90°)

⇒ 1800- 150° = 30°

**Question 16.**

**The given figure shows a circle with centre O and BCD is tangent to it at C. Show that:**

**∠ ACD + ∠ BAC = 90°.**

**Answer 16**

**Given:** A circle with centie O and BCD is a tangent at C.

**To Prove:** ∠ ACD + ∠ BAC = 90°

**Construction:** Join OC.

**Proof:** BCD is the tangent and OC is the radius

∴ OC ⊥ BD

⇒ ∠ OCD = 90°

⇒ ∠ OCA + ∠ ACD = 90° ….(i)

But in ∆ OCA

OA = OC (radii of the same circle)

∴ ∠ OCA = ∠ OAC [from (i)]
∠OAC + ∠ACD = 90°

⇒ ∠ BAC + ∠ ACD = 90° Q.E.D.

**Question 17.**

**ABC is a right triangle with angle B = 90°. A circle with BC as diameter meets hypotenuse AC at point D. Prove that **

**(i) AC x AD = AB ^{2}**

**(ii) BD**

^{2}= AD x DC.**Answer 17**

**Given:** A circle with BC as diameter meets the hypotenuse of right ∆ ABC with ∠ B = 90° meets at D. BD is joined.

**To Prove:**

(i) AC x AD = AB^{2}

(ii) BD^{2} = AD x DC

**Proof:**

(i) In ∆ABC, ∠ B = 90° and BC is the diameter of the circle.

**Question 18.**

**In the given figure. AC = AE.**

**Show that :**

**(i) CP = EP**

**(ii) BP = DP.**

**Answer 18**

**Given:** In the figure, AC = AE

**To Prove:** (i) CP = EP (ii) BP = DP

**Proof:** In ∆ ADC and ∆ ABE,

AC = AE (given)

∠ ACD = ∠ AEB (Angles in the same segment)

∠ A = ∠ A (Common)

∆ ADC ≅ ∆ ABE (ASA postulate)

AB = AD (C.P.C.T.)

But AC = AE (given)

∴ AC – AB = AE – AD

⇒ BC = DE

Now in ∆ BPC and ∆ DPE,

BC = DE (proved)

∠ C = ∠ E (Angles in the same segment)

∠ CBP = ∠ CDE (Angles in the same segment)

∴ ∆ BPC ≅ ∆ DPE (S.A.S. postulate)

∴ BP=DP (C.P.C.T.)

CP = PE (C.P.C.T.) Q.E.D.

**Question 19.**

**ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120°.**

**Calculate :**

**(i) ∠ BEC**

**(ii) ∠ BED**

**Answer 19**

(i) In cyclic pentagon, O is the centre of circle.

Join OB, OC

AB = BC = CD (given)

and ∠ ABC = 120°.

∴ ∠ BCD = ∠ ABC = 120°

OB and OC are the bisectors of ∠ ABC and ∠ BCD respectively.

∴ ∠ OBC = ∠ BCO = 60°

In ∆ BOC,

**Question 20.**

**In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ∠ACO=30°, find**

**(i) ∠BCO**

**(ii) ∠AOB**

**(iii) ∠APB**

**Answer 20**

**Given:** In the fig. O is the centre of the circle CA and CB are the tangents to the circle from C. ∠ACO = 30°

P is any point on the circle. PA and PB are joined.

**To find:**

(i) ∠BCO

(ii) ∠AOB

(iii) ∠APB

**Proof:**

(i) In ∆ OAC and ∆OBC,

OC=OC (common)

OA = OB (radius of the circle)

CA = CB (tangents to the circle)

∴ ∆OAC ≅ ∆OBC (SSS axion)

∴ ∠ACO = ∠BCO = 30°

(ii) ∴ ∠ACB = 30° + 30° = 60°

∴ ∠AOB + ∠ACB = 180°

⇒ ∠AOB+ 60° =180°

∴ ∠AOB = 180° – 60° = 120°

(iii) Arc AB, subtends ∠AOB at the centre and ∠APB is in the remaining part of the circle

∴ ∠APB = ∠AOB = x 120° = 60°

**Question 21.**

**ABC is triangle with AB = 10 cm, BC = 8 cm and AC = 6 cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles. (2011)**

**Answer 21**

**Given:** ABC is a triangle with AB = 10 cm, BC = 8 cm, AC = 6 cm. Three circle are drawn with centre A, B and C touch each other at P, Q and R respectively

**Question 22.**

**In a square ABCD, its diagonals AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and the bisector of angle ABD meets AC at N and AM at L.**

**Show that :**

**(i) ∠ ONL + ∠ OML = 180°**

**(ii) ∠ BAM = ∠ BMA**

**(iii) ALOB is a cyclic quadrilateral**

**Answer 22**

**Question 23.**

**The given figure shows a semi-circle with centre O and diameter PQ. If PA = AB and ∠ BCQ = 140°; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.**

**Answer 23**

Join PB

In cyclic quad. PBCQ,

∠ BPQ + ∠ BCQ = 180°

⇒ ∠ BPQ + 140° = 180°

∴ ∠ BPQ = 180° – 140° = 40°

Now, in ∆ PBQ,

∠ PBQ + ∠ BPQ + ∠ BQP = 180°

⇒ 90° + 40° + ∠ BQP = 180°

(∠ PBQ = 90° angle in a semicircle)

⇒ 130° + ∠ BQP = 180°

∴ ∠ BQP = 180° – 130°= 50°

Nowin cyclic quad. PQBA,

∠ PQB + ∠ PAB = 180°

⇒ 50° + ∠ PAB = 180°

∴ ∠ PAB = 180° – 50° = 130°

(ii) Now, in ∆ PAB,

∠ PAB + ∠ APB + ∠ ABP = 180°

⇒ 130° + ∠ APB + ∠ ABP = 180°

⇒ ∠ APB + ∠ ABP = 180° – 130° = 50°

But ∠ APB = ∠ ABP = 25°

∴ (PA = AB)

∠ BAQ = ∠ BPQ = 40°

(Angles in the same segment)

Now, in ∆ ABQ,

∠ AQB + ∠ QAB + ∠ ABQ = 180°

and ⇒ ∠ AQB + 40° + 115° = 180°

so ⇒ ∠ AQB + 155° = 180°

therefore ⇒ ∠ AQB = 180° – 155° = 25°

(iii) Arc AQ subtends ∠ AOQ at the centre and ∠APQ at the remaining part of the circle,

∠ AOQ = 2 ∠ APQ = 2 x 65° = 130°

Now, in ∆ AOQ,

∠ OAQ = ∠ OQA

(∵ OA = OQ radii of the same circle)

But ∠ OAQ + ∠ OQA + ∠ AOQ = 180°

⇒ ∠ OAQ + ∠ OAQ + 130° = 180°

2 ∠ OAQ = 180° – 130° = 50°

∴ ∠ OAQ = 25°

∵ ∠ OAQ = ∠ AQB (each = 25°)

But these are alternate angles.

∴ AO || BQ.

Q.E.D.

**Question 24.**

**The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°. Calculate**

**(i) angle QTR**

**(ii) angle QRP**

**(iii) angle QRS**

**(iv) angle STR**

**Answer 24**

**Question 25.**

**In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segemcnt, show that :**

**(i) ∠ BAP = ∠ ADQ**

**(ii) ∠ AOB = 2 ∠ ADQ**

**(iii) ∠ ADQ = ∠ ADB**

**Answer 25**

**Given:** PAT is the tangent to the circle with centre O, at A. Chord BC || PAT is drawn.

CDQ is a line segment which intersects the circle at C and D and meets the tangent PAT at Q.

**To Prove:**

(i) ∠ BAP = ∠ ADQ (ii)∠ AOB = 2 ∠ ADQ

(iii) ∠ ADQ = ∠ ADB

**Proof:**

(i) ∵ PAT || BC

∴ ∠ PAB = ∠ ABC (Alternate angles) ….(i)

In cyclic quad. ABCD,

Ext. ∠ADQ = ∠ABC ….(ii)

∴ ∠ PAB = ∠ ADQ (from (i) and (ii))

(ii) Arc AB subtends ∠ AOB at the centre and ∠ADB at the remaining part of the circle.

∴ ∠ AOB = 2 ∠ ADB

= 2 ∠ PAB (In the alt. segment)

= 2 ∠ ADQ [proved in (i)]
(iii) ∵ ∠ BAP = ∠ ADB

(Angles in the alt. segment)

Bui ∠ BAP = ∠ ADQ (Proved in (i))

∴ ∠ ADQ = ∠ ADB Q.E.D.

**Question 26.**

**AB is a line segment and M is its mid-point. Three semi-circles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semi-circles. Show that: AB = 6 x r**

**Answer 26**

**:**

**Given:** A line segment AB whose mid-point is M. Three circles are drawn on AB, AM and MB as diameler A circle with radius r is drawn which touches (he three circles externally at L, R and N respectively. M, P, Q are the centres of the three circles.

**To Prove:** AB= 6r

**Construction:** Join OP and OQ.

**Proof:** OM = ON = r

**Question 27.**

**TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.**

**Answer 27**

**Given:** A circle with centre C. From a point T outside th circle, TA and TB are two tangent to the circle OT intersects the circle at P, AP and AB are joined.

**To Prove:** AP is the bisector of ∠ TAB

**Construction:** Join PB.

**Question 28.**

**Two circles intersect in points P and Q. A secant passing through P intersects the circles in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.**

**Answer 28**

**Given:** Two circles intersect each other at P and Q. From P, a secant intersects the circles at A and B respectively. From A and B tangents are drawn which intersect each other at T.

**To Prove:** A. Q, B and T on a circle.

**Construction:** Join PQ.

**Proof:** AT is the tangent and AP is chord

∴ ∠ TAP = ∠ AQP (Angles in all. segment) …(i)

Similarly ∠ TBP = ∠ BQP ….(ii)

Adding (i) and (ii),

∠ TAP + ∠ TBP = ∠ AQP + ∠ BQP = ∠ AQB …..(iii)

Now, in ∆ TAB,

∠ ATB + ∠ TAP + ∠ TBP = 180°

⇒ ∠ ATB + ∠ AQB = 180° (from (iii)

∴ AQBT is a cyclic quadrilateral.

Hence A, Q, B and T lie on the same circle. Q.E.D.

**Question 29.**

**Prove that any four vertices of a regular pentagon are concyclic (lie on the same circle).**

**Answer 29**

ABCDE is a regular pentagon.

**To Prove:** Any four vertices lie on the same circle.

**Construction:** Join AC.

**Proof:** Each angle of a regular pentagon

**Question 30.**

**Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm, calculate the length of CD. [2000]**

**Answer 30**

Let CD = x

∴ chords AB and CD intersect each other at outside the circle.

∴ AX.XB = CX.XD

and ⇒(4+6)x6 = (x + 5)x5

so ⇒ 10 x 6 = 5x + 25

hence ⇒ 60 = 5x + 25

⇒ 5x = 60 – 25 = 35

∴ x = = 7

CD = 7 cm

**Question 31.**

**in the given figure. find TP if AT = 16 cm AB = 12 cm.**

**Answer 31**

In the figure.

PT is the tangent and TBA is the secant of the circle.

∴ TP^{2} = TA x TB = 16 x (16 – 12) = 16 x 4 = 64 = (8)^{2}

Hence, TP = 8 cm.

**Question 32.**

**In the following figure, a circle is inscribed in the quadrilateral ABCD. If BC = 38 cm. QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle. (1990)**

**Answer 32**

A circle with centre is inscribed (see the fig.)

in a quadrilateral ABCD. BC = 38 cm, QB = 27cm,

DC = 25 cm and AD ⊥ BC.

Join OP and QS.

and ∵ OP and OS are the radii of the circle

so ∴ OP ⊥ AD and OS ⊥ CD

hence∴ OPDS is a square

∴ OP = OS – DP = DS.

Let length of radius of the circle = r

then DP = DS = r

and ∴ CS = 25 – r

so ∵ EQ = BR = 27 cm (tangents to the circle from B)

therefore ∴ CR = BC – BR = 38 – 27 = 11 cm

Similarly CR = CS

∴ 25 – r = 11 ⇒ r = 25 – 11 = 14

∴ Radius of the circle = 14 cm

**Question 33.**

**In the given figure, XY is the diameter of the circle and PQ is a tangent to the circle at Y.**

**If ∠AXB = 50° and ∠ABX = 70°, find ∠BAY and ∠APY.**

**Answer 33**

In the above figure,

XY is a diameter of the circle PQ is tangent to the circle at Y.

∠AXB = 50° and ∠ABX = 70°

(i) In ∆AXB,

∠XAB + ∠ABX + ∠AXB = 180° (Angles of a triangle)

and ⇒ ∠XAB + 70° + 50° = 180°

so ⇒ ∠XAB + 120° =180°

threfore ⇒ ∠XAB = 180° – 120° = 60°

But, ∠XAY = 90° (Angle in a semicircle)

∴ ∠BAY = ∠XAY-∠XAB=90°- 60° = 30°

(ii) Similarly ∠XBY=90°(Angle in a semicircle) and ∠CXB = 70°

∴ ∠PBY = ∠XBY-∠XBA =90° – 70° = 20°

∵ ∠BYA = 180° – ∠AXB ( ∵ ∠BYA + ∠AYB = 180°) = 180°- 50° = 130°

∠PYA =∠ABY (Angles in the alternate segment) = ∠PBY = 20°

and ∠PYB = ∠PYA + ∠AYB

= 20° + 130° = 150°

∴ ∠APY = 180°-(∠PYA + ∠ABY)

= 180° -(150° +20°) =180° – 170° = 10°

**Question 34.**

**In the given figi QAP is the tangent point A and PBD is a straight line. If ∠ACB = 36° and ∠APB = 42°, find:**

**(i) ∠ BAP**

**(ii) ∠ABD**

**(iii) ∠ QAD**

**(iv) ∠ BCD**

**Answer 34**

In the given figure, QAP is the tangent to the circle at A and PBD is a B straight line.

**Question 35.**

**In the given figure, AB is a diameter. The tangent at C meets AB produced at Q. If ∠ CAB = 34°,**

**find :**

**(i) ∠CBA**

**(ii) ∠CQB**

**Answer 35**

In ∆ ABC,

we have ∠ ACB = 90°

[Angle in a semicircle is 90°]
(i) Also ∠ CBA + ∠ CAB + ∠ ACB = 180° [Angle sum property of a ∆ ]
⇒ ∠ CBA =180°- ∠ CAB – ∠ ACB = 180°-34°-90° = 180°-124° = 56°

(ii) CQ is a tangent at C and CB is a chord of the circle.

⇒ ∴ ∠ QCB = ∠ BAC = 34° [Angles in the alternate segments]
∠ CBQ =180°- ∠ ABC [Linear pair]
⇒ ∠ CBQ = 180°- 56° = 124° [From (i)]
In ∆ BCQ, we have

⇒ ∠ CQB = 180° – (∠ QCB + ∠ CBQ) [Angle sum property of a ∆ ]
= 180° -(34° + 124°) = 180°- 158° = 22°

Hence, ∠CQB = 22°

**Question 36.**

**In the given figure, A O is the centre of the circle. The tangents at B and D intersect each other at point P.**

**If AB is parallel to CD and ∠ ABC = 55**

**find :**

**(i) ∠ BOD**

**(ii) ∠ BPD**

**Answer 36**

In the given figure,

O is the centre of the circle AB || CD,

∠ ABC = 55° tangents at B and D are drawn which meet at P.

∵ AB || CD

∴ ∠ ABC = ∠ BCD (Alternate angles)

∴ ∠ ABC = 55° (Given)

(i) Now arc BD subtands ∠ BOD at the centre and ∠ BCD at the remaining part of the circle.

∴ ∠BOD = 2∠BCD = 2 x 55° = 110°

(ii) In quad. OBPD,

∠OBP = ∠ODP = 90° (∵ BP and DP are tangents)

∴ ∠BOD + ∠BPD = 180°

⇒ 110° +∠BPD =180°

⇒ ∠BPD =180°-110°= 70°

Hence, ∠BOD = 110° and ∠BPD = 70°

**Question 37.**

**In the figure given below PQ = QR, ∠RQP = 68°, PC and CQ are tangents to the circle with centre O. Calculate the values of:**

**(i) ∠QOP**

**(ii) ∠QCP**

**Answer 37**

**Question 38.**

**In two concentric circles, prove that all chords of the outer circle, which touch the inner circle, are of equal length.**

**Answer 38 **

**Given:** Two concentric circles with centre O AB and CD are two cords of outer circle which touch the inner circle at P and Q respectively

**To prove:** AB = CD

**Construction :** Join OA, OC, OP and OQ

**Proof:** ∵ OP and OQ are the radii of the inner circle and AB and CD are tangents

∴ OP ⊥ AB and OQ ⊥ CD

and P and Q are the midpoints of AB and CD Now in right AOAP and OCQ,

Side OP = OQ (radii of the inner circle)

Hyp. OA = OC (radii of the outer circle)

∴ ∆OAP = ∆OCQ (R.H.S. axiom)

∴ AP = CQ (c.p.c.t.)

But AP = AB and CQ = CD

∴ AB = CD Hence proved.

**Question 39.**

**In the figure, given below, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively.**

**Given that AB = 5 cm , Calculate PQ.**

**Answer 39**

In the figure, two circles with centres P and Q and radii 6 cm and 3 cm respectively

ABC is the common transverse tangent to the two circles. AB = 8 cm

Join AP and CQ

∵ AC is the tangents to the two circles and PA and QC are the radii

**Question 40.**

**In the figure, given below, O is the centre of the circumcircle of triangle XYZ. Tangents at X and Y intersect at point T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.**

**Answer 40**

In the figure, a circle with centre O, is the circumcircle of ∆XYZ.

∠XOZ =140° (given)

Tangents from X and Y to the circle meet at T such that ∠XTY = 80°

∵ ∠XTY = 80°

∴ ∠XOY= 180°-80°= 100°

But ∠XOY + ∠YOZ + ∠XOZ = 360° (Angles at a point)

and ⇒ 100°+∠YOZ+ 140o = 360o

so ⇒ 240o+∠YOZ =360°

hence ⇒ ∠YOZ =360°- 240°

⇒∠YOZ =120°

Now arc YZ subtends ∠YOZ at the centre and ∠YXZ at the remaining part of the circle

∴ ∠YOZ = 2 ∠YXZ

⇒ ∠YXZ= ∠YOZ ⇒ ∠YXZ = x 120° = 60°

**Question 41.**

**In the given figure, AE and BC intersect each other at point D. If ∠CDE = 90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, findAE.**

**Answer 41 **

In the given circle,

Chords AE and BC interesect each other at D at right angle i.e., ∠CDE = 90°, AB is joined AB = 5cm, BD = 4 cm, CD = 9 cm

Now we have to find AE.

Let DE=xm

Now in right ∆ABD,

AB^{2 }= AD^{2} + BD^{2} (Pythagoras Theorem)

⇒ (5)^{2}=AD^{2} + (4)^{2}

⇒ 25 = AD^{2} + 16

⇒ AD^{2} = 25-16 = 9 = (3)^{2}

∴AD = 3cm

∵ Chords AE and BC intersect eachothcr at D inside the circle

∴ AD x DE = BD x DC

⇒ 3 x x = 4 x 9

⇒ x= = 12cm;

∴ AE=AD + DE = 3 + 12 = 15 cm

**Question 42.**

**In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.**

**Answer 42**

ABCD is a cyclic quadrilateral

∴ ∠ABC + ∠ADC = 180°

100° +∠ADC = 180°

∠ADC = 180°- 100° = 80°

In ∆ADC

∠ACD + ∠CDA + ∠D AC =180°

40° + 80° +∠D AC = 180°

∠D AC = 180° – 80° – 40° = 60°

Now ∠DAC = 60°

⇒ ∠DCT = 60° [angle in alt. segment]

**Question 43.**

**In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x,y and z. (2015)**

**Answer 43**

In the given figure,

O is the centre of the circle.

SP is tangent ∠SRT =65°.

To find the values of x, y and z

(i) In ∆STR,

∠S = 90° (∵ OS is radius and ST is tangent)

∴ ∠T + ∠R = 90°

⇒ x + 65° = 90°

⇒ x = 90° – 65° = 25°

(ii) Arc CQ subtends ∠SOQ at the centre and

∠STQ at the remaining part of the circle.

∠y = ∠QOS = 2∠T = 2∠x = 2 x 25° = 50°

(iii) In ∆OSP,

∠S = 90°

∴ ∠SOQ or ∠SOP + ∠P = 90°

and ⇒ y+z=90o

so ⇒ 50° + z = 90°

hence ⇒ z = 90°-50° = 40°

x = 25°, y = 50° and z = 40°

Return to :- Concise Selina Maths Solutions for ICSE Class-10

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