# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 18 - Tangents and Intersecting Chords

## Selina Textbook Solutions Chapter 18 - Tangents and Intersecting Chords

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## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 18 - Tangents and Intersecting Chords Page/Excercise 18(B)

i) Since two chords AB and CD intersect each other at P.

ii) Since two chords AB and CD intersect each other at P.

iii) Since PAB is the secant and PT is the tangent

i) PAQ is a tangent and AB is the chord.

(angles in the alternate segment)

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

iii) BD is the diameter.

(angle in a semi-circle)

Now in

PQ is a tangent and OR is the radius.

But in

OT = OR (Radii of the same circle)

In

Join OC.

(angles in alternate segment)

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

Now in

Now in

DE is the tangent to the circle at P.

DE||QR (Given)

Since the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment

(DE is tangent and PQ is chord)

from (i) and (ii)

Hence, triangle PQR is an isosceles triangle.

Join OA, OB, O'A, O'B and O'O.

CD is the tangent and AO is the chord.

(angles in alternate segment)

In

OA = OB (Radii of the same circle)

From (i) and (ii)

Therefore, OA is bisector of BAC

Draw a tangent TS at P to the circles given.

Since TPS is the tangent, PD is the chord.

Subtracting (i) from (ii)

But in

TAS is a tangent and AB is a chord

But these are alternate angles

Therefore, TS||BD.

Join OC, OD and AC.

i)

ii)

PCT is a tangent and CA is a chord.

But arc DC subtends at the centre and at the

remaining part of the circle.

Join AB, PB and BQ

TP is the tangent and PA is a chord

(angles in alternate segment)

Similarly,

Adding (i) and (ii)

But they are the opposite angles of the quadrilateral

Therefore, PBQT are cyclic.

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

i) PA is the tangent and AB is a chord

( angles in the alternate segment)

AD is the bisector of

In

Therefore, is an isosceles triangle.

ii) In

Join AB.

PQ is the tangent and AB is a chord

(angles in alternate segment)

Similarly,

Adding (i) and (ii)

From (iii) and (iv)

Hence, and are supplementary.

Join AB.

i) In Rt.

Chords AE and CB intersect each other at D inside the circle

AD x DE = BD x DC

3 x DE = 4 x 9

DE = 12 cm

ii) If AD = BD .......(i)

We know that:

AD x DE = BD x DC

But AD = BD

Therefore, DE = DC .......(ii)

Adding (i) and (ii)

AD + DE = BD + DC

Therefore, AE = BC

Join AB and AD

EBM is a tangent and BD is a chord.

(angles in alternate segments)

(Vertically opposite angles)

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

Therefore, CE = BD

AB is a straight line.

AB i.e. DB is tangent to the circle at point B and BC is the diameter.

Now, OE = OC (radii of the same circle)

(vertically opposite angles)

In

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 18 - Tangents and Intersecting Chords Page/Excercise 18(C)

Given: A circle with centre O and radius r. . Also AB > CD

To prove: OM < ON

Proof: Join OA and OC.

In Rt.

Again in Rt.

From (i) and (ii)

Hence, AB is nearer to the centre than CD.

i) Radius = 10 cm

In rhombus OABC,

OC = 10 cm

In Rt.

Area of rhombus =

ii) Area of rhombus =

But area of rhombus OABC = 2 x area of

Where r is the side of the equilateral triangle OAB.

Therefore, radius of the circle = 8 cm

If two circles touch internally, then distance between their centres is equal to the difference of their radii. So, AB = (5 - 3) cm = 2 cm.

Also, the common chord PQ is the perpendicular bisector of AB. Therefore, AC = CB = AB = 1 cm

In right ACP, we have AP^{2} = AC^{2} + CP^{2}

5^{2} = 1^{2} + CP^{2}

CP^{2} = 25 -; 1 = 24

CP =

Now, PQ = 2 CP

= 2 x cm

Given: AB and AC are two equal chords of C (O, *r*).

To prove: Centre, O lies on the bisector of BAC.

Construction: Join BC. Let the bisector of BAC intersects BC in P.

Proof:

In APB and APC,

AB = AC (Given)

BAP = CAP (Given)

AP = AP (Common)

(SAS congruence criterion)

BP = CP and APB = APC (CPCT)

APB + APC = 180 (Linear pair)

=> 2APB = 180 (APB = APC)

APB = 90

Now, BP = CP and APB = 90

AP is the perpendicular bisector of chord BC.

AP passes through the centre, O of the circle.

AB is the diameter and AC is the chord.

Draw

Since and hence it bisects AC, O is the centre of the circle.

Therefore, OA = 10 cm and AL = 6 cm

Now, in Rt.

Therefore, chord is at a distance of 8 cm from the centre of the circle.

ABCD is a cyclic quadrilateral in which AD||BC

(Sum of opposite angles of a quadrilateral)

Now in

Now in

Since ABCD is a cyclic quadrilateral, therefore, BCD + BAD = 180

(since opposite angles of a cyclic quadrilateral are supplementary)

BCD + 70 = 180

BCD = 180 - 70 = 110

In BCD, we have,

CBD + BCD + BDC = 180

30 + 110 + BDC = 180

BDC = 180 - 140

BDC = 40

ABCD is a cyclic quadrilateral.

Similarly,

Hence,

Join AD.

AB is the diameter.

ADB = 90º (Angle in a semi-circle)

But, ADB + ADC = 180º (linear pair)

ADC = 90º

In ABD and ACD,

ADB = ADC (each 90º)

AB = AC (Given)

AD = AD (Common)

ABD ACD (RHS congruence criterion)

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

Hence, the circle bisects base BC at D.

Join ED, EF and DF. Also join BF, FA, AE and EC.

In cyclic quadrilateral AFBE,

(Sum of opposite angles)

Similarly in cyclic quadrilateral CEAF,

Adding (ii) and (iii)

Join OB.

In

In

Since DOC is a straight line

Join OL, OM and ON.

Let D and d be the diameter of the circumcircle and incircle.

and let R and r be the radius of the circumcircle and incircle.

In circumcircle of

Therefore, AC is the diameter of the circumcircle i.e. AC = D

Let radius of the incircle = r

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

(Tangents from the point outside the circle)

Now,

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

= AN+r+r+CN+CA

= AN+CN+2r+CA

= AC+AC+2r

= 2AC+2r

= 2D+d

Join AP and BP.

Since TPS is a tangent and PA is the chord of the circle.

(angles in alternate segments)

But

But these are alternate angles

From P, AP is the tangent and PMN is the secant for first circle.

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

From (i) and (ii)

Therefore, P is the midpoint of AB.

i) PQ is tangent and CD is a chord

(angles in the alternate segment)

ii)

iii) In

Join OC.

BCD is the tangent and OC is the radius.

Substituting in (i)

i) In

and BC is the diameter of the circle.

Therefore, AB is the tangent to the circle at B.

Now, AB is tangent and ADC is the secant

ii) In

From (i) and (ii)

Now in

In

(angles in the same segment)

AC = AE (Given)

(Common)

(ASA Postulate)

AB = AD

but AC = AE

In

(angles in the same segment)

BC = DE

(angles in the same segment)

(ASA Postulate)

BP = DP and CP = PE (cpct)

i) Join OC and OB.

AB = BC = CD and

OB and OC are the bisectors of and respectively.

In

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

ii) In cyclic quadrilateral BCDE,

In the given fig, O is the centre of the circle and CA and CB are the tangents to the circle from C. Also, ACO = 30

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

To find: (i)

(ii)

(iii)

Proof:

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

We need to find the radii of the three circles.

ABCD is a square whose diagonals AC and BD intersect each other at right angles at O.

i)

In

But, (vertically opposite angles)

Now in

Adding (i) and (ii)

ii)

and

iii) In quadrilateral ALOB,

Therefore, ALOB is a cyclic quadrilateral.

Join PB.

i) In cyclic quadrilateral PBCQ,

Now in

In cyclic quadrilateral PQBA,

ii) Now in

iii) Arc AQ subtends at the centre and APQ at the remaining part of the circle.

We have,

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

Now in

But these are alternate angles.

Hence, AO is parallel to BQ.

Join PQ, RQ and ST.

i)

Arc RQ subtends at the centre and QTR at the remaining part of the circle.

ii) Arc QP subtends at the centre and QRP at the remaining part of the circle.

iii) RS || QT

iv) Since RSTQ is a cyclic quadrilateral

(sum of opposite angles)

i) Since PAT||BC

(alternate angles) .........(i)

In cyclic quadrilateral ABCD,

from (i) and (ii)

ii) Arc AB subtends at the centre and at the remaining part of the circle.

iii)

Let O, P and Q be the centers of the circle and semicircles.

Join OP and OQ.

OR = OS = r

and AP = PM = MQ = QB =

Now, OP = OR + RP = r + (since PM=RP=radii of same circle)

Similarly, OQ = OS + SQ = r +

OM = LM -; OL = - r

Now in Rt.

Hence AB = 6 x r

Join PB.

In TAP and TBP,

TA = TB (tangents segments from an external points are equal in length)

Also, ATP = BTP. (since OT is equally inclined with TA and TB) TP = TP (common)

TAP TBP (by SAS criterion of congruency)

TAP = TBP (corresponding parts of congruent triangles are equal)

But TBP = BAP (angles in alternate segments)

Therefore, TAP = BAP.

Hence, AP bisects TAB.

Join PQ.

AT is tangent and AP is a chord.

(angles in alternate segments) ........(i)

Similarly, .......(ii)

Adding (i) and (ii)

Now in

Therefore, AQBT is a cyclic quadrilateral.

Hence, A, Q, B and T lie on a circle.

ABCDE is a regular pentagon.

In AED,

AE = ED (Sides of regular pentagon ABCDE)

EAD = EDA

In AED,

AED + EAD + EDA = 180º

108º + EAD + EAD = 180º

2EAD = 180º - 108º = 72º

EAD = 36º

EDA = 36º

BAD = BAE - EAD = 108º - 36º = 72º

In quadrilateral ABCD,

BAD + BCD = 108º + 72º = 180º

ABCD is a cyclic quadrilateral

We know that XB.XA = XD.XC

Or, XB.(XB + BA) = XD.(XD + CD)

Or, 6(6 + 4) = 5(5 + CD)

Or, 60 = 5(5 + CD)

Or, 5 + CD = = 12

Or, CD = 12 - 5 = 7 cm.

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

Therefore, TP^{2} = TA x TB

TP^{2} = 16 x (16-12) = 16 x 4 = 64 = (8)^{2}

Therefore, TP = 8 cm

From the figure we see that BQ = BR = 27 cm (since length of the tangent segments from an

external point are equal)

As BC = 38 cm

CR = CB - BR = 38 - 27

= 11 cm

Again,

CR = CS = 11cm (length of tangent segments from an external point are equal)

Now, as DC = 25 cm

DS = DC - SC

= 25 -11

= 14 cm

Now, in quadrilateral DSOP,

PDS = 90 (given)

**OSD = 90, OPD = 90 (since tangent is perpendicular to the **

radius through the point of contact)

DSOP is a parallelogram

OP||SD and PD||OS

Now, as OP = OS (radii of the same circle)

OPDS is a square. DS = OP = 14cm

radius of the circle = 14 cm

In AXB,

XAB + AXB + ABX=180 [Triangle property]

XAB + 50 + 70 = 180

XAB=180 - 120 = 60

XAY=90 [Angle of semi-circle]

BAY=XAY -XAB = 90 - 60 = 30

and BXY = BAY = 30 [Angle of same segment]

ACX = BXY + ABX [External angle = Sum of two interior angles]

= 30 + 70

= 100

also,

XYP=90 [Diameter ⊥ tangent]

APY = ACX -CYP

APY=100 - 90

APY=10

PAQ is a tangent and AB is a chord of the circle.

i) (angles in alternate segment)

ii) In

iii) (angles in the same segment)

Now in

iv) PAQ is the tangent and AD is chord

i) AB is diameter of circle.

In

ii) QC is tangent to the circle

Angle between tangent and chord = angle in alternate segment

ABQ is a straight line

i)

ii) Since, BPDO is cyclic quadrilateral, opposite angles are supplementary.

i) PQ = RQ

(opposite angles of equal sides of a triangle)

Now, QOP = 2PRQ (angle at the centre is double)

ii) PQC = PRQ (angles in alternate segments are equal)

QPC = PRQ (angles in alternate segments)

Consider two concentric circles with centres at O. Let AB and CD be two chords of the outer circle which touch the inner circle at the points M and N respectively.

To prove the given question, it is sufficient to prove AB = CD.

For this join OM, ON, OB and OD.

Let the radius of outer
and inner circles be *R* and *r* respectively.

AB touches the inner circle at M.

AB is a tangent to the inner circle

OMAB

BM = AB

AB = 2BM

Similarly ONCD, and CD = 2DN

Using Pythagoras theorem in OMB and OND

Since AC is tangent to the circle with center P at point A.

In

Also in Rt.

From (i) and (ii),

In the figure, a circle with centre O, is the circum circle of triangle XYZ.

Tangents at X and Y intersect at point T, such that XTY = 80

From Rt.

Now, since the two chords AE and BC intersect at D,

AD x DE = CD x DB

3 x DE = 9 x 4

Hence, AE = AD + DE = (3 + 12) = 15 cm

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 18 - Tangents and Intersecting Chords Page/Excercise 18(A)

OP = 10 cm; radius OT = 8 cm

Length of tangent = 6 cm.

AB = 15 cm, AC = 7.5 cm

Let 'r' be the radius of the circle.

OC = OB = r

AO = AC + OC = 7.5 + r

In ∆AOB,

AO^{2}
= AB^{2} + OB^{2}

Therefore, r = 11.25 cm

From Q, QA and QP are two tangents to the circle with centre O

Therefore, QA = QP .....(i)

Similarly, from Q, QB and QP are two tangents to the circle with centre O'

Therefore, QB = QP ......(ii)

From (i) and (ii)

QA = QB

Therefore, tangents QA and QB are equal.

From Q, QA and QP are two tangents to the circle with centre O

Therefore, QA = QP .......(i)

Similarly, from Q, QB and QP are two tangents to the circle with centre O'

Therefore, QB = QP .......(ii)

From (i) and (ii)

QA = QB

Therefore, tangents QA and QB are equal.

OS = 5 cm

OT = 3 cm

In Rt. Triangle OST

By Pythagoras Theorem,

Since OT is perpendicular to SP and OT bisects chord SP

So, SP = 8 cm

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

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

r_{1} + r_{3} = 8

r_{3} + r_{2} = 9

r_{2} + r_{1} = 6

Adding

r_{1} + r_{3} + r_{3} + r_{2} + r_{2} + r_{1} = 8+9+6

2(r_{1} + r_{2} + r_{3}) = 23

r_{1} + r_{2} + r_{3 }= 11.5 cm

r_{1} + 9 = 11.5 (Since r_{2} + r_{3 }= 9)

r_{1} = 2.5 cm

r_{2} + 6 = 11.5 (Since r_{1} + r_{3 }= 6)

r_{2} = 5.5 cm

r_{3} + 8 = 11.5 (Since r_{2} + r_{1 }= 8)

r_{3 }= 3.5 cm

Hence, r_{1} = 2.5 cm, r_{2} = 5.5 cm and r_{3 }= 3.5 cm

Let the circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.

Since AP and AS are 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,

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

AB + CD = AD + BC

Hence, AB + CD = AD + BC

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

Therefore, AP = AS.......(i)

Similarly, we can prove that:

BP = BQ .........(ii)

CR = CQ .........(iii)

DR = DS .........(iv)

Adding,

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

AB + CD = AD + BC

Hence, AB + CD = AD + BC

But AB = CD and BC = AD.......(v) Opposite sides of a ||gm

Therefore, AB + AB = BC + BC

2AB = 2 BC

AB = BC ........(vi)

From (v) and (vi)

AB = BC = CD = DA

Hence, ABCD is a rhombus.

Since from B, BQ and BP are the tangents to the circle

Therefore, BQ = BP ………..(i)

Similarly, we can prove that

AP = AR …………..(ii)

and CR = CQ ………(iii)

Adding,

AP + BQ + CR = BP + CQ + AR ………(iv)

Adding AP + BQ + CR to both sides

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

2(AP + BQ + CR) = AB + BC + CA

Therefore, AP + BQ + CR = x (AB + BC + CA)

AP + BQ + CR = x perimeter of triangle ABC

Since, from A, AP and AR are the tangents to the circle

Therefore, AP = AR

Similarly, we can prove that

BP = BQ and CR = CQ

Adding,

AP + BP + CQ = AR + BQ + CR

(AP + BP) + CQ = (AR + CR) + BQ

AB + CQ = AC + BQ

But AB = AC

Therefore, CQ = BQ or BQ = CQ

Radius of bigger circle = 6.3 cm

and of smaller circle = 3.6 cm

i)

Two circles are touching each other at P externally. O and O’ are the centers of the circles. Join OP and O’P

OP = 6.3 cm, O’P = 3.6 cm

Adding,

OP + O’P = 6.3 + 3.6 = 9.9 cm

ii)

Two circles are touching each other at P internally. O and O’ are the centers of the circles. Join OP and O’P

OP = 6.3 cm, O’P = 3.6 cm

OO’ = OP - O’P = 6.3 - 3.6 = 2.7 cm

i) In

AP = BP (Tangents from P to the circle)

OP = OP (Common)

OA = OB (Radii of the same circle)

ii) In

OA = OB (Radii of the same circle)

(Proved )

OM = OM (Common)

Hence, OM or OP is the perpendicular bisector of chord AB.

Draw TPT' as common tangent to the circles.

i) TA and TP are the tangents to the circle with centre O.

Therefore, TA = TP ………(i)

Similarly, TP = TB ………..(ii)

From (i) and (ii)

TA = TB

Therefore, TPT' is the bisector of AB.

ii) Now in

Similarly in

Adding,

In quadrilateral OPAQ,

In triangle OPQ,

OP = OQ (Radii of the same circle)

From (i) and (ii)

Join OP, OQ, OA, OB and OC.

In

OA = OC (Radii of the same circle)

OP = OP (Common)

PA = PC (Tangents from P)

Similarly, we can prove that

(Sum of interior angles of a transversal)

Now in

In

LBNO is a square.

LB = BN = OL = OM = ON = x

Since ABC is a right triangle

The incircle touches the sides of the triangle ABC and

i) In quadrilateral AROQ,

ii) Now arc RQ subtends at the centre and at the remaining part of the circle.

Join QR.

i) In quadrilateral ORPQ,

ii) In

OQ = QR (Radii of the same circle)

iii) Now arc RQ subtends at the centre and at the remaining part of the circle.

In

OB = OC (Radii of the same circle)

Now in

BQ and BR are the tangents from B to the circle.

Therefore, BR =BQ = 27 cm.

Also RC = (38 -; 27) = 11cm

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

Therefore, CS = CR = 11 cm

So, DS = (25 - 11) = 14 cm

Now DS and DP are the tangents to the circle

Therefore, DS = DP

Now, (given)

and

therefore, radius = DS = 14 cm

(angles in alternate segment)

But OS = OR (Radii of the same circle)

But in

From (i) and (ii)

Join AT and BT.

i) TC is the diameter of the circle

(Angle in a semi-circle)

ii)

(Angles in the same segment of the

circle)

(Angles in the same segment of the circle)

iii) (Angles in the same segment)

Now in

Join OC.

Therefore, PA and PA are the tangents

In quadrilateral APCO,

Now, arc BC subtends at the centre and at the remaining part of the circle

## Selina Concise Mathematics X Class 10 Chapter Solutions

- Chapter 1 - Value Added Tax
- Chapter 2 - Banking (Recurring Deposit Accounts)
- Chapter 3 - Shares and Dividends
- Chapter 4 - Linear Inequations (in one variable)
- Chapter 5 - Quadratic Equations
- Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations)
- Chapter 7 - Ratio and Proportion (Including Properties and Uses)
- Chapter 8 - Remainder And Factor Theorems
- Chapter 9 - Matrices
- Chapter 10 - Arithmetic Progression
- Chapter 11 - Geometric Progression
- Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points)
- Chapter 13 - Section and Mid-Point Formula
- Chapter 14 - Equation of a Line
- Chapter 15 - Similarity (With Applications to Maps and Models)
- Chapter 16 - Loci (Locus and its Constructions)
- Chapter 17 - Circles
- Chapter 18 - Tangents and Intersecting Chords
- Chapter 19 - Constructions (Circles)
- Chapter 20 - Cylinder, Cone and Sphere (Surface Area and Volume)
- Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)
- Chapter 22 - Heights and Distances
- Chapter 23 - Graphical Representation (Histograms and Ogives)
- Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)
- Chapter 25 - Probability

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