SELINA Solutions for Class 10 Maths Chapter 18 - Tangents and Intersecting Chords
Chapter 18 - Tangents and Intersecting Chords Exercise Ex. 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,
AO2 = AB2 + OB2
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 r1, r2 and r3 respectively.
r1 + r3 = 8
r3 + r2 = 9
r2 + r1 = 6
Adding
r1 + r3 + r3 + r2 + r2 + r1 = 8+9+6
2(r1 + r2 + r3) = 23
r1 + r2 + r3 = 11.5 cm
r1 + 9 = 11.5 (Since r2 + r3 = 9)
r1 = 2.5 cm
r2 + 6 = 11.5 (Since r1 + r3 = 6)
r2 = 5.5 cm
r3 + 8 = 11.5 (Since r2 + r1 = 8)
r3 = 3.5 cm
Hence, r1 = 2.5 cm, r2 = 5.5 cm and r3 = 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
∠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
Chapter 18 - Tangents and Intersecting Chords Exercise Ex. 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
Chapter 18 - Tangents and Intersecting Chords Exercise Ex. 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 AP2 = AC2 + CP2
52 = 12 + CP2
CP2 = 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
2
EAD = 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, TP2 = TA x TB
TP2 = 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 = 2
PRQ (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
OM
AB
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
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