Chapter 18 - Tangents and Intersecting Chords Exercise Ex. 18(C)

Question 1

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

Solution 1

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.

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 , find the radius of the circle.

Solution 2

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

Question 3

Two circles with centers 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.

Solution 3

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

Question 4

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

Solution 4

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.

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?

Solution 5

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.

Question 6

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

Solution 6

ABCD is a cyclic quadrilateral in which AD||BC

(Sum of opposite angles of a quadrilateral)

Now in 

Now in 

Question 7

In the given figure, C and D are points on the semicircle described on AB as diameter.

Given angle BAD = 70and angle DBC = 30, calculate angle BDC

Solution 7

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

Question 8

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

Solution 8

ABCD is a cyclic quadrilateral.

Similarly,

Hence, 

Question 9

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

Solution 9

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.

Question 10

Bisectors of vertex A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF =

Solution 10

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)

Question 11

In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If , find angle AOD.

Solution 11

Join OB.

In 

In 

Since DOC is a straight line

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.

Solution 12

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

Question 13

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

Solution 13

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

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.

Solution 14

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.

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 and , find:

i) 

ii) 

iii) 

Solution 15

i) PQ is tangent and CD is a chord

 (angles in the alternate segment)

ii)

iii) In 

Question 16

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

Solution 16

Join OC.

BCD is the tangent and OC is the radius.

Substituting in (i)

Question 17

ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D.

Prove that -

i) AC x AD = AB²

ii) BD² = AD x DC.

Solution 17

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 

Question 18

In the given figure AC = AE.

Show that:

i) CP = EP

ii) BP = DP

Solution 18

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)

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) 

ii)

Solution 19

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,

Question 20

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

(i) angle BCO

(ii) angle AOB

(iii) angle APB

Solution 20

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:

Question 21

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

Solution 21

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.

Question 22

In a square ABCD, its diagonal AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and bisector of angle ABD meets AC at N and AM at L. Show that -

i) 

ii) 

iii) ALOB is a cyclic quadrilateral.

Solution 22

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.

Question 23

The given figure shows a semicircle with centre O and diameter PQ. If PA = AB and ; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

Solution 23

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.

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

Solution 24

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)

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 segment, show that:

Solution 25

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)

Question 26

AB is a line segment and M is its midpoint. Three semicircles 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 semicircles. Show that: AB = 6 x r

Solution 26

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

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.

Solution 27

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.

Question 28

Two circles intersect in points P and Q. A secant passing through P intersects the circle 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.

Solution 28

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.

Question 29

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

Solution 29

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

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.

Solution 30

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.

Question 31

In the given figure, find TP if AT = 16 cm and AB = 12 cm.

Solution 31

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

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.

Solution 32

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

Question 33

In the figure, XY is the diameter of the circle, PQ is the tangent to the circle at Y. Given that and . Calculate  and .

Solution 33

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

Question 34

In the given figure, QAP is the tangent at point A and PBD is a straight line. If and ; find:

i) 

ii)

iii) 

iv) 

Solution 34

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

Question 35

In the given figure, AB is the diameter. The tangent at C meets AB produced at Q.

If

Solution 35

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

Question 36

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

If AB is parallel to CD and , find:

i) 

ii) 

Solution 36

i)

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

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

Solution 37

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)

Question 38

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

Solution 38

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

Question 39

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

Given that AB = 8 cm, calculate PQ.

Solution 39

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

In 

Also in Rt.

From (i) and (ii),

Question 40

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

Solution 40

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

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, find AE.

Solution 41

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

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.

 

 

Solution 42

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.

 

 

Solution 43