Lesson: Circles

Exercise 9.1 (10) (Multiple Choice Questions and

Answers)

Choose the correct answer from the given four

options:

Question: 1

If radii of two concentric circles are 4 cm and 5 cm,

then the length of each chord of one circle which is

tangent to the other circle is

a. 3 cm

b. 6 cm

c. 9 cm

d. 1 cm

Solution

(b)

Let C1, C2 be concentric circles with centre O. PQ be

the chord of circle C2 that touches C1 at R.

OR is radius and PQ is tangent at R.

Thus,

OR PQ

From the figure,

and

R 90 ,

OP 5cm

OR 4cm

In , by Pythagoras theorem,

OPR

2 2 2

OP OR PR

2 2

2

5 4 PR

2

PR 25 16

2

PR 9

PR 3

The chord is bisected by the perpendicular from the

origin.

Thus,

PQ 2 PR

2 3

6

Hence, length of the chord is 6 cm.

Question: 2

In the adjoining figure, if , then

AOB 125

COD

is equal to:

a.

62.5

b.

45

c.

35

d.

55

Solution

(d)

We know that supplementary angles are subtended

at the centre of a circle by a quadrilateral

circumscribing that circle.

Thus,

AOB COD 180

125 COD 180

COD 180 125

COD 55

Hence, is .

COD

55

Question: 3

In the adjoining figure, AB is a chord of the circle

and AOC is its diameter such that . If AT

ACB 50

is the tangent to the circle at the point A, then

is equal to

BAT

a.

65

b.

60

c.

50

d.

40

Solution

(c)

From the given figure,

B 90 Anglein semicircle

BAC 180 C B Anglesumproperty

BAC 180 50 90

BAC 40

AT is a tangent at the point A and OA is a radius,

Thus, and

AT OA

OAT 90

OAB BAT 90

40 BAT 90

BAT 90 40

BAT 50

Hence, is .

BAT

50

Question: 4

From a point, P which is at a distance of 13 cm from

the centre O of a circle of radius 5 cm, the pair of

tangents PQ and PR to the circle are drawn. Then

the area of the quadrilateral PQOR is

a.

2

60cm

b.

2

65cm

c.

2

30cm

d.

2

32.5cm

Solution

(a)

According to the question,

PQ is a tangent at point Q and QO is a radius.

Thus,

PQO 90

In , by the Pythagoras theorem,

OPQ

2 2 2

PQ OP OQ

2 2

2

PQ 13 5

2

PQ 169 25

2

PQ 144

PQ 12

By SSS criterion of congrOPQ OPR uence

Thus,

area of OPQ area of OPR

Area of the quadrilateral PQOR

2 area of OPR

Area of the

quadrilateral PQO

1

2 base alti

2

R

tude

RP OR

12 5

60

Hence, the area of the quadrilateral PQOR is .

2

60cm

Question: 5

At one end A of a diameter AB of a circle of radius 5

cm, tangent XAY is drawn to the circle. The length of

the chord CD parallel to XY and at a distance 8 cm

from A is

a.

4 cm

b.

5cm

c.

6cm

d.

8cm

Solution

(d)

AO is a radius and XAY is a tangent to the circle at

point A.

Thus,

OAY 90

From the figure,

AO 5cm

There is a chord CD at a perpendicular distance of 8

cm from A. CD intersects AB at M and .

CMD||XAY

Join OD.

OD 5cm Radiusof circle

OM 8 5

3cm

OMD OAY 90

In , by Pythagoras theorem,

OMD

2 2 2

MD OD MO

2 2

2

MD 5 3

2

MD 25 9

2

MD 16

MD 4

The chord CD is bisected by the perpendicular from

centre O of circle.

Thus,

CD 2 MD

2 4

8

So,

CD 2MD 2 4 8 cm

Hence, the length of chord CD is 8 cm.

Question: 6

In the adjoining figure, AT is a tangent to the circle

with centre O such that OT 4 cm and .

OTA 30

Then AT is equal to

a.

4 cm

b.

2cm

c.

2 3 cm

d.

4 3 cm

Solution

(c)

Join OA.

From the figure,

AT is a tangent at the point A and OA is a radius.

Thus,

OAT 90

and

OT 4 cm

OTA 30

Now, in ,

OTA

Base

cos30

Hypotenuse

AT

cos30

4

3 AT 3

cos30

2 4 2

AT 2 3

Hence, the length of AT is .

2 3 cm

Question: 7

In the adjoining figure, if O is the centre of a circle,

PQ is a chord and the tangent PR at P makes an

angle of with PQ, then is equal to

50

POQ

a.

100

b.

80

c.

90

d.

75

Solution

(a)

From the figure,

PR is a tangent to the circle at point P.

OP is a radius of the circle.

Thus,

OPR 90

OPQ 50 90

OPQ 90 50

OPQ 40

In ,

OPQ

OP OQ Radii of samecircle

Thus,

Angles opposite to equal Q OPQ 40 sides

We know that, the sum of the interior angles in a

triangle is .

180

Thus, in ,

OPQ

POQ P Q 180

POQ 180 P Q

POQ 180 40 40

POQ 100

Hence,

POQ 100

Question: 8

In the adjoining figure, if PA and PB are tangents to

the circle with centre O such that , then

APB 50

is equal to

OAB

a.

25

b.

30

c.

40

d.

50

Solution

(a)

In ,

OAB

OA OB Radii of samecircle

Angles opposite to equal OAB OBA sides

PA is a tangent to the circle at point A and OA is a

radius.

Thus,

OAP 90

Similarly,

OBP 90

We know that in a quadrilateral, the sum of all the

interior angles is .

360

Thus, in quadrilateral PAOB,

P A O B 360

50 90 O 90 360

O 360 90 90 50

O 130

We know that, the sum of all the interior angles of a

triangle is .

180

Thus, in ,

OAB

O OAB OBA 180

130 OAB OAB 180 OAB OBA

2 OAB 180 130

2 OAB 50

50

OAB

2

OAB 25

Hence, is .

OAB

25

Question: 9

If two tangents inclined at an angle are drawn to

60

a circle of radius 3 cm, then length of each tangent is

equal to

a.

3

3 cm

2

b.

6cm

c.

3cm

d.

3 3 cm

Solution

(d)

In the figure given below,

LM and LN are tangents at points M and N of the

circle respectively and OM is the radius of the circle.

OM 3cm

Thus,

LMO 90

The line from the centre of the circle bisects the

angle formed by the two tangents.

Thus,

1

OLM NLM

2

1

60

2

30

In ,

LMO

OM

tan30

LM

1 3

LM

3

LM 3 3

Similarly,

LN 3 3

Hence, the length of each tangent is .

3 3 cm

Question: 10

In the adjoining figure, if PQR is the tangent to a

circle at Q whose centre is O, AB is a chord parallel

to PR and , then is equal to

BQR 70

AQB

a.

20

b.

40

c.

35

d.

45

Solution

(b)

Given:

AB PQRA

Alternate interior aB BQR 70 ngles

Alternate interior aOQR AMQ ngles

PQR is a tangent at point Q and OQ is a radius of

the circle.

Thus,

OQR 90

OQB BQR 90

OQB 70 90

OQB 90 70

OQB 20

QM is a perpendicular to the chord AB that bisects

AB.

Thus, and

AMO 90

MA MB

QMA QMB Each 90

MQ MQ Common side

Thus,

By SAS criterion of congrQMA QMB uence

A B

A 70 B 70

We know that the sum of the interior angles in a

triangle is .

180

Thus,

A AMQ AQM 180

70 90 AQM 180

AQM 180 70 90

AQM 20

AQB AQM OQB

AQB 20 20

AQB 40

Hence, is equal to .

AQB

40

Exercise 9.2 (10)

Write ‘True’ or ‘False’ and justify your answer in

each of the following:

Question: 1

If a chord AB subtends an angle of at the centre

60

of a circle, then angle between the tangents at A and

B is also .

60

Solution

False.

In the figure,

AP and PB are two tangents to a circle and AB is a

chord.

It is clear that PBOA is a cyclic.

Thus,

AOB APB 180 Supplimentary angles

60 APB 180

APB 180 60

APB 120

Hence, the angle between the tangents at A and B is

not .

60

Question: 2

The length of tangent from an external point on a

circle is always greater than the radius of the circle.

Solution

False.

From the figure,

LM is a tangent and OM is a radius.

Thus,

OML 90

Let

OL 5cm and OM 4 cm

In , by Pythagoras theorem,

OML

2 2 2

OL OM ML

2 2

2

5 4 ML

2

ML 25 16

2

ML 9

ML 3

It is clear that .

ML OM

Hence, tangent is less than the radius.

Question: 3

The length of tangent from an external point P on a

circle with centre O is always less than OP.

Solution

True.

From the figure,

PM is a tangent and OM is a radius.

Thus,

OMP 90

Fig. Exm_9.2_3

Since, in a right-angled triangle, hypotenuse is the

longest side.

Thus,

In , .

OMP

OP PM

Question: 4

The angle between two tangents to a circle may be

.

0

Solution

False.

The angle between two tangents to a circle cannot be

because in this condition, two tangents will

0

become one.

Question: 5

If angle between two tangents drawn from a point P

to a circle of radius and centre O is , then

a

90

.

OP a 2

Solution

True.

From the figure,

PL and PM are tangents and OQ is a radius.

In ,

OPQ

OPQ 45

OQ

sin OPQ

OP

a

sin 45

OP

1 a

OP

2

OP a 2

Question: 6

If angle between two tangents drawn from a point P

to a circle of radius and centre O is , then

a

60

.

OP a 3

Solution

False.

From the figure,

PL and PM are tangents and OL and OM are radius.

LPM

OPL

2

60

2

30

In ,

OLP

OL

sin OPL

OP

a

sin 30

OP

1 a

2 OP

OP 2a

Question: 7

The tangent to the circumcircle of an isosceles

triangle ABC at A, in which , is parallel to

AB AC

BC.

Solution

True.

From the figure,

In ,

ABC

AB AC Given

Isosceles triC B angle

Anglesin alternate

CAD B

segment

CAD C

These angles are alternate interior angles.

Thus,

AD||BC

Question: 8

If a number of circles touch a given line segment PQ

at a point A, then their centres lie on the

perpendicular bisector of PQ.

Solution

True.

From the figure,

PQ is a line segment.

There are two circles with centres that

O andO

touches PQ on A.

PQ is a tangent and radius of the two

OA andO A

circles.

Thus,

OAQ O AQ 90

It is possible only when and lie on the same line.

O

O

Hence, the centres of all circles lie on the

perpendicular bisector of PQ.

Question: 9

If a number of circles pass through the end points P

and Q of a line segment PQ, then their centres lie on

the perpendicular bisector of PQ.

Solution

True.

From the figure,

A and B are the centres of the two circles that passes

through the end points P and Q of the line segment

PQ.

In and ,

ACP

ACQ

AP AQ

AC AC

ACP ACQ 90

Thus,

ACP ACQ

Thus, .

CP CQ

Hence, the centres of the circles lie on the

perpendicular bisector of PQ.

Question: 10

AB is a diameter of a circle and AC is its chord such

that . If the tangent at C intersects AB

BAC 30

extended at D, then .

BC BD

Solution

True.

From the figure,

We know that the angle made by a segment on the

centre of a circle is two times of the angle made by

any part of the circle.

Thus,

BOC 2 BAC

2 30

60

In ,

CDO

CDO 180 60 90 Anglesumproperty

30

CBD CAB ACB Exterior Angleproperty

120

BAC BDC 30

Thus,

BC BD

Exercise 9.3 (10)

Question: 1

Out of the two concentric circles, the radius of the

outer circle is 5 cm and the chord AC of length 8 cm

is a tangent to the inner circle. Find the radius of the

inner circle.

Solution

From the figure,

and

OA 5cm

AC 8cm

AC is a tangent at the point B and OB is a radius.

Thus, and .

OBA 90

OB AC

We know that the perpendicular from the centre

bisects the chord.

Thus,

AC 2AB

AC

AB

2

8

AB

2

AB 4

In ,

AOB

2 2 2

OA AB OB

2 2

2

5 4 OB

2

OB 25 16

2

OB 9

OB 3

Hence, the radius of the inner circle is 3 cm.

Question: 2

Two tangents PQ and PR are drawn from an external

point to a circle with centre O. Prove that QORP is a

cyclic quadrilateral.

Solution

From the figure,

PQ and PR are tangents to the circle and OQ and OR

are radius.

Thus,

OQP ORP 90

In quadrilateral QORP, by the angle sum property,

OQP ORP QOR QPR 360

90 90 QOR QPR 360

QOR QPR 360 90 90

QOR QPR 180

Hence, quadrilateral QORP is cyclic.

Question: 3

If from an external point B of a circle with centre O,

two tangents BC and BD are drawn such that

, prove that i.e.

DBC 120

BC BD BO

.

BO 2 BC

Solution

From the figure,

DBC 120

BC and BD are two tangents to the circle and OC

and OD are radius of the circle.

ODB 90

We know that the tangents are inclined equally to

the line joining the centre of the circle to the external

point.

Thus,

CBD

OBD

2

120

2

60

In ,

ODB

BD

cos60

BO

1 BD 1

cos60

2 BO 2

BO 2BD

BO BD BD

The tangents are from same external point.

Thus,

BC BD

Hence,

BO BC BD

Question: 4

Prove that the centre of a circle touching two

intersecting lines lies on the angle bisector of the

lines.

Solution

From the figure,

AB and are the tangents to the circle

A B

intersecting at L. and are radius.

O N

O M

Let LP does not pass through .

O

In and ,

O ML

O NL

O ML O NL 90

O L O L Common

O M O N Radius

Thus,

O ML O NL

Thus,

MLO O LN

This condition is possible when lies on LP.

O

Thus, LP passes through the centre that lies on

O

the angle bisector of the tangents.

Question: 5

In the adjoining figure, AB and CD are common

tangents to two circles of unequal radii. Prove that

AB CD.

Solution

From the figure,

AB and CD are common tangents to the two circles.

Extend AB and CD that meet at E.

Now, EA and EC are two tangents from the same

point.

Thus,

EA EC ...... 1

Similarly, EB and ED are two tangents from the same

point.

Thus,

EB ED ...... 2

Subtract equation (2) from (1).

EA EB EC ED ...... 3

From the figure,

and

EA EB AB

EC ED CD

Put the above result in equation (3).

Hence,

AB CD

Question: 6

In Question 5 above, if radii of the two circles are

equal, prove that AB CD.

Solution

From the figure,

AB and CD are common tangents to the two circles.

A B 90

Thus,

OA O BA

OA O B Equal radii of two circles

Thus, quadrilateral is a parallelogram.

AOO B

Similarly, quadrilateral is a parallelogram.

COO D

AB OO Oppositesidesof parallelogram

CD OO Oppositesidesof parallelogram

Thus,

AB CD

Hence proved.

Question: 7

In the adjoining figure, common tangents AB and

CD to two circles intersect at E. Prove that AB CD.

Solution

From the given figure,

AB and CD are common tangents to two circles that

intersect at E.

EA and EC are two tangents from the same points.

Thus,

EA EC ...... 1

Similarly, EB and ED are two tangents from the same

points.

Thus,

EB ED ...... 2

Add equations (1) and (2).

EA EB EC ED ...... 3

From the given figure,

and

EA EB AB

EC ED CD

Put the above result in equation (3).

AB CD

Hence proved.

Question: 8

A chord PQ of a circle is parallel to the tangent

drawn at a point R of the circle. Prove that R bisects

the arc PRQ.

Solution

From the figure,

PQ is a chord of the circle and TR is a tangent at the

point R.

Join PR and QR.

Now,

TRP Q Anglesin alternatesegment

TRP P PQ TR A

Thus,

P Q

RQ RP Converseof isosceles property

A

A

RQ RP

Hence, R bisects the arc PRQ.

Question: 9

Prove that the tangents drawn at the ends of a chord

of a circle make equal angles with the chord.

Solution

From the figure,

LM is a chord and XL and YM are two tangents at

points L and M.

Locate a point N on arc LM and join LN and MN.

YML N Anglesin alternatesegment

XLM N

From the above results,

YML XLM

Hence, the tangents drawn at the ends of a chord of

a circle make equal angles with the chord.

Question: 10

Prove that a diameter AB of a circle bisects all those

chords which are parallel to the tangent at the point

A.

Solution

From the figure,

AB is a diameter of a circle and AP is a tangent at

point A.

L is a chord parallel to AP.

PAB 90 radius tangent

PAB 1 AP L

Thus,

AB L

We know that the perpendicular from the centre of

the circle to the chord bisects the chord.

Hence, AB bisects all those chords which are parallel

to AP.

Exercise 9.4 (14)

Question: 1

If a hexagon ABCDEF circumscribe a circle, prove

that .

AB CD EF BC DE FA

Solution

From the figure, ABCDEF is a hexagon

circumscribing a circle.

We know that the tangents from the same external

points are equal.

Thus,

AP AU

BP BQ

CR CQ

DR DS

ET ES

FT FU

Add all the above results.

AP BP CR DR ET FT AU BQ CQ DS ES FU

AB CD EF BC DE FA

Hence Proved.

Question: 2

Let s denote the semi-perimeter of a triangle ABC in

which , , . If a circle touches the

BC a

CA b

AB c

sides BC, CA, AB at D, E, F, respectively, prove that

.

BD s– b

Solution

From the figure,

ABC is a triangle and a circle touches the sides BC,

CA, AB at D, E, F, respectively.

We know that the tangents from the same external

points are equal.

Thus,

BD BF x

AF AE c x

CD CE a x

AC AE CE

b c x a x

b 2x a c

Add on both sides.

b

2b 2x a b c

a b c

b x

2

a b c

b x s s

2

x s b

BD s b BD x

Hence,

BD s– b

Question: 3

From an external point P, two tangents, PA and PB

are drawn to a circle with centre O. At one point E on

the circle tangent is drawn which intersects PA and

PB at C and D, respectively. If , find the

PA 10cm

the perimeter of the triangle PCD.

Solution

From the figure,

PA and PB are two tangents drawn to a circle from

external point P and .

PA 10cm

Perimeter of the

PC CD PD

triangle PCD

PC CE ED PD CD CE ED

We know that the tangents from same external

points are equal.

Thus, and .

CE CA

DE DB

So,

Perimeter of the

PC CA DB PD

triangle PCD

PA PB PA PC CA andPB DB PD

PA PA PA PB

2PA

2 10

20

Hence, the perimeter of the triangle PCD is 20 cm.

Question: 4

If AB is a chord of a circle with centre O, AOC is a

diameter and AT is the tangent at A as shown in

figure. Prove that .

BAT ACB

Solution

From the figure,

AB is a chord of the circle and AT is the tangent at

the point A.

B 90 Anglein semicircle

In triangle ABC, by the angle sum property,

B CAB ACB 180

90 CAB ACB 180

CAB ACB 180 90

CAB ACB 90 ...... 1

Now,

CAT 90 radius tangent

CAB BAT 90 ...... 2

From equations (1) and (2).

CAB ACB CAB BAT

ACB BAT

BAT ACB

Hence proved.

Question: 5

Two circles with centres O and of radii 3 cm and 4

O

cm, respectively intersect at two points P and Q such

that OP and are tangents to the two circles. Find

O P

the length of the common chord PQ.

Solution

From the figure,

OP OQ 3cm

O P O Q 4 cm

Thus, quadrilateral is a kite.

OPO Q

and are tangents.

OP

O P

PQ OO Diagonalsof akite

PSO 90

O PO 90 Radius tangent

In triangle ,by the Pythagoras theorem,

O PO

2 2 2

OO O P OP

2 2 2

OO 4 3

2

OO 16 9

2

OO 25

OO 25

OO 5

We know that in a triangle,

a b c

s

2

Here, and

a 3,

b 4

c 5

Thus,

3 4 5

s

2

12

2

6

Area of

triangleO PO

s s a s b s c

6 6 3 6 4 6 5

6 3 2 1

36

6

1

Area of O PO OO PS

2

1

6 5 PS

2

2

PS 6

5

12

PS

5

Similarly,

12

QS

5

Now,

PQ PS QS

12 12

5 5

24

5

Hence, PQ is cm.

24

5

Question: 6

In a right triangle, ABC in which , a circle is

B 90

drawn with AB as diameter intersecting the

hypotenuse AC and P. Prove that the tangent to the

circle at P bisects BC.

Solution

From the figure,

B 90

In triangle ABC, by the angle sum property,

A B C 180

A 90 C 180

A C 180 90

...A C 90 ... 2

5 90 Anglein semicircle

Now in triangle APB, by the angle sum property,

A 1 5 180

A 1 90 180

A 1 180 90

A 1 90 ...... 2

From equations (1) and (2).

A C A 1

..C 1 .... 3

1 3 Anglesin alternatesegment

...... 4

3 4 Oppositeangles

...... 5

From equations (3), (4) and (5).

C 4

Now, by the converse of isosceles triangle property,

MP MC

We know that tangents from the same points are

equal.

Thus,

MP MB

Thus,

MB MC

Hence, the tangent to the circle at P bisects BC.

Question: 7

In the adjoining figure, tangents PQ and PR are

drawn to a circle such that . A chord RS

RPQ 30

is drawn parallel to the tangent PQ. Find .

RQS

Solution

From the figure,

and

RPQ 30

RS QPA

In ,

PQR

PQ PR Tangentsfrom samepoint

PRQ RQP Isosceles triangleproperty

Now by the angle sum property,

PRQ RQP RPQ 180

PRQ PRQ RPQ 180

2 PRQ 30 180

180 30

PRQ

2

PRQ 75

Thus,

PRQ RQP 75

RQP QRS 75 Alternateinterior angles

PRQ RSQ 75 Anglesin alternatesegment

In triangle RSQ, by the angle sum property,

RQS 75 75 180

RQS 180 75 75

RQS 30

Hence, is .

RQS

30

Question: 8

AB is a diameter and AC is a chord of a circle with

centre O such that . The tangent at C

BAC 30

intersects extended AB at a point D. Prove that

.

BC BD

Solution

From the figure,

BAC 30

1 BAC 30 Anglesin alternatesegment

2 90 Anglein semicircle

4 2 BAC Exterior angleproperty

4 90 30

4 120

In triangle BDC, by the angle sum property,

1 4 BDC 180

30 120 BDC 180

BDC 180 30 120

BDC 30

1 BDC 30

Now, by the converse of isosceles triangle property,

.

BD BC

Question: 9

Prove that the tangent drawn at the mid-point of an

arc of a circle is parallel to the chord joining the end

points of the arc.

Solution

From the figure,

SR is a tangent and Q is the midpoint of arc TP.

We know that angles in alternate segment are equal.

Thus,

..1 4 .... 1

..2 3 .... 2

Since, Q is the midpoint of arc TP.

Arc TQ Arc QP

Thus, QT QP

In triangle QPT,

QT QP

2 4 Isoscelestriangleproperty

...... 3

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

1 2

We know that and are alternate interior

1

2

angles.

Thus,

PT||SR

Hence proved.

Question: 10

In the adjoining figure, the common tangent, AB and

CD to two circles with centres O and intersect at

O

E. Prove that the points O, E, are collinear.

O

Solution

From the figure,

AB and CD are two tangents common to the two

circles.

AED CEB Vertically oppositeangles

We know that the tangents are equally inclined to

the line that joins the centre to the external point.

Thus,

AEO OEC

DEO O EB

We know that the sum of angles around a point is

.

360

Thus,

AED CEB OEC AEO O EB DEO 360

AED AED AEO AEO DEO DEO 360

2 AED AEO DEO 360

AED AEO DEO 180

OEO 180

Hence, the points O, E and are collinear.

O

Question: 11

In the adjoining figure, O is the centre of a circle of

radius 5 cm, T is a point such that and

OT 13cm

OT intersects the circle at E. If AB is the tangent to

the circle at E, find the length of AB.

Solution

From the figure,

, AB is a tangent at the point E of the

OT 13cm

circle of radius

r 5cm

AEO OPA 90 Radius tangent

In triangle OPT, by Pythagoras theorem,

2 2 2

PT OT OP

2 2 2

PT 13 5

2

PT 169 25

2

PT 144

PT 144

PT 12

Now,

OP OE 5cm

ET OT OE

13 5

8

AET 180 AEO

180 90

90

We know that tangents from same external point are

equal.

Let

AP AE x

In , by Pythagoras theorem,

AET

2 2 2

AT AE ET

2 2

2

12 x x 8

2 2

144 x 2 12x x 64

24x 144 64

80

x

24

10

x

3

Similarly,

10

BE

3

AB AE BE

10 10

3 3

20

3

Hence,

20

AB cm

3

Question: 12

The tangent at a point C of a circle and a diameter

AB when extended intersect at P. If ,

PCA 110

find CBA.

Solution

According to the question,

PCA 110

Thus,

PCB BCA 110

PCB 110 BCA

PCB 110 90 BCA 90

PCB 20

CAB PCB 20 Anglesin alternatesegment

In triangle BCA, by the angle sum property,

BCA CAB CBA 180

90 20 CBA 180

CBA 180 90 20

CBA 70

Hence, is

CBA

70

Question: 13

If an isosceles triangle ABC, in which

, is inscribed in a circle of radius 9

AB AC 6cm

cm, find the area of the triangle.

Solution

From the figure,

AB AC 6cm

OB OC OA 9cm

Thus, quadrilateral ABOC is a kite.

We know that the diagonals of a kite are

perpendicular to each other.

Thus,

AO BC

Let

AD x

In triangle ABD, by Pythagoras theorem,

2 2 2

AB AD BD

2 2 2

BD AB AD

2 2 2

BD 6 x

2

2

BD 36 x ...... 1

In triangle BDO, by the Pythagoras theorem,

2 2 2

BD 9 9 x

2

2

BD 81 81 x 2 9x

2

2

BD 81 81 x 18x

2

2

BD x 18x ...... 2

From (1) and (2).

2 2

36 x x 18x

36

x

18

x 2

Put the value of in equation (1).

x

2

2

BD 36 x

2 2

BD 36 2

2

BD 36 4

2

BD 32

BD 4 2

Similarly,

CD 4 2

BC BD CD

4 2 4 2

8 2

Area of ABC

1

BC AD

2

1

BC AD

2

1

8 2 2

2

8 2

Hence, the area of the triangle is .

2

8 2 cm

Question: 14

A is a point at a distance 13 cm from the centre O of

a circle of radius 5 cm. AP and AQ are the tangents

to the circle at P and Q. If a tangent BC is drawn at a

point R lying on the minor arc PQ to intersect AP at

B and AQ at C, find the perimeter of the .

ABC

Solution

From the figure,

OP 5cm

OA 13cm

BC is a tangent to the circle at point R.

BP is a tangent and OP is a radius.

Thus,

BPO 90

In triangle OPA, by the Pythagoras theorem,

2 2 2

OA OP AP

2 2 2

AP OA OP

2 2 2

AP 13 5

2

AP 169 25

2

AP 144

AP 144

AP 12

Perimeter of trian ABgl Be AB CC CA

AB BR CR CA

We know that tangents from same external point are

equal.

Thus, and

BR BP

CR CQ

So,

Perimeter of triang AB BP Cle AB QC CA

AP AQ

AP AP

2AP

2 12

24

Hence, the perimeter of the is .

ABC

24 cm