188

After learning this chapter you will be able to* Recognise similiar triangles.* Understand the properties of similar triangles.* Know the properties of Right angled triangle.* Solve the problems on similar triangles.* Understand the properties of tangents to the circle.* Understand the properties of touching circles.* Solve the problems on tangents to the circles

1. Concept of SimilarityWe see a number of objects around us, such as a twenty-five paise coin and

a fifty paise coin, photograph of an object and its enlarged copy, a marble and a rubberball, mini model of a building and the original building etc., all these pairs of objectshave same shape but they differ in size. All these pairs of objects are said to be similar.

Activity : 1Cut out some geometrical figures like triangles, quadrilaterals etc., from a piece

of card board. Hold these figures, one by one between a point source of light and

8THEOREMS ON TRIANGLES

AND CIRCLES

(Srinivasa Ramanujam) (Srinivasa Ramanujam)

189

a wall. Observe the shadow cast by each figure on the wall. The shadows have thesame shape, as the original figures, but are larger in size. The shadows are said tobe similar to the original figures.

Observe the following similar figures, note that they have same shapes, althoughthey are different in size.

190

Thales, a Greek mathematician (660 B.C.) is said to have introducedgeometry in Greece is believed to have found the heights of the pyramidsin Egypt, using shadows and the principle of similar triangles. The use ofsimilar triangles has made possible the measurements of heights anddistances.

Activity : 2Observe the pairs of figures given below. Record your observations with respect

to the angles and sides. Write your conclusion.Figures Corresponding Corresponding Similar/

angles sides notequal/unequal proportional/ similar

not proportional

191

From the above observations, it can be concluded that,Two polygons having the same number of sides are similar, If and only if,(i) The angles of one triangle are equal to the corresponding angles of the other and(ii) The sides of one triangle are proportional to the corresponding sides of the other.

ProportionalityIf two ratios are equal, then they are said to be in proportion.

If a:b = c:d or dc

ba= . Then a, b, c, and d are in proportion.

2. Similarity of Triangles.

Consider the triangles ABC and DEF. The corresponding vertices can bediagrammatically shown as,

A B C D E F

Corresponding angles

EDFBAC

DEFABC

DFEACB

Think!When will two right angledtriangles become similar?

Note:To prove the similarity of twotriangles, it is important toidentify the corresponding sidesand corresponding angles.

D

E F

192

Corresponding sides are AB DE

BC EF

AC DF

Also, DEAB

= EFBC

= FDCA

ABC is similar to DEF. This can be written symbolically asABC ||| DEF(The symbol ||| means Similar to)In case of triangles, let us examine whether both the conditions are required for

the triangles to be similar.In order to prove the conditions for triangles to be similar, you require the

knowledge of the following results.

3. Basic Proportionality theorem. [Thales theorem.]Activity : 1Construct ABC with BC = 9 cm, AB = 8 cm and CA = 10cms.Mark a point X on AB, such that BX = 2cms, as shown onthe adjoining figure. Through X draw XY parallel to BC.Measure AY and YC. Write the measurements.

Sides Ratios

AX =... BX = ... AY = .... YC = .... BXAX

= YCAY

=

What is your conclusion?

Activity : 2Construct ABC, with BC = 5cms. AB = 4cms, and AC = 3cms.Produce BA to X such that AX = 4cms. (Shown in the figure).through X draw XY parallel to BC to meet CA. produced atY. Measure BX, AY and CY and tabulate as shown.

Sides Ratios

AX =... BX = ... AY = .... CY = .... BXAX

= CYAY

=

What is your conclusion?

Think :Are Congruenttriangles similar?

193

Activity : 3Construct PQR with the measurements given in theadjoining figure. Produce PQ to S, such that QS = 2cms.Draw ST parallel to QR to meet PR produced at T.

Measure PT and PR. Write the ratios RTPR

and QSPQ

Also write your conclusion.From the above activities, you can conclude that,A straight line drawn parallel to a side of a triangle, divides the othertwo sides proportionately.This is known as Basic proportionality theorem or Thales theorem.Think!What is the Converse of Basic proportionality theorem?

4. Corollary of basic proportionality theorem.

ActivityConstruct PQR, given PQ = 8cms, QR = 6cms andPR = 10cms. Mark a point A on QP, Such that QA = 4cms.Through A draw AB parallel to QR. Measure the sides of thetriangle PAB. Tabulate the measurements as shown.

Sides Ratios

PQ = QR = PR = PA = AB = PB = PQPA

= QRAB

= PRPB

=

Examine the relation among the ratios. Write your conclusion?From the above activity, it can been seen that,

PQPA

= QRAB

= PRPB

This relationship can be stated as follows :

If a line is drawn parallel to a side of a triangle, then the sides of thenew triangle formed are proportional to the sides of the given triangle.This is the Corollary of Basic proportionality theorem.

RT

P Q S50 40

O

O

6 2

194

Worked Examples :1. In ABC, DE is parallel to BC AD = 5.7cms, BD = 9.5cms. EC = 6cms. find

AE.Solution : DE || BC (given)

ECAE

DBAD

= (B.P.T)

6AE

5.97.5= (on substitution)

5.9

6x7.5AE =

AE = 3.6 cms.

2. In the adjoining figure XY || BC. AX = p 3, BX = 2p 2, and 41

CYAY

= find p.

Solution : XY || BC (given)

CYAY

BXAX

= (B.P.T.)

41

2p23P=

(on substitution)

4(p 3) = 1(2p 2)4p 12 = 2p 2 2p = 10

p = 5

3) In the adjoining figure EF || CA and FG || AB. Prove GBDG

ECDE

=

Solution : In DAC, EF || CA (given)

FADF

ECDE

= ........ (i) (B.P.T.)

Similarly in DAB, FG || AB.

GBDG

FADF

= ...... (ii) (B.P.T.)

From (i) and (ii), GBDG

ECDE

=

195

4) In the figure XY || BC Prove, (i) ACAY

ABAX

= (ii) CY

ACBXAB

=.

Solution: XY || BC (given)

CYAY

BXAX

= (B.P.T)

AYCY

AXBX

=

1AYCY1

AXBX

+=+ (adding 1 to both sides)

AYAYCY

AXAXBX +

=

+

AYAC

AXAB

= (From the figure)

AC

AYABAX

=

(ii) ACAY

ABAX

= (Proved)

ACAY1

ABAX1 = (Subtracting 1 from both sides)

ACAYAC

ABAXAB

=

ACCY

ABBX

= (From the figure)

CYAC

BXAB

=

196

5) In the figure XY || BC, show that BCXY

ACAY

ABAX

==

Solution: Drawn XZ || ACXY || BC given

ACAY

ABAX

= ...... (i) cor. B.P.T.

XZ || AC (Construction)

BCCZ

ABAX

= Cor. B.P.T.

But CZ = XY ( Opp. sides of parallelogram XZCY)

BCXY

ABAX

= .......... (ii)

From (i) and (ii), BCXY

ACAY

ABAX

==

Exercise : 8.1

1. Study the adjoining figure. Write the ratios inrelation to basic proportionality theorem and itscorollary, in terms of a, b, c and d.

2. In the adjoining figure, DE || AB,AD = 7, CD = 5 and BC = 18 cms.Find (i) BE (ii) CE

3. A 6 mts Pole casts a shadow of 8 mts. at a certain time of the day. Find thelength of the shadow cast by a 4.5 mts. tower at the same time.

4. Which of the following sets of data make FG || BC?(i) AB = 14, AF = 6, AC = 7, AG = 3.(ii) AB = 12, FB = 3, AC = 8, AG = 6.(iii) AF = 6, FB = 5, AG = 9, GC = 8.

197

5. In the adjoining figure ABCD is aparallelogram. P is a point on BC,DP and AB are produced to meet at L.Prove that DP : PL = DC : BL

6. Show that in a trapezium the line joining the midpoints of non-parallel sides is parallelto the parallel sides.

5. Theorems on similar triangles.

Activity : 1Construct PQR having QR = 4cms, Q = 400 and R = 500 Construct

another XYZ, with YZ = 8 cms, Y = 400, and Z = 500.Measure the sides and remaining angles of both triangles Tabulate the measurements

as shown below.

Triangle Angle Sides Ratios PQR P =..... PQ =... QR =... RP =...XYZ X =.... XY =... YZ =... ZX =...

Observe the relationship between the angles and the ratio of sides of the trianglesPQR and XYZ.

What is your conclusion?

Activity : 2Construct ABC in which AB = 8cms, BC = 10cms, and AC = 6 cms. Mark a point D on AC,Such that AD = 2cms, as shown in the figure.Through D draw DE || AB.Measure the sides and the angles of ABC andDEC. Tabulate the measurements as shown below.

Triangle Sides Angles Ratios

ABC AB =... BC =... CA =.....=A ..=B ..=C

DEC DE =... EC =... CD =......D = ..=E ..=C

A

D C

B

P

L

A

E

C

D

8 cm

10 cm

2cm

B

..=

XYPQ

..=

YZQR

..=

ZXRP

..=

DCAC

..=

ECBC

..=

DEAB

198

Observe the relation between the angles and ratio of Corresponding sides of bothtriangles.

You find that (i) the angles of ABC are equal to Corresponding angles of DEC.

and (ii) CDCA

ECBC

DEAB

==

Theorem.1

If two triangles are equiangular, then their corresponding sides areproportional.

Data : In triangles ABC and DEFEDFBAC =DEFABC =

and DFEACB =

To prove : DFAC

EFBC

DEAB

==

Construction : Mark X on AB and Y on AC, such that AX = DE and AY = DF.Join XY.

Proof : Statement ReasonsIn AXY and DEF, DA = DataAX = DE, AY = DF Construction

DEFAXY SAS Postulate XY = EF and DEFAXY = Congruent triangle property

ABCDEFAXY == Data

199

i.e, ABCAXY = XY || BC ... (i) AXY and ABC are corresponding angles.

XYBC

AYAC

AXAB

== Basic proportionality theorem and corollary.

i.e EFBC

DFAC

DEAB

== From Construction and (i)

Converse of Theorem 1

Activity :Construct ABC in which AB = 2 cms, BC = 4 cms. and CA = 3 cms. Construct

another triangle PQR with PQ = 4 cms, QR = 4 cms, and PR = 6 cms. Measurethe angles of the triangles. Tabulate the angles measured and the ratios as shown.

ABC..=A ..=B ..=C

PQR..=P ..=Q ..=R

Observe the table and write your Conclusion.If the Corresponding sides of two triangles are proportional, then thetriangles are equiangular.

Worked Examples :1) In the figure PQR ||| NMR. Identify the

corresponding vertices, corresponding sides andtheir ratios.Solution :

PQR || NMR.NP PQ NMMQ QR MRRR PR NR

Ratios NRPR

MRQR

NMPQ

==

.=

PQAB

.=QRBC

..=

PRAC

200

2) In ABC, D and E are the midpoints on AB and AC such that DE || BC,

show that AEAC

DEBC

ADAB

==

Solution :

In ABC, DE || BC ABCADE =

ACBAED = (corresponding angles) DACBAC = (common angle)

ABC and ADE are equiangular.

Hence AEAC

DEBC

ADAB

==

3) In the adjoining figure AD = 5.6cmsAB = 8.4 cms, AE = 3.8 cms. andAC = 5.7 cms. Show that DE || BC.Solution :

AD = 5.6 cms (given)BD = 8.4 - 5.6 = 2.8 cms (From figure)AE = 3.8 cms (given)EC = 5.7 3.8 = 1.9 cms (from figure)

12

8.26.5

BDAD

==

12

9.18.3

ECAE

==

Hence ECAE

BDAD

=

DE || BC. (B.P.T.)

201

4) In the parallelogram ABRQ, the diagonals QB and segment AF intersect at H, asshown in the figure. Prove that, AH.HB = HQ.FH.Solution :

In the parallelogram ABRQ,AQ || BR and AB || QR.

Now in AHQ and FHB.HBFAQH = (AQ || BF)HFBQAH = (Alternate angles)FBHAHQ = (Vertically opposite angles)

AHQ and FBH are equiangular

HBHQ

FHAH

= i.e., AH : HB = HQ. FH

5) If one of the diagonals of a trapezium divides the other in the ratio 2:1. Provethat one of the parallel sides is twice the other.Solution :Let ABCD be a trapezium in which AB || CD and AE = 2 CE.In ABE and CDE, you have

EBACDE =EABDCE =BEADEC = (vertically opposite angles)

ABE and CDE are equiangular.

HenceCEAE

CDAB

=

CECE2

CDAB

= ( AE = 2CE)

21

CDAB

=

Hence AB = 2CD

alternate angles}

202

6) A man whose height is 1.5mts standing 8mts. from a lamp post, observes that hisshadow cast by the light is 2m. in length. How high is the lamp above the ground?Solution :Let the lamp post be LM. and the height of the man = OP and length of theshadow = ON.In LMN and PON

090PONLMN ==NPOPLM = (corresponding angles)PNOLNM = (common angle)

LMN and PON are equiangular

Hence ONMN

POLM

=

210

5.1LM

=

LM = 25.1x10

LM = 7.5 mts

The lamp is 7.5 mtrs from the ground

Exercise : 8.21) Construct triangle ABC in which AB = 3 cms, BC = 3.5 cms and CA = 5 cms.

Construct DEF similar to ABC, having 10cms as its longest side.2) Select the sets of numbers in the following, which can form similar triangles.

(i) 3, 4, 6, (ii) 9, 12, 18 (iii) 8, 6, 12 (iv) 3, 4, 9 (v) 2, 41/2, 4.3) Study the following figures and find out in each case whether the triangles are similar.

Give reason.1) 2)

Fig 1 Fig 2

203

Fig 3Fig 4

3) 4)

4) In the figure, Write the(i) Corresponding vertices and(ii) Corresponding sides of the APQand ABC. Hence show that AP.AB = AQ.AC.

5) ABC is right angled at B and D is any pointon AB. DE is perpendicular to AC. If AD = 4cmsAB = 16 cms, AC = 24 cms find AE.

6) In the trapezium ABCD, AD || BC and the diagonals

intersect at O. Prove that OAOD

OCOB

=

7) A vertical pole of 10 mts. casts a shadow of 8mts. at certain time of the day.What will be the length of the shadow cast by a tower standing next to the pole,if its height is 110 mts?

8) Prove that two Isosceles triangles having an angle of one equal to the correspondingangle of the other are similar.

9) Construct two similar triangles, taking the measurements of your choice. Comparethe perimeters of these two triangles and write your conclusion.

10) Triangle ABC has sides of length 5, 6 and 7 units, while triangle PQR has a perimeterof 360 units when will ABC similar to PQR? and hence find the sides of thetriangle PQR.

204

6. Areas of similar triangles.

Activity : 1Construct ABC in which AB = 2 cms, BC = 4 cms and CA = 3 cms. Draw

AD BC, find the area of the ABC.Construct a similar triangle PQR in which PQ = 4 cms, QR = 8 cms and

RP = 6 cms. Draw PSQR, find the area of PQR.

Tabulate the measurements as shown below:

Triangles Area in Square of Ratio Sq. units corresponding side

PQR = sq. units QR2 =ABC = sq. units BC2 =

Activity : 2Construct two similar triangles taking the measurements of your choice. Tabulate

the measurements as shown above.What relation do you find between the areas of the triangles and the squares of

the corresponding sides?

Theorem 2The areas of similar triangles are proportional to the squares of the correspondingsides.

Data : Let ABC, DEF be similar triangles, in which BC and EF arecorresponding sides.

=

ABCPQR

D

E M F

=2

2

BCQR

205

To prove : 22

EFBC

DEFofAreaABCofArea

=

Construction : Draw AL BC and DM EF.

Proof : Statement Reason

DM.EF21

AL.BC21

DEFofAreaABCofArea

=

Area of triangle is 21

base x height

DMAL

xEFBC

= ....(i)

In ALM and DME DEMABL = Data DMEALB = Construction EDMBAL =

ALB ||| DME Triangles are equiangular.

DEAB

DMAL

= Corresponding sides of similar triangles

EFBC

DEAB

DMAL

== From the similar ABC and DEF.

i.e. EFBC

DMAL

= ........ (ii)

EFBC

.

EFBC

DEFofAreaABCofArea

=

Substituting (ii) in (i)

2

2EFBC

DEFofAreaABCofArea

=

206

Activity :Draw similar triangles ABC and PQR. Draw BXAC and Draw QY PR. Prove

that, 2

2PRAC

PQRofAreaABCofArea

=

Worked Examples :

1) ABC ||| DEF. 55.2

EFBC

= If the area of ABC = 120 sq.cms. find the area

of DEF.

Solution : 22

EFBC

DEFABC

=

(theorem)

2

2

5)5.2(

DEFABC

=

2525.6

DEF120

=

25.625x120DEF = Area of DEF = 480 sq. cms

2) ABC ||| PQR. Area of ABC is 36 cm2 and area of PQR is 25 cm2.If QR = 6 cms, find the length of the side of ABC corresponding to QR.Solution : Side corresponding to QR in ABC is BC.

2

2QRBC

PQRofAreaABCofArea

=

(Theorem 2)

2

2

6BC

2536

=

BC2 = 2

256x36

BC = 2536x36

=

536

56x6

= ; BC = 7.2 cms

207

A

D

B

C

X

X

3) A trapezium ABCD has its sides AB || CD and its diagonals intersect at O. Ifside AB is twice CD, find the ratio of the triangle AOB to the triangle COD.Solution : Let ABCD be a trapezium, in which AB || CD and AB = 2CD.In AOB and COD

CODAOB = vertically opposite angles

OCDOAB = (AB || CD alternateangles)

ODCOBA = (AB || CD alternate angles) Triangles are equiangular and hence AOB ||| COD.

2

2

CDAB

CODAOB

=

(theorem)

2

2CD)CD2(

CODAOB

=

= 2

2

CDCD4

14

CODAOB

=

AOB : COD = 4 : 1

4) ABC is a right angled triangle with A = 900. ADBC. Show that 22

ACAB

ACDABD

=

Solution : In ABC, BAC = 900(data)

090ADBADC == ACD90ABC 0 =

i.e ACD90ABD 0 = ............ (i)

ABC and ABD are one and the sameIn ACD, ADC = 900

CAD = 900 ACD ........... (ii)In ABD and ACD

090ADCADB == (data)CADABD = [From (i) and (ii)]

ABD and ACD are equiangular and hence similar

2

2

ACAB

ACDABD

=

(Theorm - 2)

208

5) If the areas of two similar triangles are equal, prove that they are congruent.Solution:Let ABC ||| PQRArea of ABC = Area of PQRABC ||| PQR (data)

2

2

2

2

2

2

PRAC

QRBC

PQAB

PQRofAreaABCofArea

===

........... (i)

1PQRofAreaABCofArea

=

(Area of ABC = Area of PQR)

1PRAC

QRBC

PQAB

2

2

2

2

2

2

=== From (i)

i.e. AB2 = PQ2, BC2 = QR2 and AC2 = PR2 AB = PQ, BC = QR and AC = PR ABC PQR [SSS congruence]

Exercise : 8.31) Two corresponding sides of similar triangles are 3.6cms, and 2.4cms, respectively.

If the area of the bigger triangle is 45 sq. cms. Find the area of the smaller triangle.

2) In ABC, D and E are the mid points of AB and AC, If the area ofABC = 60 sq.cms. Show that the area of ADE = 15 sq.cms.

3) Two similar triangles have areas 392.sq.cm and 200 sq.cms., respectively; find theratio of any pair of corresponding sides.

4) The area of the triangle ABC is 25.6 sq.cms XY is drawn parallel to BC andit divides AB in the ratio 5:3. Find the area of the triangle AXY.

5) In ABC, DEAC and CFAB. BE and CF intersect at O. Show that

2

2

CEBF

COEBOF

=

6) In the figure BC || DE, area of ABC = 25sq.cms, area of trapezium BCED = 24 sq.cmsand DE = 14cms. Calculate the length of BC.

209

7) XY is drawn parallel to BC, the base of ABC. If AXY : TrapeziumXBCY = 4:5. Show that, AX : XB = 2:1.

8) Prove that the areas of similar triangles have the same ratio as the squares ofcorresponding altitudes.

9) Prove that the areas of similar triangles have the same ratio as the squares ofcorresponding medians.

10) Prove that the areas of similar triangles have the same ratio as the squares of theradii of their circum circles.

RIGHTANGLED TRIANGLE

Contribution of Indian mathematicians.The Geometry of the Vedic period in India originated with the construction of

various kinds of altars and fire places (Homa Kundas) for performing Vedic rites. Thenecessary measurements for constructing altars were done with the help of a rope calledsulba. The sulbasutras composed during the period 800 BC-500 B.C, contain a wealthof information regarding the knowledge of Vedic seers.

The Baudhayana Sulbasutras, contain a clear statement about a right angled triangle,which states that, The diagonal of the rectangle produces both areas which its lengthand breadth produce separately. It is called Baudhayana theorem.

It is worthwhile mentioning the names of Apastamba and Katyayana, who gavethe above theorem in identical terms. In Taittiriya Samhita (2000 B.C), an instanceof the kind 392 = 362 + 152 is mentioned.

Bhaskaracharya I (1114 A.D) has given a dissection proof using similar triangles.Pythagoras introduced formal definitions into geometry, stated the above theorem

as,

In a right angled triangle, the square on the hypotenuse in equalto the sum of the squares on the remaining sidesThis is universally accepted as Pythagoras theorem.

1. Pythagoras Theorem :

Activity : 1* Construct a right angled triangle ABC, right

angled at C on a card board as shown in thefigure.

210

* Draw Squares on AC, BC and on the hypotenuse AB as shown below.

* Mark the middle point of the square, drawn on the longest side containing the rightangle (ie., BC) as P (Middle point of the square on BC can be obtained by joiningthe diagonals).

* Through P Draw DE || AB.* and through P Draw FGAB* Mark the quadrilaterals formed as 1,2,3, and 4 as shown in the square on BC.

Check out whether they are congruent).* Mark the square on AC as 5.* Cut the quadrilaterals 1, 2, 3, 4, and 5 separately.* Arrange these quadrilaterals on the square drawn on the hypotenuse AB. (Shown

as shaded part).What is your conclusion about the sides of the right angled triangle?

Activity : 2Construct a right angle PQR, in which, PQ = 3cms, QR = 4cms, and Q = 900.

Measure side PR. Tabulate the squares of the lengths of the sides as shown below.

PQ2....... QR2....... PR2....... PQ2 + QR2 = .......

Observe the table and write your conclusion?The above activities point out that, the square on the hypotenuse is equal to the

sum of the squares on the other two sides.

A

BCD

G

FE

P

5

54 2

3

1

1

2

3

4

211

Theorem : 3 (Pythagoras theorem)In a right angled triangle, the Square on the hypotenuse is equal to the

sum of the squares on the other two sides.Data :

In ABC, BAC = 900

To prove :BC2 = AB2 + AC2

Construction : Draw AD BC.Proof :

Statement ReasonIn ABC and DBA

090BDABAC == Data and construction.

ABDABC = Common

ABC ||| DBA triangles are equiangular

BABC

DBAB

= Ratio of corresponding sides of similar triangles.

i.e., BC.DB = AB2 ...... (i)In ABC and DAC

090ADCBAC == data and construction.

ACDACB = Common

ABC ||| DAC triangles are equiangular.

DCAC

ACBC

= Ratio of corresponding sides of similar triangles.

i.e., BC.DC = AC2 ...... (ii)BC.DB + BC.DC = AB2 + AC2 adding (i) and (ii)BC (BD + DC) = AB2 + AC2BC . BC = AB2 + AC2 From fig BD + DC = BCie., BC2 = AB2 + AC2.

212

Converse of Pythagoras theoremActivity:

Construct ABC, given AB = 2.5cms, BC = 6cms and AC = 6.5 cms.

Measure the angles of ABC. Square the length of the sides. Tabulate themeasurements as shown below.

=A .... =B .... =C ..... AB2 = ...... BC2 = ..... AC2 = ..... AB2 + BC2

Observe the measurement you have recorded. What is your conclusion?

Activity:Repeat the above activity by constructing another triangle PQR, in which PQ =

6 cms, QR = 8cms, and RP = 10cmsFrom the activities and your conclusion, can you write the converse of pythagoras

theorem?

Converse of Pythagoras theorem.If the square on one side of a triangle is equal to the sum of the squares

on the other two sides, then those two sides contain a right angle.

2. Pythagorean tripletsConsider the triplet of natural numbers 3, 4, 5. The triangle whose sides are 3,

4, and 5 units is a right angled triangle, since 52 = 42 + 32. Another such triplet is 5,12, and 13, since 132 = 122 + 52. Such triplets of natural numbers are called Pythagoreantriplets. Write some more examples for Pythagorean triplets.

The sides of a right angled triangle have a special relationship amongthem. This relation is widely used in many branches of mathematics,such as mensuration and trigonometry.

Worked Examples :1) In ABC, B = 900, AB = 5cms

BC = 12cms find ACSolution : AC2 = AB2 + BC2

= 52 + 122

= 25 + 144 = 169 Sq.cms

AC = 169 =13 cms

213

2) In a right angled ABC, B = 900, AC = 17cm, AB = 8cms find BC.In ABC, B = 900 AC2 = AB2 + BC2 172 = 82 + BC2

BC2 = 172 82 = 289 64 = 225 sq.cms

BC = 225 =15 cms

3) The sides of a triangle are 7, 24 and 25 units. Prove that it is a right angled triangle.Name the right angle.Solution :Let PQR be a triangleand PQ = 7 units, QR = 24 units and RP = 25 units.Now PQ2 = 72 = 49 Sq. units

QR2 = 242 = 576 Sq. unitsRP2 = 252 = 625 Sq. units (on squaring)

PQ2 + QR2 = 49 + 576 = 625 Sq. unitsPQ2 + QR2 = RP2. unitsTriangle is a right angled triangle (Converse of Py. Theorem. )

Q is a right angle (RP is the longest side.)4) A ladder whose foot is 6 mts. from the wall in front of a building, reached a

windowsill 8 m above the ground. What is the length of the ladder?Solution :Let the height of the window sill be AB = 8 mts. Distancebetween the house and the foot of the ladder = BC = 6 mts.Let AC be the ladder.Then ABC = 900

AC2 = AB2 + BC2 Pythagoras theorem= 82 + 62= 64 + 36

AC2 = 100 AC = 100 AC = 10. The length of the ladder is 10 mts

P

Q R

257

24

214

5) An insect 8m away from the foot of a lamp post which is 6m tall, crawls towardsit. After moving through a distance, its distance from the top of the lamp post isequal to the distance it has moved. How far is the insect away from the foot ofthe Lamp post? [Bhaskaracharyas Leelavathi]Solution :Distance between the insect and the foot of theLamp post = BD = 8mts. The height of theLamp post = AB = 6mts.After moving a distance. Let the insect be at C,Let AC = CD = x mts. BC = (8 x) mts.In ABC, B = 900. Pythagoras theorem

AC2 = AB2 + BC2

x2 = 62 + (8 x)2 x2 = 36 + 64 16x + x2 (Transposing) 16x = 100

x = 16100

= 6.25 (Simplifying)

BC = 8 x BC = (8 6.25)

= 1.75 mts.The insect is 1.75 mts. away from the foot of the lamp post.

Exercise : 8.41) The sides of a right angled triangle are 5 cms and 12cms, find the hypotenuse.2) Find the length of the diagonal of a square of side 12 mts.3) The length of the diagonal of a rectangular playground is 125 mts and the length

of one side is 75 mts. Find the length of other side.4) Given below are the sides of a triangle. In which cases are the triangles right angled?

(i) 8, 15, 17 (ii) 9, 10, 14, (iii) 7, 24, 25.

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5) A ladder 5m. long rests against a wall at a height of 4.8m. from the ground. Calculatethe distance of the foot of the ladder from the wall.

6) Two poles of height 6m and 11m stand vertically on the ground such that the distancebetween them is 12m. find the distance between their tops.

7) A man walks 8 km due North, then 5 km to East and 4 km to North. How faris he from the starting point?

8) A tree 32 m tall broke due to a gale and its top fell at a distance of 16m. fromits foot. At what height above the ground did the tree break?

9) In an equilateral triangle ABC, ANBC. Prove that AN2 = 3BN2.10) In ABC, AB = AC and BDAC prove that BD2 + CD2 = 2AC.CD11) AD is the altitude through A in the ABC and DB : CD = 3:1 Prove that

BC2 = 2 (AB2 AC2).12) A lotus is 20cms above the water surface in a pond and its stem is partly below

the water surface. As the wind blew, the stem is pushed aside so that the lotustouched the water at 40cms away from the original position of the stem. Originallyhow much of the stem was below the water surface? (From Leelavathi ofBhaskaracharya).

TOUCHING CIRCLESYou are already familiar with circles and their properties. Do the following activities

and find out more properties.

Activity : 11) Draw two congruent circles. Join their centres.2) Draw two intersecting circles with radii 4.5 cms and 3cms. Join their centres.3) Draw two circles with radii 5.4cms and 3.2cms such that,

(i) They touch each other externally(ii) They touch each other internally

Mark the point, where the circles touch each other. Join the centres of these circles,and the point of contact.

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PA B

Q

R

Write your observations.(i) regarding the line joining the centres of the circle.(ii) regarding the line joining the centres and the point of contact.Theorem 4.

If two circles touch each other, the point of contact and the centres ofthe circles are collinear.

Data : Two circles with centres A and B touch each other at the pointP, externally (in fig 1) and internally (in fig 2) respectively.

To prove : A, B, and P are collinear.Construction : Draw the common tangent RPQ at P. Join AP and BP.Proof : Statement Reasonfig (1)

090APQ = ........... (i) Radius through the point of contact isperpendicular to the tangent.

090BPQ = ........... (ii) Same as above0180BPQAPQ =+ Adding (i) and (ii)

APB is a straight line APQ and BPQ is a linear pair. A, B and P are collinearfig (2)AP PQ ............... (iii) Radius through the point of contact is

perpendicular to the tangent.

Fig (ii)Fig (i)

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BP PQ ............. (iv) (Same as above)AP and BP are both perpendicularto the same line PQ From (iii) and (iv)AP and BP lie on the same line.

ABP is a line.

A, B and P are collinear.

1. Distance between the centre of touching circlesActivity :

Draw two circles of radii 3.6 and 2.2 cms touching, (i) Externally (ii) Internally.Measure the distance between the centres in both cases. What is your observation?You find that,(1) If two circles touch each other externally, the distance between their centres is equal

to the sum of their radii. [d = R + r](2) If two circles touch each other internally, the distance between their centres is equal

to the difference of their radii [d = R r]

Note : Radii of the circles are denoted as R and r.

Worked Examples :1. Three circles of radii 3cms, 4 cms, and 5 cms, with centers A, B, and C respectively

touch externally as in the figure. Find the perimeter of the ABC.Solution :AB = AP + BP Externally touching circles. = 3 + 4 = 7 cmBC = BQ + QC Externally touching circles = 4 + 5 = 9 cmsCA = CR + RA = 5 + 3 = 8 cmsThe Perimeter of the ABC = AB + BC + CA

= 7 + 9 + 8 = 24 cms

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2) Three circles with centres A, B, and C, touch each other as shown in the figure.If the radii of these circles are 8 cm, 3 cm and 2 cms respectively, find the perimeterof the triangle ABC.

Solution :

In ABCAB = 8 BP (Internally touching circles) = 8 3 = 5 cms.BC = BQ + QC (Externally touching circles) = 3 + 2 = 5 cmsCA = 8 CR = 8 2 = 6 cms

Perimeter of ABC = AB + BC + CA = 5 + 5 + 6

= 16 cms

3) A straight line drawn through the point of contact of two circles whose centresare A and B intersects the circles at P and Q, respectively, show that AP andBQ are parallel.

Solution :Consider two circles with centres A and B,touching externally as shown in the figure.In AMP AM = AP (radii of the same circle) AMPAPM = (i)

BMQAMP = (ii) (Vertically opposite angles)

In BMQ,BM = QM, (Radii of the same circle)

BQMBMQ = (iii)APM = BQM From (i) (ii) and (iii)

AP || BQ ( APM and BQM are alternate angles)

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Exercise : 8.51) Draw two circles of radii 3.4 cms and 1.8 cms with their centres 5.2 cms apart.

How do these circles touch each other? and why?2) If the circles of radii 3.4 cms and 1.8 cms are drawn with their centres 1.6 cms

apart, How do they touch each other? Why?3) Draw triangle ABC, in which BC = 8cm, AC=7cms and AB = 6cms with A, B,

and C as centres. Draw circles of radii 2.5 cms, 3.5 cms and 4.5 cms respectivelyand show that these circles touch in pairs.

4) A red and a blue cardboard discs of radii 4.5 cms and 2 cms are fixed to a straightstring of length 10 cms; as shown in the figure. What is the radius of another disc,which touches the circular discs at P and Q.

5) Two circles of radii 8 cms and 5cms with their centers A and Brespectively touch externally asshown in the figure. Calculate thelength of direct common tangentPQ.

6) In the given figure AB = 8cms. M is the midpoint of AB.Semi circles are drawn on AB, AMand MB as diameters. A circle withcentre O touches all three semicircles as shown. Prove that the

radius of this circle is 311

cms.

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2. Properties of tangents drawn to a circle from an external point.You have already learnt the construction of tangents to a given circle, from an

external point. Now, do the following activities and know more about the propertiesof tangents.

Activity : 1Study the figure given below. Construct two tangents to a circle of radius 3cms,

from a point 5 cms away from the centre.* Measure the lengths of the tangents AP and AQ.* Measure the angles between the tangents, and

the line joining the centre (O) and the externalpoint (A). i.e. PAO and QAO .

* Measure the angles between the radius and theline joining the centre and the external point. i.e.,

POA and QOA .* Tabulate the measurements as shown below:

Length of Tangents Angle between the line joining the centre and external pointand tangents and radius

PA = .... QA = ..... PAO = .... QAO =... POA =... QOA =...Activity : 2

Draw a circle of radius 5 cms with centre O. Construct two tangents PA andPB from an external point P at a distance of 13 cms from O.* Measure, tangents PA, and PB.* Measure APO , BPO and AOP and BOP .* Record the measurement in the table as shown.

tangents AnglesPA = ........................ APO = ....... AOP = .......PB = ....................... BPO = ....... BOP = .......

From the activities performed what conclusion would you draw?You find that, (i) The tangents are equal(ii) The tangents make equal angles with the line joining the centre and the external

point.(iii) The angles between the radius and the line joining the centre and the external point

are equal.

P

Q

AMO

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Theorem : 5The tangents drawn to a circle from an external point are,(i) equal(ii) equally inclined to the line joining the external point and the centre(iii) subtend equal angles at the centre.

Data:

O is the centre of the circle. PA and PB are the tangents drawn from an externalpoint P. OA, OB and OP are joined.To prove.

(i) PA = PB(ii) APO = BPO(iii) AOP = BOP

Proof :

Statement Reason

In AOP and BOP, OA = OB Radii of the same circle.

090OBPOAP == Radius drawn at the point of contact isperpendicular to the tangent.

OP = OP Common

AOP BOP RHS PostulateHence (i) PA = PB (ii) BPOAPO = Properties of congruent triangles. (iii) BOPAOP =Worked Examples :1) A is an external point 26cms, away from the centre of the circle and the length

of the tangent drawn from A to the circle is 24 cms. Find the radius of the circle.Solution :Distance between the centre and externalpoint = OA = 26 cms.length of the tangent drawn from A to thecircle = AP = 24cms.

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Join OP.OPAP (radius drawn at the point of contactto a tangent is perpendicular.

OPA = 900

OP2 = OA2 PA2

= 262 242

= (26 + 24) (26 24) = 50 x 2 = 100

OP = 100 OP = 10 cmsThe radius of the circle is 10 cms

2) AC, CE, and EH are tangents drawn to the circle,at B, D and F respectively. Prove CB + EF = CE.CD = CB. Tangents drawn from external point C.DE = EF Tangents from the external point E.CD + DE = CB + EF By addingCB + EF = CE (CD + DE = CE)

3) In the figure AP, AX and AY are the tangents drawn to the circles, prove thatAY = AX.Solution :AP = AX....... (1)Tangents drawn from

the external point Ato the circle withcentre C.

AP = AY....... (2) Tangents drawn fromthe external point Ato the Circle withcentre D.

AY = AX From (1) and (2)

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4) Find the length of the tangent drawn to a circle of radius 8 cms., from a pointwhich is at a distance of 10cms, from the centre of the circle.Solution :OAAP Radius at the point of contact and the tangent are perpendicular.

OAP = 900

OP2 = OA2 + AP2 (Pythagoras theorem) AP2 = OP2 OA2

= 1002 82

= 100 64 = 36

AP = 36 = 6 cms

The length of the tangent = 6 cms.

5) In the figure APB is a tangent at P to the circle with centre QPB = 600,find POQSolution : APB is a tangent OPB = 900

QPB = 600 OPQ = 900 600 = 300

In OPQ OPQ = OQP = 300 (OP = OQ radii of the same circle) POQ = 1800 (300 + 300)

POQ = 1200

6) O is the centre of the circle. AB, BC and CA touch the circle at L, M and Nrespectively. If B = 700, C = 600, Calculate LOM , LON and MONSolution :

OLB = 900 AB is a tangent and OL isperpendicular

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A BP

O

Also OMB = 900 BC is a tangent and OMis perpendicular

In the quadrilateral OLBM,

LOM = 1800 700

LOM = 1100

Similarly, MON = 1800 600

MON = 1200

LON = 3600 (1100 + 1200) ( sum of angles at O is 3600) = 3600 2300

LON = 1300

7) Two concentric circles are of radii 13 cms and 5 cms. Find the length of the chordof the outer circle which touches the inner circle.Solution :Let O be the centre of the concentric circles.Chord AB touches the inner circle at P. OA,OB and OP are joined.In OAP OPAB.

OA2 = AP2 + OP2 (Pythagoras theorem) i.e. AP2 = OA2 OP2

= 132 52

= 169 25 = 144 AP = 144 = 12 cms

Similary in OBP BP2 = OB2 OP2

= 132 52 = 144 BP = 144 = 12 cms

Chord AB = AP + BP = 12 +12 = 24 cms

The length of the chord AB = 24 cms.

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8) In the figure XY and PC are common tangents to two touching circles. Prove thatXPY = 900

Solution :CX = CP (Tangents from C) == CPYPXC x0 ......... (i)Similarly CY = CP

== CPXPYC y0 ..... (2)In PXY,

=++ XPYPYXPXY 1800

x0 + y0 + (x0 + y0) = 1800x0 + y0 + x0 + y0 = 1800

2x0 + 2y0 = 1800

x + y = 2180

= 900

i.e. XPY = 900

Exercise : 8.6

1) From the figure, name(a) Tangents drawn from A to circle C1.(b) Tangents drawn from A to circle C2.(c) Tangents drawn from B to circles C1

and C2(d) Sets of equal tangents.

2) From the figure find the length of the tangentCD given AP = 3cm, and PC = 8cm.

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3) In the figure PA and PB are tangents,AOB = 1400 what is the measure of APB ?

4) In the figure PAO = 400. What measure ofthe POA makes AP a tangent?

5) AB is a tangent to a circle with centre O and A is the point of contact. IfOBA = 450. Prove that AB = OA.

6) Tangents PQ and PR are drawn to the circle from an external point P. IfPQ = 9 cms, and PQR = 600. Find the length of the chord QR.

7) In the figure PQ and PR are tangents to thecircle with centre O. If QPR = 900, Showthat PQOR is a square.

8) In two concentric circles of radii 6cms and 10cms with centre O. OP is theradius of the smaller circle, OPAB which cuts the outer circle at A and B. Findthe length of AB.

9) In the figure AT and BT are the tangents to acircle with centre O. Another tangent PQ isdrawn such that TP = TQ. Show that TAB ||| TPQ

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10) ABC is an Isosceles triangle in whichAB = AC and sides of the triangle touchthe circle at P, Q and R. Prove that Qis the mid point of the base BC.

11) The sides of a quadrilateral ABCD are tangents to the circle with centre O. IfAB = 8 cms, and CD = 5cms find AD + BC.

12) TP and TQ are tangents drawn to a circle withcentre O. Show that PTQ = 2 OPQ .

ThalesThales (640-546 B.C), a Greek mathematician,was the first who initiated and formulated theTheoretical Study of Geometry to make astronomya more exact science.

PythagorasPythagoras (540 B.C.) a Greek mathematician,was a pupil of Thales. He proved the well-knownand tremendously useful theorem credited afterhis name : The Theorem of Pythagoras.