But the Following Shapes Are Not Similar

Two figures having the same shape not necessarily the same size are called similar figures. Similar figures have the shape but not the same size as circles of different radii and square of different sides.

But the following shapes are not similar.

Use of similarity

The heights of mountains (e.g. Mount Everest) or distances of some long distant objects (eg. Moon) can not be measured directly with the help of a measuring tape. In fact all these heights and distances are found out using the idea of indirect measurements, which is based on the principle of similarity of figures.

Congruent figures

Two figures are said to be congruent, if they have the same shape and same size.

Since all of them do not have the same radius. They are not congruent to each other. Note that some are congruent and some are not, but all of the have the same shape. So they all are, what we call, Similar.

Two similar figures have the same shape but not necessarily the same size. Therefore, all circles are similar.

From the above, we can say that all congruent figures are similar but the similar figures need not be congruent.

Not similar: a circle and a square cannot be similar. Similarly, a triangle and a square cannot be similar. This is evident just by looking at figures.

Definition:

Let’s take two quadrilaterals ABCD and PQRS

These figures appear to be similar but cannot be certain about it. Therefore, we must have some definition of similarity. This will decide whether the figures are similar or not.

Essence of the similarity of two figures.

1.  Let us is look at the photographs below.

We will are once say that they are the same photographs of the same person but are in different sizes. We would say, therefore, that the three photographs are similar.

2.  Look at the two photographs of the same person at the age of 10 years and the other at the age of 40 years. The photographs are of the same size but not of the same shape. So they are not similar.

3.  Let us see what a photographer does when she prints photographs of different sizes from the same negative. We have heard about the stamp size, passport size and postcard size photographs.

She generally takes a photograph on a small size film, say of 35mm and then enlarges it into bigger size, say 45mm(or 55mm), its corresponding line segment in the bigger photograph will be 45/35 (or 55/35) of that of the line segment. This really every line segment of the smaller photograph is enlarged into the ratio of 35:45 (or 55:35).further, if we consider inclinations (or angles) are always equal.

This is the essence of similarity of two figures and in particular of two polygons.

Similar figures can be obtained by magnifying or shrinking the original figure.

Rule:

Two polygons of the same number of sides are similar:

1.  Their corresponding sides are in the same ratio.

2.  Their corresponding angles are equal.

Same ratio of the corresponding sides is referred are as SCALE FACTOR. (or the Representative Fraction) for the polygons.

Scale factor =new measurement

old measurement

Old measurement x SF = new measurement

- Scale factor more than 1 => shape gets

bigger

- Scale factor less than 1 => shape gets

smaller

1.We can easily say that the polygons below are similar.

Remark: we can verify that if one polygon is similar to another polygon and this second polygon is similar to a third polygon, Then the first polygon is similar to the third one.

2.  We may note that in the two quadrilaterals (a square and a rectangle) shown below, corresponding angles are equal, but their corresponding sides are not in the same ratio.

Similarly, we may note that in that in the two quadrilaterals (a square and a rhombus) shown below, corresponding sides are in the same ratio, but their corresponding angles are not equal. So again, the two polygons are not similar.

A B D

A C

D C B

SIMILARITY OF TRIANGLES.

Two triangles are said to be similar, if their corresponding angles are equal and corresponding sides are proportional.

If ∆s ABC and DEF are similar, then ÐA= ÐD, ÐB= ÐE and ÐC=ÐF

And AB = BC = AC

DE EF DF

A D

B C E F

THREE SIMILARITY POSTUALATERS FOR TRAINGLES.

1)  SAS-postulate: if two triangles have a pair of corresponding angles equal and the sides including them proportional, then the triangles are similar.

if in s ABC and DEF

ÐA=ÐD

And AB = BC

DE EF

THEN ABC ∆DEF

AAA-postulate: (Angle-Angle-Angle) This is when, two triangles have the pairs of corresponding angles equal, the triangles are similar.

ÐA= ÐD AND ÐB= ÐE

Then ∆ABC~ ∆DEF

Proof: take two triangles ABC and DEF such that ÐA = ÐD, ÐB = ÐE and ÐC = ÐF

Cut DP=AB and DQ= AC and join PQ.

IN ∆’S ABC and DPQ,

AB=DP

AC=DQ

ÐBAC =ÐPDQ

∆ABC @ ∆DPQ

│ SAS congruence criterion.

This gives

ÐB = ÐP

ÐB= ÐE

ÐP = ÐE

But these form a pair of corresponding angles.

PQ║EF

DP = DQ

PE QF

By basic proportionality Theorem

PE = QF TAKING RECIPROCALS

DP DQ

PE +1 = QF +1

DP DQ

PE + DP = QF + DQ

DP DQ

DE= DF

DP DQ

DE= DF

AB AC

AB = AC

DE DF

SIMILARLY,

AB = BC and so AB = BC = AC

DE EF DE EF DF

THEREFORE ∆ABC ~ ∆DEF

SSS-postulate: if two triangles have their three pairs of corresponding sides proportional, the triangles are similar.

IF IN ∆S ABC and DEF

AB = BC = AC

DE EF DF

THEN ∆ABC~∆ DEF

DIFFERNCES BETWEEN CONGURANT AND SIMILAR TRIANGLES.

Congruent triangles are equal in all respects i.e. their angles are equal, their sides are equal and their areas are equal. Congruent triangles are always similar.

But similar triangles need not be congruent.

To establish the similarity of two triangles, it’s sufficient to satisfy one condition.

1.  Their corresponding angles are equal.

2.  Their corresponding sides are proportional

If the corresponding angles are equal, the triangles are equiangular

Truth relating two equiangular triangles.

A famous Greek mathematician Thales gave an important truth relating two equiangular triangles which is as follows:

The ratio of any two corresponding sides in any two equiangular triangles is always equal.

It is believed that he had used the Basic proportionality Theorem (now known as Thales Theorem) for the same.

BASIC PROPOTIONALITY THEOREM.

If a line is drawn parallel to one side of the triangle to intersect the other two sides is distinpoints, the other two sides are divided in the same ratio.

Given: A triangle ABC in which a line parallel to side BC intersects the other two sides AB and AC at D and E respectively.

To prove: AD = AE

DB EC

CONSTRUCTION: Join BE and CD and then draw DM ^ AC and EN ^ AB

Proof: Area of ∆ADE= 1 Base x height

2

=1AD X EN

2

ar(ADE)= 1 DB x EN

2

ar( ADE)=1EC x DM

2

Therefore, ar(ADE)

ar(BDE)

1AD x EN

2 = AD

1AD x EN DB

2

CRITERIA FOR SIMILARITY OF TRIANGLES

1.  SYMBOL OF SIMILARITY

Here we can see that A corresponds to E and C corresponds to F. Symbolically, we write similarity of the two triangles as ‘∆ABC~ ∆DEF’ and read it ‘as triangle ABC is similar to Triangle DEF’. So the symbol ~ stands for ‘is similar to’.

Minimum essential requirements of the two triangles.

Now, we will examine that for checking the similarity of two triangles, say ABC and DEF, whether we should always look for all the equality relations for the corresponding angles.

ÐA= ÐD, ÐB =ÐE, ÐC=ÐF

AB = BC = CA

DE EF FE

We may recall that, we have some criteria for congruency of two triangles

involving only two pairs of corresponding parts or elements of the two triangles. Here also, we shall try to arrive certain criteria for the similarity of two triangles involving relationship between less numbers of pairs of corresponding part of the two triangles, instead of all six pairs of corresponding parts.

EXAMPLE: draw the line segments BC and EF of two different lengths, 3cm and 5 cm respectively. Then, at the points B and C respectively, construct angles PBC and QCB of some must measure, say, 60º and 40°. Also at the points E and F, construct angles REF and SFE of 60º and 40º respectively.

Let rays BP and CQ intersect each other at ÐA and rays ÐER and ÐFS intersect each other at D. In the two triangles ABC and DEF, we see that ÐB= ÐE, ÐC=ÐF AND ÐA =ÐD

That is corresponding angles of these two triangles are equal. Regarding CA are equal to 0.6.

FD

BA= 3 = 0.6

EF 5

Thus AB = BC = CA

DE EF DF

ILLUSTRATIVE EXAMPLES

Example 1. State which pairs of triangles in figure are similar. Write down the similarity criterion used by you for answering the questions and also write pairs of similar triangles in symbolic form.

i.  iv.

ii. 

iii. 

Solution. (i) In ∆ABC and PQR

ÐA = ÐP ∆ABC ~ ∆PQR

ÐB = ÐQ AAA similarity criterion.

ÐC =ÐR

ii. In ∆ABC and ∆QPR

AB = BC = CA

QR RP PQ

iii. in ∆MNL and ∆QPR

ML = MN

QR QP

ÐNML =ÐPQR

And

∆MNL ~ ∆PQR

SAS similarity criterion

iv.in ∆DEF and ∆PQR

ÐD = ÐP (=70º)

ÐE = ÐQ (=80º)

ÐF = ÐR (=30º)

Example 2. S and T are points on PR and QR of ∆PQR such that ÐP=ÐRTS.

Show that ∆ RPQ ~ ∆RTS.

Solution.

S and T are points on the sides PR and QR of ∆PQR such that ÐP =ÐRTS.

To prove: ∆RPQ ~ ∆RTS

Proof: in ∆RPQ and RTS

ÐRPQ = ÐRTS ∆RPQ ~ ∆RTS

ÐQRP = ÐSRT

Two chords AB and CD intersect each other at the point P. prove that

1.  ∆APC ~ ∆DPB

2.  AP. PB = CP. DP.

Solution: Two chords AB and CD intersect each other at the point P.

To prove: (i) ∆APC ~ ∆DPB

(ii) AP. PB = CP. DP

Proof: (i) ∆APC and ∆DPB

ÐAPC = ÐDPB

│Vert. opp. Ðs

ÐCDP = ÐBDP

│Angles in the same segment

\ ∆APC ~ ∆DPB

│AA similarity criterion

(ii) ∆APC ~ ∆DPB │ Proved above in (1)

\ AP = CP

DP BP

Corresponding sides of two similar triangles are proportional.

AP.BP = CP.DP

Þ AP. PB = CP.DP

EXAMPLE: Two chords AB and GD of a circle intersect each other at the point P (when produced ) outside the circle.

Prove that

1.  ∆PAC ~ ∆PDB

2.  PA.PB = PC.PD

Solution: Two chords AB and CD of a circle intersect each other at the point P (when produced) out the circle.

To Prove: 1. ∆PAC ~ ∆PDB

2. PA. PB = PC. PD

Proof: (1). We know that in a cyclic quadrilaterals the exterior angle is equal to the interior opposite angle.

Therefore,

ÐPAC = ÐPDB ….(1)

And ÐPCA = ÐPCB ….(2)

In view of (1) and (2),

∆PAC ~ ∆PBD

│ AA similarity criterion

(2). ∆PAC ~ ∆PDB │Proved above in (1)

\ PA = PC

PD PB

│ Corresponding side of the similar triangles are proportional.

Þ PA. PB ~ PC. PD.

AREAS OF SIMLAR TRIANGLES

THEOREM: the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given: two triangles ABC and PQR such that ∆ABC ~ ∆PQR

To prove.

ar(ABC) = AB 2 = BC 2= CA 2

ar(PQR) PQ QR RP

ar ( ABC)=1 BC x AM

2

And

ar(PQR)=1 QR x PN

2

So,

ar (ABC) = 1 x BC x AM

2

ar (PQR) = 1 x QR x PN

2

= BC x AM

QR x PN

Now, in ∆ABM and ∆PQN,

ÐB = ÐQ (As ∆ABC - ∆ PQR)

ÐM = ÐN (Each is of 90º)

So, ∆ABM - ∆PQN

│ AA similarity criterion

Therefore, AM = AB

PN PQ ….(2)

│\Corresponding sides of two similar triangles are proportional

Also, ∆ABC - ∆PQR (Given)

So, AB = BC = CA

PQ QR RP …(3)

│. . Corresponding sides of two similar triangles are proportional

Therefore, ar(ABC) = AB x AM

ar (PQR) = PQ PN

[From (1) and (3)]

= AB x AB

PQ PQ [From (2)]

= AB 2

PQ

Now using (3), we get

ar (ABC) = AB 2 x BC 2 x CA 2

ar (PQR) PQ QR RP

ILLUSTRATIVE EXAMPLES

Example 1. Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm2 and 121 cm2.

Solution: ∆ABC ~ ∆PQR │ Given

\ ar (∆ABC) = BC 2

Ar (∆PQR) QR

│The ration of the areas of two similar triangles is equal to the square of the ratio of their

Corresponding sides.