DONE BY:

VIRALI DTTANI

11B

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 figuresTwo 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.

A

B C

D

P S

RQ

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 measurementOld measurement x SF = new measurement- Scale factor more than 1 => shape getsbigger- Scale factor less than 1 => shape getssmaller

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 A= D, B= E and C=FAnd 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

A=D 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.

A= D AND B= E

Then ∆ABC~ ∆DEF

AA

B C E

D

F

Proof: take two triangles ABC and DEF such that A = D, B = E and C = F

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

IN ∆’S ABC and DPQ,AB=DPAC=DQ BAC =PDQ

∆ABC ∆DPQ│ SAS congruence criterion.

This gives B = PB= EP = E

But these form a pair of corresponding angles.PQ║EF

DP = DQPE QF

By basic proportionality TheoremPE = QF TAKING RECIPROCALS DP DQ

PE +1 = QF +1 DP DQ

PE + DP = QF + DQ DP DQDE= DFDP DQ

DE= DFAB AC

AB = ACDE DF

A

B C

D

P

E F

Q

SIMILARLY,

AB = BC and so AB = BC = ACDE 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 DFTHEN ∆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,

A

B C E

D

F

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

40ºº60º

R

40ºº

P

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.

A= D, B =E, C=F

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 A and rays ER and FS intersect each other at D. In the two triangles ABC and DEF, we see that B= E, C=F AND A =D

A

B C E F

D

60º

Q

B C E F

SD

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

A = P ∆ABC ~ ∆PQR

B = Q AAA similarity criterion.

60º

80 404080

60º

3

2.5

2

6

4

5

6

10

7070 52.5

70 70

C =R

ii. In ∆ABC and ∆QPR

AB = BC = CA QR RP PQ

iii. in ∆MNL and ∆QPR ML = MN QR QP NML =PQRAnd∆MNL ~ ∆PQR

SAS similarity criterion

iv.in ∆DEF and ∆PQR D = P (=70º) E = Q (=80º) F = R (=30º)

Example 2. S and T are points on PR and QR of ∆PQR such that P=RTS.Show that ∆ RPQ ~ ∆RTS.

Solution.S and T are points on the sides PR and QR of ∆PQR such that P =RTS.

To prove: ∆RPQ ~ ∆RTS

Proof: in ∆RPQ and RTS

RPQ = RTS ∆RPQ ~ ∆RTS

QRP = SRT

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

1. ∆APC ~ ∆DPB2. 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

APC = DPB

│Vert. opp. s

CDP = BDP

│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

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.

P

D

BCC

A

Prove that1. ∆PAC ~ ∆PDB2. 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,

PAC = PDB ….(1)

And PCA = PCB ….(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. 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

B

D

CP

A

To prove.

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

ar(PQR) PQ QR RP

ar ( ABC)=1 BC x AM 2

Andar(PQR)=1 QR x PN

2So, ar (ABC) = 1 x BC x AM

2ar (PQR) = 1 x QR x PN

2= BC x AM QR x PN

Now, in ∆ABM and ∆PQN, B = Q (As ∆ABC - ∆ PQR) M = N (Each is of 90º)So, ∆ABM - ∆PQN │ AA similarity criterionTherefore, 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 proportionalTherefore, 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.

64 = BC 2

121 15.4 2 2

8 = BC

11 15.4 Taking square root on both sides BC = 8 x 15.4 11 BC = 11.2 cm.

Example 2. D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Find the ration of the areas ∆DEF and ∆ABC.

Solution : D, E and F are respectively the mid-points of side AB, BC and CA of ∆ABC.

= 1 and AB = BC = CA 2 BC DE EF FD

B . . The corresponding sides of two similar triangles Are proportional.

Form first two of (2),

AB = BC . . AM and DN are the medians.DE EF

= 2BM 2EN

= BN EN ….(3)

Also,ABM = DEF ….(4)

Example :Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution ; Give ABCD is a square whose one diagonal is AC. ∆APC and ∆BQC are two equilateral triangles described on the diagonal AC and side BC of the square ABCD.

To prove. ar (∆BQC) = 1 ar (∆APC) 2

Proof : . . ∆APC and ∆BQC are both equilateral triangles

∆APC ~ ∆BQC │ AAA similarity criterion ar(∆APC) = AC 2

ar(∆BQC) BC

The ratio of the areas of two similar triangles is

A

D C

Q

P

B

equal to the square of the ratio of their corresponding sides.

= AC2 BC2

= √ 2 BC 2 BC │ Diagonal = √2 side = 2

ar(∆BQC) =1 ar(∆APC). 2

Example 3: A line PQ is drawn parallel to the base BC of ∆ABC which meets sides AB and AC at points P and Q respectively. If

AP = 1 PB; find the value of : 3 Area of ∆ ABC Area of ∆ABC

(1) Area of ∆APQ (2) Area of trapezium PBCQ

Solution : AP=1 PB 3AP:PB= 1:3AP+PB = AB

AP= 1 AB 1+3

= 1 AB4

In ∆s APQ and ABC, we have

APQ= corresponding ABC

AQP = corresponding ACB

By AA-similarity

∆APQ ~ ∆ ABC the ratio of the corresponding of similar triangles

A

P

B C

Q

i. Area of ∆ABC = AB2 to the ratio of their corresponding sides.Area of ∆APQ AP2

Area of ∆ABC = AB 2 = 16Area of ∆APB 1AB2 1

16

Area of ∆APQ = AP 2 Area of ∆ABC AB2

1AB2

16 = 1AB2 16

Area of ∆APQ = 1 Area of ∆ABC-Area of ∆APQ 16-1

Area of ∆APQ = 1Area of trapezium PBCQ

THE WORLD OF 3D OBJECTS3D OBJECTS ARE THOSE OBCTS THAT HAS 3 DIFFERENT DIMENSIONS. Almost everything around us is 3d objects.

ARE 3D SHAPES EVER GEOMETRICALLY SIMILAR?This is one question that mathematicians have been working on, and finally they reached a point where it was proved that Yes, 3D shapes are similar.

SIMILARITIES OF 3D OBJECTSTwo figures having the same shape not necessarily the same size are similar figures. Similar figures have the SAME shape but not the same size as spheres of different radii and cubes and cuboids of different sides.

We say these types of object 3D objects because the have sides which are equal to each other. For example: a cube and a cuboid has got 6 sides but its length, width and height are repetitions of their opposite sides.

Same applies to a sphere because it has the repetitions of its radii.

Similar shapes can also be congruent. We know that all congruent figures are similar but not all similar figures are congruent.e.g

CONGURENCE OF TRIANGULAR PRISMS.

ABCD is a square and DEC is an equilateral triangle. Prove that AE=BE

DE= CE AD= BCEDA= ECB

Sides of an equilateral triangle.Sides of a square.

EDC= ECD and ADC = BCDEDC + ADC= ECD + BCD

EDA = ECBExample 2: In an isosceles triangle ABC, with AB= AC, the bisectors of B and C intersect each other at O. Join A to O. show that:

I. OB=OC

C

BA

D

E

AB= ACB=C angles of the opposite side of the triangle are always equal.1B=1C2 2OCB = OBC

Example 3: In triangles ABC and DBC on the same base BC such that AB=AC and DB=DC. Prove that ABD =ACD.

In ∆ABC AB= AC

ACB = ABC opposite angles in triangle are equal.In ∆DBC

DB= DCDCB= DBC

ACB - DCB= ABC-DBCACD=ABDABD=ACD.SURFACE AREAS OF SIMILAR OBJECTSTHEOREM: the ratio of the surface areas of two similar 3D objects is equal to the square of the ratio of their corresponding sides.

As mentioned before 3D objects have repetitive sides.

O

CB

A

D

CB

A

4cm

3cm8cm

8cm

6cm

AB

Lets first find out the surface areas first and then we will compare them and find out whether they are similar or not.

Surface area of a cuboid = S = 2lb + 2lh + 2hb

Surface area of cuboid A = 2 × (3×4) + 2 × (4×8) + 2 × (3×8) S = 2×12 + 2×32 + 2×24 S = 24 + 64 + 48 S = 136cm2

Surface area of cuboid A = 2 × (6×8) + 2 × (8×16) + 2 × (6×16) S = 2×48 + 2×128 + 2×96 S = 96 + 256 + 192

S=544cm2

Ok now we will find the scale factor of one of the Scale factor =new measurement old measurement = 136 =1 544 4

=1:4Therefore the scale factor (k) = 1:4Let check whether the cuboids above are similar.

Surface area of cuboid A = k2

Surface area of cuboid B

136cm 2 = 1 2

544cm2 4 1 = 1 2

4 4

A cube is a very simple form of cuboid which has edges that are all the same length. All the faces are therefore identical squares. Dice are good examples of cubes.

If l=3 If l=6

16cm

we got this formula of surface area because the opposite rectangles making the cuboid are equal. So we multiply the by 2

12cm

S = 6l2 S = 6l2S= 6(6) S=6(36)S= 36cm2 S=216 cm2

Scale factor =new measurement old measurement = 3 =1 6 2

=1:2Therefore the scale factor (k) = 1:2

Surface area of cube 1= k2

Surface area of cube 236 cm2 = 1 2

216 cm2 2 1= 1 2

4 2

A cone is any object that tapers to a point (or apex). So a pyramid is a type of cone, as is a similar object with a 5, 7, or even 9-sided base.

As for a prism, there are many different cones, so there are many different formulae for calculating the surface area. However, the formulae for the standard cone, with a circular base, and for a pyramid, with a square base, will be given here.

Try this out.

Surface area of a cone = S = Πrl + Πr2

Easy right!!

R=10cm R=5cm

6cm

k=2:1therefore the ratio of their areas must be 4:1

s of A= Πrl + Πr2

=377.1+314.2 =691.3 cm2

s of B= Πrl + Πr2

94.29 + 78.6 =172.9 cm2.

VOLUMES OF SIMILAR OBJECTS.

If the scale factor of a pair of 3-dimensional figures is a: b, then the ratio of their volumes is a3: b3

THEOREM: the ratio of the VOLUMES of two similar objects is equal to the cube of the ratio of their corresponding sides.

A sphere is an object shaped like a tennis ball. It looks circular when viewed from any direction. This is a very strange object, mathematically, because it is so complex while being extremely simple.

V = 4/3Πr3

R=2cm R=1cmA B

V of sphere A=4 Πr3

3 =4 x 22 x 23

3 7 =33.5cm3

V of sphere B=4 Πr3

3=4 x 22 x 13

3 7=4.2 cm3

Scale factor (k) =2:1Volume of sphere A= k3

Volume of sphere B33.5=23

4.27.99=8

R=10cm R=10cm R=5cm

V = hb v= hb 3 3= 12 x 15 =5 x 6 3 3 =120 =30

3 3

=4Ocm3 =10 cm3

Scale factor=10= 2:1 5

12cm 6cm

Volume of cone A = k3

Volume of cone B 40= 2 3

10 1 4=8

NOT SIMILAR!!!!

The volume of spare is equal to the volume of a cylinder whose height s 4cm If their radius are unknown, find their volume.

4Πr2 = 2 Πr2h

3

3 x 4r = 2 x 4 x 3 r = 64 3 6

Capacity In litters of a bucket 24cm in diameter.

V= 1 x 3.14 x r2 x h 3

12= 18+X 8 X12X=8(18+X)12X=144+8X4X=144X=36cm

Volume of larger cones=1 x 3.14 x 144 x 54 3

24cm

18cm 16cm

x

12cm

18+X

8cm

X

E

=8138.8 cm3

Volume of smaller cone=1 x 3.14 x 64 x 36 3 =2411.5 cm3

Volume of bucket= volume of larger cone – volume of smaller cone. = 8138.8 – 2411.5 =5727.3cm3

SUMMARY

Similarity of Triangles : Three PostulatesSAS – Postulate: If two triangles have a pair of corresponding angles equal and the sides including them proportional, than the triangles are similar.AA – Postulate: If two triangles have two pairs of corresponding angles equal, the triangles are similar.SSS-Postulate: If two triangles have their three pairs of corresponding sides proportional, the triangles are similar.

Relation between the Areas of Two Similar Triangles:

Basic Proportionality Theorem and Conversely: A line drawn parallel to any side of a triangle, divides the other two sides proportionally. (Basic Proportionality Theorem) In the give figure, DE ║ BC

AD = AE DE ║ BC BD CE

Conversely, If a line divides two sides of a triangle proportionally, the line is parallel to the third side.Similarity as a size Transformation: In a size transformation, a given figure is enlarged (or reduced) by a scale factor K, such that the resulting figure is similar to the given figure.If k > 1, then the transformation is the given figure.If k < 1, then the transformation is reduction and If k = 1, then the transformation is an identity transformation.

A

B C

D

For a scale factor k, the Measure of the length of the resulting plane figure = k times the corresponding length of the given plane figureArea of the resulting plane figure = k2 x (area of the given plane figure)In case of solids: = k3 x (volume of the given solid)