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Similarity:Figures which have the same shape are called similar figures. Two shapes are Similar if the only difference is size (and possibly the need to turn or flip one around).

Example:

One shape can become another using Resizing (also called dilation, contraction, compression, enlargement or even expansion), then the shapes are Similar:

These Shapes are Similar!

There may be Turns, Flips or Slides, Too!

Sometimes it can be hard to see if two shapes are Similar, because you may need to turn, flip or slide one shape as well as resizing it.

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Rotation Turn!

Reflection Flip!

Translation Slide!

Examples

These shapes are all Similar:

Resized Resized and Reflected Resized and Rotated

When two shapes are similar, then there are some special properties:

1. Their corresponding sides are in the same ratio.2. Their corresponding angles are equal

This can make life a lot easier when solving geometry puzzles, as in this example:

 

Example: What is the missing length here?

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Notice that the red triangle has the same angles as the main triangle ...

... they both have one right angle, and a shared angle in the left corner

 

In fact you could flip over the red triangle, rotate it a little, then resize it and it would fit exactly on top of the main triangle. So they are similar triangles.

So the line lengths will be in proportion, and we can calculate:

? = 80 × (130/127) = 81.9

The polygons below are similar.

We can verify that if one polygon is similar to another polygon and this second polygon is similar to the first polygon, Then the first polygon is similar to the second one.

In the two quadrilaterals (a square and a rectangle) shown below, the 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 BSIMILARITY OF TRIANGLES.

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

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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

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.

AA

B C E F

D

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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

SIMILARLY,

AB = BC and so AB = BC = ACDE EF DE EF DFTHEREFORE ∆ABC ~ ∆DEF

A

B C

P

E F

Q

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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

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

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)

A

B C E

D

F

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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

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

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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

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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 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 3

A

D C

Q

PB

A

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AP: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

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

DIFFERNCES BETWEEN CONGURANT AND SIMILAR TRIANGLES.

Congruent triangles are equal in all respects i.e. their angles are equal, their sides are

P

B C

Q

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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

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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.

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

A

B C E F

D

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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

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

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

60ºB

Q

C E

80

60º

3

F

2.5

80

2

4

65

40

S

40

610

D

60º

7070 52.5

70 70

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B = Q AAA similarity criterion.

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)

Areas of similar shapes:

The following rectangles are similar and their ratio of corresponding sides is y. WXYZ is of length b and width a.

area of WXYZ=a × b

area of ABCD= ya× yb= y2 ab

∴ area ABCDarea WXYZ

= y2 abab

= y2

“If two figures are similar and their sides are in the ratio y then their areas will be in the ratio y2”.

Example:

∆ ACE ∆ BCD

If the area of ∆ BCD=6cm2, find the area of∆ ACE.

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Solution:

Ratio of corresponding sides = BCAC

=23

Ratio of areas = ¿ Area ∆ ACE=x

49=6

x

x=6 ×9

4=13 1

2

Perimeters of similar shapes:

The following rectangles are similar and their ratio of corresponding sides is 2.

Perimeter of ABCD = (2+3 ) ×2=10 cm

Perimeter of WXYZ = (4+6 )×2=20 cm

∴ Perimeter WXYZPerimeter ABCD

=2010

=2

“If the ratio of corresponding sides of two similar figures is y, then the ratio of their perimeters is also y.”

More Examples:

1. Ally washes a napkin in hot water and it shrinks by 10%. What is the percentage reduction in its area?

Solution:

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Ratio of sides = 100−10

100= 90

100= 9

10

Ratio of areas = ( 910

)2

= 81100

∴The reduction in area = 100−81=19 %

2. On heating, a metallic strip expands by 30%. What is the percentage of increase in the surface area?

Solution:

Ratio of corresponding sides = 130100

=1310

Ratio of areas = ( 1310

)2

=169100

Percentage increase in area =169−100=69 %

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 they have sides which are equal to each other. For example: a cube and a cuboids 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.

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

SURFACE AREAS OF SIMILAR 3D OBJECTSTHEOREM: the ratio of the surface areas of two similar 3D objects is equal to the square of the ratio of their corresponding sides.

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

Now we will find the scale factor of one of the

4cm

3cm8cm

8cm

6cm16cm

AB

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

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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

S = 6l2 S = 6l2

S= 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

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Surface area of cube 236 cm2 = 1 2

216 cm2 2 1= 1 2

4 2

Volumes of similar 3-D shapes:

A and B are two similar 3-D shapes. Their ratio of corresponding sides is k .

volumeof A=abc

volume of B=ka× kb × kc=k3 abc

volumeof Bvolume of A

= k 3abcabc

=k 3

“If the ratio of the corresponding sides of two 3-D objects is k , then the ratio of their volumes is k3.”

Example:

There are two similar boxes. The height of the small box is 3 cm and the height of the big one is 7cm. If the volume of the smaller box is 120 cm3, find the volume of the bigger box?

Ratio of sides = 37

Ratio of volumes = ( 37)

3

= 27343

volume of bigger box = x

27343

=120x

K

L

M X

Y

Z

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x=343 ×12027

=1524 49

cm3

Congruency:

Figures which have the same size and shape are called Congruent.

Two shapes are congruent if you can Turn, Flip and/or Slide one so it fits exactly on the other.In this example the shapes are congruent (you only need to flip one over and move it a little)

Congruent or Similar?

The two shapes need to be the same size to be congruent. (If you need to resize one shape to make it the same as the other, the shapes are called Similar)

If you ...   Then the shapes are ...

... only Rotate, Reflect and/or Translate  Congruent

... need to ResizeSimilar

Congruent? Why such a funny word that basically means "equal"? Probably because they would only be "equal" if laid on top of each other. Anyway it comes from Latin congruere, "to agree". So the shapes "agree"

Congruency in triangles:

For two triangles to be congruent there are certain conditions.

1. Side, Side, Side (SSS)

If three sides of a triangle are equal with three sides of another triangle, then the two triangles are congruent.

Q

U

ZB

A

S

Q

P RA

B

C

L

K MC

A

X

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∆ KLM ≡ ∆ XYZ (SSS )

2. Side, Angle, Side (SAS)

If two sides from one triangle are equal to two sides from another triangle and the angle between these two pairs of sides are equal, then the two triangles are congruent.

∆ BAS ≡∆ QUZ (SAS)

3. Angle, Side, Angle (ASA)

If two angles from a triangle are equal to two corresponding angles from another triangle, and the side between the pairs of angles is equal in both triangles, then they are congruent.

∆ ABC ≡∆ PQR ( ASA)

4. Right Angle, Hypotenuse, Side (RHS)

In right-angled triangles, if the longest sides are equal and another side from the first triangle is equal to a side of the second triangle, then the triangles are congruent.

∆ CAX ≡ ∆ KLM (RHS )

5. Angle, Angle, Side (AAS)

R

S

TA

B

C

E

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If two angles and the non-included side of a triangle are equal to the corresponding parts of another triangle, then the two triangles are congruent.

∆ ABC ∆ RST ( AAS)

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.

A

B C

D

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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.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)