Conditions for similarity of PolygonsTwo polygons of the same number of sides are similar, if

(i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion).

Any circles are similar

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Ratio of sides of similar triangles

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AB C

DE DE ED

A A

FG

Ar AED =12 AD*EF

Ar EDC =12DC*EF

Ar AED/Ar EDC = 12 AD*EF12 DC *EF

= ADDC

Ar AED =12 AE*DG

Ar EDB =12 EB*DG

Ar AED/Ar EDB = 12 AE *DG12 EB*DG

= AEEB

===

Ratio of sides of similar triangles

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AB C

DETheorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Theorem 6.1 : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio

AE

EB+ 1 =

AD

DC+ 1

AB

EB=

AC

DC

CROSS MULTIPLYINGEBAB =

DCAC =

BC EDBC

A

E D

A

B C

DEA

E D

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Theorem 6.3 (AAA or AA)Corresponding angles are equal, then their corresponding sides are in the same ratio . ---AAA criterionIf two angles are same then the third has to be same, hence AA similarity is enough to get AAA similarity

Theorem 6.4 (SSS ) Corresponding sides are in the same ratio then their corresponding angles are equal and hence the two triangles are similar.---SSS Criterion

Theorem 6.5 (SAS) If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

xy

wzAA

A = AXY =

WZ

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Ratio of areas of similar trianglesP

Q R

A

B CM N12 BC *AM12QR*PN

=

B = Q (Given that ABC PQR )M = Q = 90ABPQ =

AMPN =

BCQR

=12 BC *AM12QR*PN

BC *BCQR*QR= =

BC2

QR2

=Square of ratio of corresponding sides

Area ABC

Area of PQR

Theorem 6.6 : The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

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Baudhayan (Pythagoras) Theorem

Perpendicular divides Yellow triangle into two triangles pink and blue Pink is similar to the original yellow

Pink and blue are similar to themselves Blue is similar to yellow

A

B CD

D

B

AD

A

C

A

B C

A

B C

D

B A

D

A

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Proof of Baudhayan (Pythagoras) theorem B

CD

A

DA B

B

C

ADAB =

ABAC

AD.AC = AB2

AB

CA

D

B

CDBC =

BCAC

CD.AC = BC2

AD.AC+CD.AC = AB2 +BC2AC AD+CD( ) = AB2 +BC2AC2 = AB2 + BC2

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A B

O

C D

By AA similarity , OAB~OCD

AOOB =

OCOD

Ex 6.3ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO

CO =BO DO

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13. D is a point on the side BC of a triangle ABC such that ADC = BAC. Show that CA2 = CB.CD

A

B

C

D

D C

A

Prove that the triangles are similar by AA ruleWrite the corres ratios

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14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ABC~PQR.

A

B CD

P

Q RM

ABPQ

= ACPR

= ADPM

- - - - - - - - (1)

QMMR

= BDDC

QMMR

+ 1 = BDDC

+ 1

QRMR

= BCDC

- - - - - - - -(2)

from (1) and (2) , we see all the three sides of the ABC and PQRare proportional Hence by SSS rule, PQR ABC

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15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

4m28m

?

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AB CD

P

Q RM

BCQR

=12 BC12 QR

= BDQM

- - - -(1)

Given that ABPQ

= BCQR

- - - (2)

From(1) and (2), PQM ABD

Hence, PQAB

= PMAD

16. If AD and PM are medians of triangles ABC and PQR, respectively where ABC~PQR, prove that AB / AD = PQ / PM

For two similar triangles, not only their sides are in proportion, their medians

are also in same proportion

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Ex6.42. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

A B

CD

O

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4. If the areas of two similar triangles are equal, prove that they are congruent.

Let the base and height of the two triangles be b1, h1 and b2, h2 respectively

b1 b2h1 h2

Areas Same Triangles similar12b1* h1 = 1

2b2 * h2

b1* h1 = b2 * h2

b2 * h1* h1h2

= b2 * h2

h1 = h2Hence, b1 = b2Hence, the triangles are congruent to each other

b1h1 =

b2h2

b1*h2 = b2*h1

b1= b2*h1h2 @2

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AREAS OF PARALLELOGRAMS AND TRIANGLES 161

File Name : C:\Computer Station\Maths-IX\Chapter\Chap-9\Chap-9 (03-01-2006).PM65

Now, suppose ABCD is a parallelogram whose one of the diagonals is AC(see Fig. 9.20). Let ANA DC. Note that

' ADC # ' CBA (Why?)So, ar (ADC) = ar (CBA) (Why?)

Therefore, ar (ADC) = 1 ar (ABCD)2

= 1 (DC AN)2 u (Why?)

So, area of ' ADC = 12 base DC corresponding altitude ANIn other words, area of a triangle is half the product of its base (or any side) andthe corresponding altitude (or height). Do you remember that you have learnt thisformula for area of a triangle in Class VII ? From this formula, you can see that twotriangles with same base (or equal bases) and equal areas will have equalcorresponding altitudes.

For having equal corresponding altitudes, the triangles must lie between the sameparallels. From this, you arrive at the following converse of Theorem 9.2 .Theorem 9.3 : Two triangles having the same base (or equal bases) and equalareas lie between the same parallels.Let us now take some examples to illustrate the use of the above results.Example 3 : Show that a median of a triangle divides it into two triangles of equalareas.Solution : Let ABC be a triangle and let AD be one of its medians (see Fig. 9.21).You wish to show that

ar (ABD) = ar (ACD).Since the formula for area involves altitude, let usdraw ANABC.

Now ar(ABD) = 12 base altitude (of ' ABD)

= 1 BD AN2 u u

Fig. 9.20

Fig. 9.21

NCE

RT

not t

o be r

epub

lishe

dFriday, May 18, 12