Simila

rityBy :

Altafiyani

Rahmatika

Class : IX D

Cli

ck

thi

s!!

DEFINITIONSimilarity is a pair of plane figures/plane objects

that are same in shape but different in size (equivalent). Similarity is denoted by “~”

A pair of figures are called similar if they have The Requirements of Similarity, that are:

What makes a pair of figures called

similar?

All the corresponding angles are equal in measure. Example:

90 oD

A B

C

90 oN

K L

M

Two rectangles beside are known similar. In every rectangle, the magnitude of each angle is 90 o

(right angle). So, <A = <K, <B = <L, <C = <M, and <D = <N.

Similarity of Plane Figures

All the corrresponding sides are proportional. Example :

6 cm

3 cm

H

E F

GS

P Q

R

8 cm

4 cm

= =

= = =

Look at the figure! The proportion of each width :

The proportion of each length :

Because the proportion of the width and the length are same, so all the corresponding sides are proportional.

Exercise 1A

Whether ABCD is similar with KLMN?

Answer:

Because all the corresponding angles are same in measure, compare all the corresponding length! Thus:

ABCD is similar with KLMN

5 cm 7 cm

14 cm10 cm

= =

=

=

=

65o 115o

A

D C

B K

MN

L

115o65o

Exercise 1BLook at the figure!

If the both trapezoids above are known similar, determine the length of MN and QR!

Answer:

12 cm4 cm

6 cm

3 cm

12 x PS = 6 x 4

12 x PS = 24

PS =

PS = 2 cm

=

=

12 x 3 = MN x 4

36 = MN x 4

MN =

MN = 9 cm

N

K L

M

S

P Q

R

=

= QR = PS

QR = 2 cm

The first is...

Similarity of Triangles

Especially for triangles, two triangles called similar if they have satisfy the following requirements:

All the corresponding angles are equal in measure : angle, angle, angle (a.a.a). Example:

Hello! I want to explain about...

F G

H

50o

30o 90o

A B

C

90o

50o

30o

Two triangles above are known similar. Based on the picture, we can conclude: <A = <F

<B = <G<C = <H

So, the both triangles above satisfy the a.a.a requirements

==

All the corresponding sides are proportional : side, side, side (s.s.s). Example:

Two triangles above are known similar. Based on the picture, we can conclude :

Because the proportion of all the corresponding

sides are same, so the both triangles above satisfy the s.s.s requirements

The second is...

J K

L

10 cmO P

Q

5 cm

= = = =

Two of corresponding sides are proportional and the corresponding angles which flanked are same in measure: side, angle, side (s.a.s). Example:

Based on the figure above, we can conclude:

Beside that, <S = <W = 60 o. <S and <W are the corresponding angles which flanked.So, the both triangles above satisfy the s.a.s requirements.

And the last is...

W

60 o

6 cmV

X

60 o

R S

T

4 cm

= = ==

Exercise 2AWhich of these triangles that are similar?

Answer:

Use the third requirements of similarity in triangles (s.a.s) :

a. All of the corresponding angles which flanked are same in measurement: <B = <E = <J = 54o

b. The proportion of all of the corresponding sides:

ABC and EFG :

A

C

B

3 cm

5 cm

54o

E

F

G

8 cm

54o

H

I

J

15 cm

54o

= and =

EFG and HIJ :

ABC and HIJ :

and= = = =

= = and =

So, the triangles which are similar are EFG and HIJ

or EFG ~HIJ

Look at the figure!

AB is parallel with EC. If DE = 10 cm, AE = 2 cm, and AB = 6 cm. Determine the length of EC!

Answer:

Exercise 2B

E

D

C

BA

E

D

C

BA

D

BA

1

D

E C

2

So, the length of EC is 5 cm.

=

=

=

=

12 x EC = 6 x 10

EC =

= = 5 cm

Exercise 2CLook at the figure!

OPQ is a right triangle and PR as the altitude of OPQ. OR = 8 cm and QR = 2 cm. Determine the length of PR!

Answer:O P

Q

R

O P

Q

R

Because ROP is similar with RPQ, so:

PR2 = OR x QRPR =PR =PR =PR = 4 cm

O

P R

P

Q R

P

O

P RQ

=