Simila
     rity
               Cli
               ck
               thi
               s!!


By :
Altafiyani
Rahmatika
Class : IX D
DEFINITION
   Similarity is a pair of plane figures/plane objects
that are same in shape but different in size
(equivalent). Similarity is denoted by “~”
Similarity of
                                        What makes a pair
 Plane Figures                          of figures called
                                             similar?

A pair of figures are called similar if they have The
Requirements of Similarity, that are:
 All the corresponding angles are equal in measure.
  Example:
  D                     C
          90 o              Two rectangles beside are known
                            similar. In every rectangle, the
  A                     B   magnitude of each angle is 90 o
                            (right angle). So, <A = <K, <B = <L,
      N             M
             90 o           <C = <M, and <D = <N.
      K             L
All the corrresponding sides are proportional.
   Example :
                              S                         R
      H                G
   3 cm                                                 4 cm

      E                F
             6 cm             P                         Q
                                         8 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                                        N                       M
              D                        C

       5 cm                                7 cm
              65o           115o
                                                  65o           115o
       A                           B       K
                    10 cm                               14 cm          L

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
Exercise 1B
Look at the figure!
           N                  M
                                              S   3 cm   R
    6 cm

                                          P       4 cm       Q
     K             12 cm          L
If the both trapezoids above are known similar, determine the
length of MN and QR!
Answer:                 PS =             =            MN =
               =
                           PS = 2 cm                         MN = 9 cm
               =           QR = PS            =
                           QR = 2 cm   12 x 3 = MN x 4
     12 x PS = 6 x 4
     12 x PS = 24                         36 = MN x 4
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:
                             C                           Hello! I want to
                                                     H
                                                         explain about...
                                                           The first is...
                       50o                     50o

                       90o              30o   90o
            30o                    F                 G
     A                       B
    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:                      L
                Q




  O                    P       J         10 cm               K
            5 cm

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

      =     =              =   =          =      =

Because the proportion of all the corresponding
sides are same, so the both triangles above satisfy the
s.s.s requirements
Two of corresponding sides are proportional and the
corresponding angles which flanked are same in
measure: side, angle, side (s.a.s). Example:
                 T                     X




                 60 o
        R               S              60 o
             4 cm            V                 W   And the
                                   6 cm
                                                   last is...
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.
Exercise 2A
Which of these triangles that are similar?
                                   G               J
               C
                                                   54o
                          8 cm             15 cm           I
                   3 cm
           54o                   54o
  A                B      E
        5 cm
                                       F    H
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 :
                         =       and        =
EFG and HIJ :
                =     =   and   =   =


ABC and HIJ :   =     =   and   =


  So, the triangles
  which are similar
  are EFG and HIJ
    or EFG ~HIJ
Exercise 2B
Look at the figure!                D




                               E       C


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


                                                       2
    E        C                     1
                                                   E           C

                           A                   B
A                B
=                    =
                           12 x EC = 6 x 10
              =
                                EC =

              =
                                   =    = 5 cm

So, the length of EC is 5 cm.
Exercise 2C
    Look at the figure!
                                   Q
                           R               OPQ is a right triangle and PR as
                                           the altitude of OPQ. OR = 8 cm
                                           and QR = 2 cm. Determine the
       O                           P
                                           length of PR!
                                       P      O
    Answer:            Q                              Because ROP is similar
               R                                      with RPQ, so:
                               Q       R                       =
                           O
O                      P
                                               R
                                                           PR2 = OR x QR
                                   P
                               P                           PR =
                                                           PR =
                                                           PR =
           P       Q           R                           PR = 4 cm

Similarity

  • 1.
    Simila rity Cli ck thi s!! By : Altafiyani Rahmatika Class : IX D
  • 3.
    DEFINITION Similarity is a pair of plane figures/plane objects that are same in shape but different in size (equivalent). Similarity is denoted by “~”
  • 4.
    Similarity of What makes a pair Plane Figures of figures called similar? A pair of figures are called similar if they have The Requirements of Similarity, that are:  All the corresponding angles are equal in measure. Example: D C 90 o Two rectangles beside are known similar. In every rectangle, the A B magnitude of each angle is 90 o (right angle). So, <A = <K, <B = <L, N M 90 o <C = <M, and <D = <N. K L
  • 5.
    All the corrrespondingsides are proportional. Example : S R H G 3 cm 4 cm E F 6 cm P Q 8 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.
  • 6.
    Exercise 1A N M D C 5 cm 7 cm 65o 115o 65o 115o A B K 10 cm 14 cm L 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
  • 7.
    Exercise 1B Look atthe figure! N M S 3 cm R 6 cm P 4 cm Q K 12 cm L If the both trapezoids above are known similar, determine the length of MN and QR! Answer: PS = = MN = = PS = 2 cm MN = 9 cm = QR = PS = QR = 2 cm 12 x 3 = MN x 4 12 x PS = 6 x 4 12 x PS = 24 36 = MN x 4
  • 8.
    Similarity of Triangles Especiallyfor 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: C Hello! I want to H explain about... The first is... 50o 50o 90o 30o 90o 30o F G A B 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
  • 9.
    All the correspondingsides are proportional : side, side, side (s.s.s). Example: L Q O P J 10 cm K 5 cm Two triangles above are known similar. Based on the picture, we can conclude : The second is... = = = = = = Because the proportion of all the corresponding sides are same, so the both triangles above satisfy the s.s.s requirements
  • 10.
    Two of correspondingsides are proportional and the corresponding angles which flanked are same in measure: side, angle, side (s.a.s). Example: T X 60 o R S 60 o 4 cm V W And the 6 cm last is... 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.
  • 11.
    Exercise 2A Which ofthese triangles that are similar? G J C 54o 8 cm 15 cm I 3 cm 54o 54o A B E 5 cm F H 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 : = and =
  • 12.
    EFG and HIJ: = = and = = ABC and HIJ : = = and = So, the triangles which are similar are EFG and HIJ or EFG ~HIJ
  • 13.
    Exercise 2B Look atthe figure! D E C A B AB is parallel with EC. If DE = 10 cm, AE = 2 cm, and AB = 6 cm. Determine the length of EC! Answer: D D D 2 E C 1 E C A B A B
  • 14.
    = = 12 x EC = 6 x 10 = EC = = = = 5 cm So, the length of EC is 5 cm.
  • 15.
    Exercise 2C Look at the figure! Q R OPQ is a right triangle and PR as the altitude of OPQ. OR = 8 cm and QR = 2 cm. Determine the O P length of PR! P O Answer: Q Because ROP is similar R with RPQ, so: Q R = O O P R PR2 = OR x QR P P PR = PR = PR = P Q R PR = 4 cm