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  1. 1. Simila rity Cli ck thi s!!By :AltafiyaniRahmatikaClass : IX D
  2. 2. DEFINITION Similarity is a pair of plane figures/plane objectsthat are same in shape but different in size(equivalent). Similarity is denoted by “~”
  3. 3. Similarity of What makes a pair Plane Figures of figures called similar?A pair of figures are called similar if they have TheRequirements 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
  4. 4. 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 aresame, so all the corresponding sides are proportional.
  5. 5. Exercise 1A N M D C 5 cm 7 cm 65o 115o 65o 115o A B K 10 cm 14 cm LWhether 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
  6. 6. Exercise 1BLook at the figure! N M S 3 cm R 6 cm P 4 cm Q K 12 cm LIf the both trapezoids above are known similar, determine thelength 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
  7. 7. Similarity of TrianglesEspecially for triangles, two triangles called similar if they havesatisfy 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
  8. 8. All the corresponding sides are proportional : side, side,side (s.s.s). Example: L Q O P J 10 cm K 5 cmTwo triangles above are known similar. Based on thepicture, we can conclude : The second is... = = = = = =Because the proportion of all the correspondingsides are same, so the both triangles above satisfy thes.s.s requirements
  9. 9. Two of corresponding sides are proportional and thecorresponding angles which flanked are same inmeasure: 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 thecorresponding angles which flanked.So, the both trianglesabove satisfy the s.a.s requirements.
  10. 10. Exercise 2AWhich 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 HAnswer: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 = 54ob. The proportion of all of the corresponding sides: ABC and EFG : = and =
  11. 11. EFG and HIJ : = = and = =ABC and HIJ : = = and = So, the triangles which are similar are EFG and HIJ or EFG ~HIJ
  12. 12. Exercise 2BLook at the figure! D E C A BAB 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 BA B
  13. 13. = = 12 x EC = 6 x 10 = EC = = = = 5 cmSo, the length of EC is 5 cm.
  14. 14. 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 = OO P R PR2 = OR x QR P P PR = PR = PR = P Q R PR = 4 cm