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  • 1. Simila rity Cli ck thi s!!By :AltafiyaniRahmatikaClass : IX D
  • 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. 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. 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. 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. 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. 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. 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. 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. 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. EFG and HIJ : = = and = =ABC and HIJ : = = and = So, the triangles which are similar are EFG and HIJ or EFG ~HIJ
  • 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. = = 12 x EC = 6 x 10 = EC = = = = 5 cmSo, the length of EC is 5 cm.
  • 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

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