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PROPORTIONAL SEGMENTS & BASIC SIMILARITY THEOREM

PROPORTIONAL SEGMENTS & BASIC SIMILARITY THEOREM. Express each ratio in simplest form. ILLUSTRATION. Pt. X divides segment AB so that AX to XB is 3 : 2. Pt. Y divides segment CD so that CY to YD is 3 : 2. A. X. B. 12. 8. C. Y. D. 6. 4. THEOREM: PROPORTIONAL SEGMENTS

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PROPORTIONAL SEGMENTS & BASIC SIMILARITY THEOREM

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  1. PROPORTIONAL SEGMENTS&BASIC SIMILARITY THEOREM

  2. Express each ratio in simplest form.

  3. ILLUSTRATION • Pt. X divides segment AB so that AX to XB is 3 : 2. • Pt. Y divides segment CD so that CY to YD is 3 : 2 A X B 12 8 C Y D 6 4

  4. THEOREM: PROPORTIONAL SEGMENTS • “Two segments are divided proportionally if the measures of the segments of one have the same ratio as the measures of the corresponding segments of the other.” A X B 12 8 SOME PROPORTIONS AX CY 1. C Y D XB YD 6 4 12 6 8 4 3 3 2 2

  5. THEOREM: PROPORTIONAL SEGMENTS • “Two segments are divided proportionally if the measures of the segments of one have the same ratio as the measures of the corresponding segments of the other.” A X B 12 8 SOME PROPORTIONS AB CD 2. C Y D 6 4 XB YD 20 10 8 4 5 5 2 2

  6. THEOREM: PROPORTIONAL SEGMENTS • “Two segments are divided proportionally if the measures of the segments of one have the same ratio as the measures of the corresponding segments of the other.” A X B 12 8 SOME PROPORTIONS AB CD 3. C Y D 6 4 AX CY 20 10 12 6 5 5 3 3

  7. THEOREM: PROPORTIONAL SEGMENTS • “Two segments are divided proportionally if the measures of the segments of one have the same ratio as the measures of the corresponding segments of the other.” A X B 12 8 SOME PROPORTIONS AX +AB CY +CD 4. C Y D 6 4 AX CY 32 16 12 6 8 8 3 3

  8. Illustrative Examples

  9. Suppose segment AC and segment MP are divided proportionally by points B and N respectively. Then, A B C 8 12 AB MN 1. M N P 2 3 BC NP 2. AB BC MN NP

  10. Suppose segment AC and segment MP are divided proportionally by points B and N respectively. Then, A B C 8 12 AB MN 3. M N P 2 3 AC MP 4. BC NP AC MP

  11. Solution: X : 32 = 6 : 8 Applying the law of proportion 8(x) = 6( 32) 8x = 192 X = 24 Find the unknown parts assuming the segments are divided proportionally. X 32 6 8

  12. Illustrative examples • SOLVE: • One string is divided into lengths 18 cm and 15 cm. a second string is also to be divided into such that the two strings will become proportional. If the longest portion of the second string has length 60 cm, what is the length of the other portion of the second string?

  13. x • Solution: • Let x = the length of the other portion of the second string. 15 18 60 x 5 x 60 6 60 5(60) 6x 15 18 6x = 300 X = 50, the length of the other portion of the second string

  14. BASIC PROPORTIONALITY THEOREM • If a line intersects two sides of a triangle and is parallel to the third side, then it divides the first two sides proportionally.

  15. RESTATEMENT OF THE THEOREM B • If a line (EF) intersects two sides ( AB & CB) of a triangle (ABC) and is parallel to the third side( AC ), then it divides the first two sides proportionally. • Thus, E F C A

  16. OTHER PROPORTIONS B • 1. BE : EA = BF : FC • 2. BE : BA = BF : BC • 3. BA : EA = BC : FC • 4. BE : BF = EA : FC • 5. FC : EA = BC : BA • 6. EF : AC = BF : BC • 7. EF : AC = BE : BA E F C A

  17. VERIFYING A PROPORTIONS( an example) B 1. BE : EA = BF : FC • 15 : 5 = 12 : 4 • By simplifying, • 3 : 1 = 3 : 1 15 12 6 E F 4 5 8 C A

  18. VERIFYING A PROPORTIONS B 2. BE : BA = BF : BC • 15 : 20 = 12 : 16 • By simplifying, • 3 : 4 = 3 : 4 15 12 6 E F 4 5 8 C A

  19. VERIFYING A PROPORTIONS B 3. BA : EA = BC : FC • 20 : 5 = 16 : 4 • By simplifying, • 4 : 1 = 4 : 1 15 12 6 E F 4 5 8 C A

  20. VERIFYING A PROPORTIONS B 4. BE : BF = EA : FC • 15 : 12 = 5 : 4 • By simplifying, • 5 : 4 = 5 : 4 15 12 6 E F 4 5 8 C A

  21. VERIFYING A PROPORTIONS B 5. FC : EA = BC : BA • 4 : 5 = 16 : 20 • By simplifying, • 4 : 5 = 4 : 5 15 12 6 E F 4 5 8 C A

  22. VERIFYING A PROPORTIONS B 6. EF : AC = BF : BC • 6 : 8 = 12 : 16 • By simplifying, • 3 : 4 = 3 : 4 15 12 6 E F 4 5 8 C A

  23. VERIFYING A PROPORTIONS B 6. EF : AC = BE : BA • 6 : 8 = 15 : 20 • By simplifying, • 3 : 4 = 3 : 4 15 12 6 E F 4 5 8 C A

  24. Exercises • GIVEN: DE // BC, • AD = 9, AE = 12, DE = 10,DB = 18. • Find, • BC, AC and CE. A 9 12 D E 10 18 C B

  25. Solution • Find BC, • BC : DE = BA : DA • BC : 10 = 27 : 9 or • BC : 10 = 3 : 1 • Applying principle of proportion • BC(1) = 10(3) • BC = 30 A 9 12 D E 10 18 30 C B

  26. Solution • Find AC, • AC : AE = BA : DA • AC : 12 = 27 : 9 or • AC : 12 = 3 : 1 • Applying principle of proportion • AC(1) = 12(3) • AC = 36 A 9 12 D E 10 18 C B

  27. Solution • Find CE, • CE : AE = BD : DA • CE : 12 = 18 : 9 or • CE : 12 = 2 : 1 • Applying principle of proportion • CE(1) = 12(2) • CE = 24 A 9 12 D E 10 24 18 C B

  28. Solution • Another way to find CE, • CE = AC - AE • Hence, AC =36, then • CE = 36 - 12 • CE = 24 A 9 12 D E 10 24 18 C B

  29. Quiz • Solve the following problem. Show your solution.( one –half crosswise) • 1. Uncle Tom plans to divide an 80- meter rope into three pieces in the ratio 3 : 5 : 8. what will be the length of each piece?

  30. QUIZ • 2. In the figure, find the values of x and y. 30 y 15 12 x 10

  31. Assignment • EXERCISES A. • Geometry Workbook, page 95

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