Presentation is loading. Please wait.

Presentation is loading. Please wait.

PROPORTIONAL SEGMENTS & BASIC SIMILARITY THEOREM

Similar presentations


Presentation on theme: "PROPORTIONAL SEGMENTS & BASIC SIMILARITY THEOREM"— Presentation transcript:

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 4. CY +CD 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,
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,
8 12 AB MN 3. M N P 2 3 AC MP 4. BC NP AC MP

11 Applying the law of proportion 8(x) = 6( 32) 8x = 192 X = 24
Find the unknown parts assuming the segments are divided proportionally. Solution: X : 32 = 6 : 8 Applying the law of proportion 8(x) = 6( 32) 8x = 192 X = 24 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 Let x = the length of the other portion of the second string.
Solution: Let x = the length of the other portion of the second string. x 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 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 B E F C A

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

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

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

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

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

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

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

24 Exercises BC, AC and CE. 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
Applying principle of proportion BC(1) = 10(3) BC = 30 A 9 12 D E 10 18 30 C B

26 Solution AC = 36 Find AC, AC : AE = BA : DA AC : 12 = 27 : 9 or
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
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
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 2. In the figure, find the values of x and y.
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


Download ppt "PROPORTIONAL SEGMENTS & BASIC SIMILARITY THEOREM"

Similar presentations


Ads by Google