PROPORTIONAL SEGMENTS & BASIC SIMILARITY THEOREM
description
Transcript of PROPORTIONAL SEGMENTS & BASIC SIMILARITY THEOREM
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PROPORTIONAL PROPORTIONAL SEGMENTSSEGMENTS
&&BASIC SIMILARITY BASIC SIMILARITY
THEOREMTHEOREM
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Express each ratio in simplest form.
18
12
12
8
14
8
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ILLUSTRATIONILLUSTRATION
Pt. X divides Pt. X divides segment AB segment AB so that AX to so that AX to XB is 3 : 2.XB is 3 : 2.
Pt. Y divides Pt. Y divides segment CD segment CD so that CY to so that CY to YD is 3 : 2YD is 3 : 2
A
DYC
X B12 8
6 4
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THEOREM: THEOREM: PROPORTIONAL SEGMENTSPROPORTIONAL SEGMENTS ““Two segments are divided proportionally Two segments are divided proportionally
if if the measures of the segments of one the measures of the segments of one have the same ratiohave the same ratio as the measures of as the measures of the corresponding segments of the other.”the corresponding segments of the other.”
A
DYC
X B12 8
6 4AX CY
12
XB YD
6 8 4
SOME PROPORTIONS
3 3
2 2
1.
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THEOREM: THEOREM: PROPORTIONAL SEGMENTSPROPORTIONAL SEGMENTS ““Two segments are divided proportionally Two segments are divided proportionally
if if the measures of the segments of one the measures of the segments of one have the same ratiohave the same ratio as the measures of as the measures of the corresponding segments of the other.”the corresponding segments of the other.”
A
DYC
X B12 8
6 4AB CD
20
XB YD
108 4
SOME PROPORTIONS
5 5
2 2
2.
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THEOREM: THEOREM: PROPORTIONAL SEGMENTSPROPORTIONAL SEGMENTS ““Two segments are divided proportionally Two segments are divided proportionally
if if the measures of the segments of one the measures of the segments of one have the same ratiohave the same ratio as the measures of as the measures of the corresponding segments of the other.”the corresponding segments of the other.”
A
DYC
X B12 8
6 4AB CD
20
AX CY
1012 6
SOME PROPORTIONS
5 5
3 3
3.
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THEOREM: THEOREM: PROPORTIONAL SEGMENTSPROPORTIONAL SEGMENTS ““Two segments are divided proportionally Two segments are divided proportionally
if if the measures of the segments of one the measures of the segments of one have the same ratiohave the same ratio as the measures of as the measures of the corresponding segments of the other.”the corresponding segments of the other.”
A
DYC
X B12 8
6 4AX +AB CY +CD
32
AX CY
1612 6
SOME PROPORTIONS
8 8
3 3
4.
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Illustrative Examples
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Suppose segment AC and segment MP are Suppose segment AC and segment MP are divided proportionally by points B and N divided proportionally by points B and N respectively. Then,respectively. Then,
A
PNM
B C8 12
2 3AB MN
AB
BC NP
BCMN NP
1.
2.
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Suppose segment AC and segment MP are Suppose segment AC and segment MP are divided proportionally by points B and N divided proportionally by points B and N respectively. Then,respectively. Then,
A
PNM
B C8 12
2 3AB MN
BC
AC MP
NPAC MP
3.
4.
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Find the unknown parts assuming the Find the unknown parts assuming the segments are divided proportionally.segments are divided proportionally.
X 32
6 8
Solution:Solution:
X : 32 = 6 : 8X : 32 = 6 : 8
Applying the law of Applying the law of proportionproportion
8(x) = 6( 32)8(x) = 6( 32)
8x = 1928x = 192
X = 24X = 24
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Illustrative examplesIllustrative examples
SOLVE:SOLVE: One string is divided into lengths 18 cm One string is divided into lengths 18 cm
and 15 cm. a second string is also to be and 15 cm. a second string is also to be divided into such that the two strings will divided into such that the two strings will become proportional. If the longest portion become proportional. If the longest portion of the second string has length 60 cm, of the second string has length 60 cm, what is the length of the other portion of what is the length of the other portion of the second string?the second string?
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Solution:Solution: Let x = the length of the other Let x = the length of the other
portion of the second string.portion of the second string.
x 60
15 18
x18
60
6015
5x6
6x 5(60)
X = 50, the length of the other portion of the second string
6x = 300
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BASIC PROPORTIONALITY BASIC PROPORTIONALITY THEOREMTHEOREM
If a line intersects If a line intersects two sides of a two sides of a triangle and is triangle and is parallel to the parallel to the third side, then it third side, then it divides the first divides the first two sides two sides proportionally.proportionally.
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RESTATEMENT OF THE RESTATEMENT OF THE THEOREMTHEOREM
If a line (If a line (EFEF) ) intersects two sides intersects two sides ( ( AB & CBAB & CB) of a ) of a triangle (triangle (ABCABC) and is ) and is parallel to the third parallel to the third side(side( AC AC ), then it ), then it divides the first two divides the first two sides proportionally.sides proportionally.
Thus, Thus,
B
E F
CA
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OTHER PROPORTIONSOTHER PROPORTIONS
1. BE : EA = BF : FC1. BE : EA = BF : FC 2. BE : BA = BF : BC2. BE : BA = BF : BC 3. BA : EA = BC : FC3. BA : EA = BC : FC 4. BE : BF = EA : FC4. BE : BF = EA : FC 5. FC : EA = BC : BA5. FC : EA = BC : BA 6. EF : AC = BF : BC6. EF : AC = BF : BC 7. EF : AC = BE : BA7. EF : AC = BE : BA
B
E F
CA
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VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS( an example)( an example)
1. BE : EA = BF : 1. BE : EA = BF : FCFC
15 : 5 = 12 : 415 : 5 = 12 : 4 By simplifying,By simplifying, 3 : 1 = 3 : 13 : 1 = 3 : 1
B
E F
CA
15
4
12
58
6
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VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS
2. BE : BA = BF : 2. BE : BA = BF : BCBC
15 : 20 = 12 : 15 : 20 = 12 : 1616
By simplifying,By simplifying,3 : 4 = 3 : 43 : 4 = 3 : 4
B
E F
CA
15
4
12
58
6
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VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS
3. BA : EA = BC : 3. BA : EA = BC : FCFC
20 : 5 = 16 : 420 : 5 = 16 : 4
By simplifying,By simplifying,4 : 1 = 4 : 14 : 1 = 4 : 1
B
E F
CA
15
4
12
58
6
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VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS
4. BE : BF = EA : 4. BE : BF = EA : FCFC
15 : 12 = 5 : 415 : 12 = 5 : 4 By simplifying,By simplifying,
5 : 4 = 5 : 45 : 4 = 5 : 4
B
E F
CA
15
4
12
58
6
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VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS
5. FC : EA = BC : 5. FC : EA = BC : BABA
4 : 5 = 16 : 204 : 5 = 16 : 20 By simplifying,By simplifying,
4 : 5 = 4 : 54 : 5 = 4 : 5
B
E F
CA
15
4
12
58
6
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VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS
6. EF : AC = BF : 6. EF : AC = BF : BCBC
6 : 8 = 12 : 166 : 8 = 12 : 16 By simplifying,By simplifying,
3 : 4 = 3 : 43 : 4 = 3 : 4
B
E F
CA
15
4
12
58
6
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VERIFYING A VERIFYING A PROPORTIONSPROPORTIONS
6. EF : AC = BE : 6. EF : AC = BE : BABA
6 : 8 = 15 : 206 : 8 = 15 : 20 By simplifying,By simplifying,
3 : 4 = 3 : 43 : 4 = 3 : 4
B
E F
CA
15
4
12
58
6
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ExercisesExercises
GIVEN: DE // BC, GIVEN: DE // BC, AD = 9, AE = 12, AD = 9, AE = 12,
DE = 10,DB = 18.DE = 10,DB = 18.Find,Find,BC, AC and CE.BC, AC and CE.
A
D E
CB
9 12
18
10
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Solution Solution Find BC,Find BC, BC : DE = BA : DABC : DE = BA : DA BC : 10 = 27 : 9 BC : 10 = 27 : 9 oror BC : 10 = 3 : 1BC : 10 = 3 : 1 Applying principle Applying principle
of proportionof proportion BC(1) = 10(3)BC(1) = 10(3) BC = 30BC = 30
A
D E
CB
9 12
18
10
30
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Solution Solution Find AC,Find AC, AC : AE = BA : DAAC : AE = BA : DA AC : 12 = AC : 12 = 27 : 927 : 9 oror AC : 12 = 3 : 1AC : 12 = 3 : 1 Applying principle Applying principle
of proportionof proportion AC(1) = 12(3)AC(1) = 12(3)AC = 36AC = 36
A
D E
CB
9 12
18
10
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Solution Solution Find CE,Find CE, CE : AE = BD : DACE : AE = BD : DA CE : 12 = 18 : 9CE : 12 = 18 : 9 oror CE : 12 = 2 : 1CE : 12 = 2 : 1 Applying principle Applying principle
of proportionof proportion CE(1) = 12(2)CE(1) = 12(2) CE = 24CE = 24
A
D E
CB
9 12
18
10
24
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Solution Solution Another way to Another way to
find CE,find CE, CE = AC - AECE = AC - AE Hence, AC =36, Hence, AC =36,
thenthen CE = 36 - 12CE = 36 - 12 CE = 24CE = 24
A
D E
CB
9 12
18
10
24
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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?
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QUIZ
2. In the figure, find the values of x and y.
y15
10
30
12
x
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Assignment EXERCISES A. Geometry Workbook, page 95