1 sss similarity aa similarity sas similarity parallel to side of triangle parallel transversals are...

1

SSS SIMILARITY

AA SIMILARITY

SAS SIMILARITY

PARALLEL TO SIDE OF TRIANGLE

PARALLEL TRANSVERSALS

ARE POLYGONSSIMILAR?

PROBLEM 1

PROBLEM 2

PROBLEM 3

PROBLEM 4

PROBLEM 5

STANDARDS 4 and 5

END SHOW

JOINING MIDPOINTSIN A TRIANGLE

PROBLEM 6

PROPORTIONS

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

http://www.mrperezonlinemathtutor.com/

2

Standard 4:

Students prove basic theorems involving congruence and similarity.

Los estudiantes prueban teoremas básicos que involucran congruencia y semejanza.

Standard 5:

Students prove triangles are congruent or similar and are able to use the concept of corresponding parts of congruent triangles.

Los estudiantes prueban que triángulos son congruentes o semejantes y son capaces de usar el concepto de partes correspondientes de triángulos congruentes.



3

STANDARD 1.3

In a classroom there are 3 boys and 6 girls. What is the ratio of boys to girls?

3 to 6 or 3:6 or36

12

=

=0.5 (0.5)-1 0

5

0

2 1.00

....

33

(100%)= 50%



4

D

C=

A

B

How do you express the ratio of A to B?

A to B A : BA

B

How do you express the ratio of C to D?

C to D C : DC

D

Now when we equal to ratios, we get a PROPORTION:



5

AD

A and D are the EXTREMES

BC

B and C are the MEANS

Cross-multiplying:

(A)(D)=(C)(B) The product of the MEANS is equal to the product of the EXTREMES

=

A

DBC

=

A

DBC

=



6

A

BC

K

L

M

KLAB

MK

CALMBC

If A K and B L and C M thenSTANDARDS 4 and 5

Triangles are SIMILAR when the corresponding sides are proportional: SSS similarity

CAB MKL

Both triangles are similar ( )

KLAB

MKCA

LMBC

OR



7

46

X+6 2X+3A S

BR

C T

The triangles below are similar, find CA=? and TS=?

STANDARDS 4 and 5



8

4

6

X+6

2X+3

4

BC

6

RT

X+6

CA

2X+3

TS

STANDARDS 4 and 5

4(2X+3) = 6(X+6)

8X +12 = 6X + 36-12 -12

A S

BR

C T

8X = 6X + 24-6X -6X

2X = 242 2X = 12

CA = X + 6

= + 6

= 18

TS = 2X + 3

= 2( ) + 312

12

= 24 + 3

= 27

The triangles below are similar, find CA=? and TS=?



9

J R

K

S

L T

JK

X+2

X+2RS

X+6

X+6

KL

10

10

ST

20

20

STANDARDS 4 and 5

What is the value of X and JL if JKL RST. JK= X+2, KL=10, RS=X+6,

ST= 20, and JL = 5X + 2

= =

20(X+2) = 10(X+6)

20X +40 = 10X + 60-40 -40

20X = 10X + 20-10X -10X

10X = 2010 10

X = 2

JL = 5X + 2

= 5( ) + 22

= 10 + 2

= 12

5X+2



10

A

B C

X

Y Z

ABC XYZ

By

AA similarity

STANDARDS 4 and 5

andA Xif C Z then



11

A

B C

X

Y Z

ABXY

ACXZ

and A Xif then ABC XYZ

By SAS similarity

STANDARDS 4 and 5PRESENTATION CREATED BY SIMON PEREZ. All rights reserved


12

E

S

F

R

G

T

U

STANDARDS 4 and 5

2. A line perpendicular to one line is perpendicular to any line parallel to it.

First Prove that all triangles in the figure are similar among them:

1. Two lines cut by a common perpendicular transversal are parallel.

3. Two perpendicular lines form 4 right angles.

4. Alternate interior angles are congruent.

5. Corresponding angles are congruent

H

If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS.



13

E

S

F

R

G

T

U

STANDARDS 4 and 5

H


All highlighted triangles in figure are similar by AA SIMILARITY!



14

E

S

F

R

G

T

U

STANDARDS 4 and 5


25

3

15

=UR

GF

RF

EG

=3

15

RF

25

15RF = (3)(25)

15 15

15RF = 75

RF = 5

HAll highlighted triangles in figure are similar by AA SIMILARITY!



15

E

S

F

R

G

T

U

STANDARDS 4 and 5


3

=UR

GF

RF

EG

=3

15

RF

25

15RF = (3)(25)

15 15

15RF = 75

RF = 5

H

5

RF = UF + UR2 2 2

-9 -9

5 = UF + 32 22

25 = UF + 92

UF = 162

UF = 4

Applying the Pythagorean Theorem



16

E

S

F

R

G

T

U

STANDARDS 4 and 5


=UR

GF

RF

EG

=3

15

RF

25

15RF = (3)(25)

15 15

15RF = 75

RF = 5

H

5

RF = UF + UR2 2 2

-9 -9

5 = UF + 32 22

25 = UF + 92

UF = 162

UF = 4


10

25

=RT

EG

RS

GF

=5+10

25

RS

15

25RS = (15)(15)

25 25

25RS = 225

RS = 9

15



17

E

S

F

R

G

T

U

STANDARDS 4 and 5


=UR

GF

RF

EG

=3

15

RF

25

15RF = (3)(25)

15 15

15RF = 75

RF = 5

H

RF = UF + UR2 2 2

-9 -9

5 = UF + 32 22

25 = UF + 92

UF = 162

UF = 4


=RT

EG

RS

GF

=5+10

25

RS

15

25RS = (15)(15)

25 25

25RS = 225

RS = 9



18

A

B C

D E

DBAD AE

ECDE BC

STANDARDS 4 and 5

thenIf



19

A

B C

D E

DBAD AE

ECDE BC

STANDARDS 4 and 5

thenIf



20STANDARDS 4 and 5

A

S

B

C

T

6

2440

X

18

Find the value for X

=SB

CS

TA

CT

= 6

18

X

24(24) (24)

=(6)(24)

18X

= 144

18X

X = 8



21

A

B

C

D E F

DEEF

ABBC



22STANDARDS 4 and 5

Z

Y

6

7

Y + 5Z + 1

Find the values for Y and Z:=

Y+5

7

Y

67Y = 6(Y+5)

7Y = 6Y + 30

-6Y -6Y

Y = 30


23STANDARDS 4 and 5

Z

Y

6

7

Y + 5Z + 1

Find the values for Y and Z:=

Y+5

7

Y

67Y = 6(Y+5)

7Y = 6Y + 30

-6Y -6Y

Y = 30

30

30 +5

ZY

Z+1

Y+5

35Z = 30(Z+1)

35Z = 30Z +30-30Z -30Z

5Z = 305 5Z = 6

=

Z30

Z+1

30+5=

Z30

Z+1

35=


24

K

L M

R S

If KR RL KS SMand then RS LM and RS= LM12


25STANDARDS 4 and 5

K

L M

R S7

7

10

10

120

Find in the problem below the value for RS:

then

RS = (120)1

2

RS = 60

RS = LM1

2If


27

X

X

10

10

10

1515

15

Y

Y

30

30

The two irregular polygons are similar find values for X and Y:

STANDARDS 4 and 5

=

=

X 14.8

(30) (30)

=(15)(30)

10Y

=450

10Y

Y = 45

98

98

9898X

9810

15=

X =10

1598


1 sss similarity aa similarity sas similarity parallel to side of triangle parallel transversals are...

Documents