# PROPORTIONS SSS SIMILARITY PARALLEL TRANSVERSALS PROBLEM 1 PROBLEM 2

## Presentation on theme: "PROPORTIONS SSS SIMILARITY PARALLEL TRANSVERSALS PROBLEM 1 PROBLEM 2"— Presentation transcript:

PROPORTIONS SSS SIMILARITY PARALLEL TRANSVERSALS PROBLEM 1 PROBLEM 2
STANDARDS 4 and 5 PROPORTIONS SSS SIMILARITY PARALLEL TRANSVERSALS PROBLEM 1 PROBLEM 2 PROBLEM 5 AA SIMILARITY JOINING MIDPOINTS IN A TRIANGLE SAS SIMILARITY PROBLEM 6 ARE POLYGONS SIMILAR? PROBLEM 3 PARALLEL TO SIDE OF TRIANGLE PROBLEM 4 END SHOW PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Students prove basic theorems involving congruence and similarity.
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. PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

In a classroom there are 3 boys and 6 girls
In a classroom there are 3 boys and 6 girls. What is the ratio of boys to girls? . 3 3 6 1 2 = 3 to 6 or 3:6 or 5 2 1.0 (0.5) (100%) = 50% =0.5 - 1 0 STANDARD 1.3 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

A A to B A : B B C C to D C : D D A C = B D
How do you express the ratio of A to B? A B A to B A : B How do you express the ratio of C to D? C D C to D C : D Now when we equal to ratios, we get a PROPORTION: A B D C = PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

The product of the MEANS is equal to the product of the EXTREMES
= B D B C B and C are the MEANS A D A and D are the EXTREMES = A D B C = Cross-multiplying: (A)(D)=(C)(B) The product of the MEANS is equal to the product of the EXTREMES PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

STANDARDS 4 and 5 If A K and B L and C M then Triangles are SIMILAR when the corresponding sides are proportional: SSS similarity A CA AB BC B C LM MK KL KL AB MK CA LM BC OR K M CAB MKL Both triangles are similar ( ) L PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

The triangles below are similar, find CA=? and TS=?
6 4 A C T X+6 S 2X+3 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

The triangles below are similar, find CA=? and TS=?
RT B BC 6 4 A C S T CA X+6 TS 2X+3 4 6 X+6 2X+3 CA = X + 6 = 12 4(2X+3) = 6(X+6) = 18 8X +12 = 6X + 36 TS = 2X + 3 8X = 6X + 24 = 2( ) + 3 12 -6X -6X = 2X = 24 = 27 X = 12 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

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 JK X+2 RS X+6 KL 10 ST 20 J R K S L T 5X+2 = JL = 5X + 2 = 5( ) + 2 2 20(X+2) = 10(X+6) = 20X +40 = 10X + 60 = 12 20X = 10X + 20 -10X -10X 10X = 20 X = 2 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

X A Y Z C B if A X and C Z then ABC XYZ By AA similarity

X XY A AB AC Y Z C B if and A X then ABC XYZ By SAS similarity XZ

First Prove that all triangles in the figure are similar among them:
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. E S F R G T U H First Prove that all triangles in the figure are similar among them: 1. Two lines cut by a common perpendicular transversal are parallel. 2. A line perpendicular to one line is perpendicular to any line parallel to it. 3. Two perpendicular lines form 4 right angles. 4. Alternate interior angles are congruent. 5. Corresponding angles are congruent STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

All highlighted triangles in figure are similar by AA SIMILARITY! H
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. E S F R G T U All highlighted triangles in figure are similar by AA SIMILARITY! H STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

All highlighted triangles in figure are similar by AA SIMILARITY! H
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. 25 15 3 E S F R G T U All highlighted triangles in figure are similar by AA SIMILARITY! H RF UR = EG GF = 3 15 RF 25 15RF = (3)(25) 15RF = 75 RF = 5 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Applying the Pythagorean Theorem EG GF RF = UF + UR
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. 3 E S F R G T U 5 H RF UR = Applying the Pythagorean Theorem EG GF RF = UF + UR 2 = 3 15 RF 25 5 = UF + 3 2 25 = UF + 9 2 15RF = (3)(25) 15RF = 75 UF = 16 2 UF = 4 RF = 5 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Applying the Pythagorean Theorem EG GF RF = UF + UR 25RS = (15)(15) =
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. 25 15 10 E S F R G T U 5 H RS RT = GF EG RF UR = 5+10 25 RS 15 = Applying the Pythagorean Theorem EG GF RF = UF + UR 2 25RS = (15)(15) = 3 15 RF 25 5 = UF + 3 2 25RS = 225 25 = UF + 9 2 15RF = (3)(25) RS = 9 15RF = 75 UF = 16 2 UF = 4 RF = 5 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

Applying the Pythagorean Theorem EG GF RF = UF + UR 25RS = (15)(15) =
If EG= 25, GF=15, EF= 20, FT = 10, UR= 3, and Given EG RT. Find RF, UF, and RS. E S F R G T U H RS RT = GF EG RF UR = 5+10 25 RS 15 = Applying the Pythagorean Theorem EG GF RF = UF + UR 2 25RS = (15)(15) = 3 15 RF 25 5 = UF + 3 2 25RS = 225 25 = UF + 9 2 15RF = (3)(25) RS = 9 15RF = 75 UF = 16 2 UF = 4 RF = 5 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

A AD AE E D DB EC B C If DE BC then STANDARDS 4 and 5

A AD AE E D DB EC B C If then DE BC STANDARDS 4 and 5

Find the value for X B 6 S TA SB 18 = CT CS C X 6 (24) (24) = 40 24 24
(24) (24) = (6)(24) 18 X = 144 18 X X = 8 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

C BC B AB A EF DE E F D STANDARDS 4 and 5

Find the values for Y and Z: Y Y+5 = 6 7

Find the values for Y and Z: Y Y+5 = 6 7
30 +5 Y = 30 Z Z+1 = Y Y+5 30 Z 30 Z+1 30+5 = Z 30 Z+1 35 = 35Z = 30(Z+1) 35Z = 30Z +30 -30Z -30Z 5Z = 30 Z = 6 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

K S R L M and RS= LM 1 2 If KR RL KS SM and then RS LM

K S R M L Find in the problem below the value for RS: 10 1 7 LM If
120 RS = LM 1 2 If then RS = (120) 1 2 RS = 60 STANDARDS 4 and 5 PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

The two irregular polygons are similar find values for X and Y: 30 98