1. 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
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2. 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.
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3. 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
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4. 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 =
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5. 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
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6. 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
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7. The triangles below are similar, find CA=? and TS=?6 4 A C T
X+6 S
2X+3
STANDARDS 4 and 5
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8. 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
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9. 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
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10. X A Y Z C B if A X and C Z then ABC XYZ By AA similarity
STANDARDS 4 and 5
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11. X XY A AB AC Y Z C B if and A X then ABC XYZ By SAS similarity XZ
STANDARDS 4 and 5
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12. 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
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13. 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
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14. 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
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15. 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
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16. 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
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17. 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
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18. A AD AE E D DB EC B C If DE BC then STANDARDS 4 and 5
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19. A AD AE E D DB EC B C If then DE BC STANDARDS 4 and 5
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20. 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
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21. C BC B AB A EF DE E F D STANDARDS 4 and 5
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22. Find the values for Y and Z: Y Y+5 = 6 7
STANDARDS 4 and 5
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23. 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
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24. K S R L M and RS= LM 1 2 If KR RL KS SM and then RS LM
STANDARDS 4 and 5
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25. 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
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26. The two irregular polygons are similar find values for X and Y: 30 9810 Y X 15
STANDARDS 4
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27. The two irregular polygons are similar find values for X and Y: 30 9810 10 Y X X Y 15 15 = X 98 10 15 = 98 15
(30) (30) =
X = 10 15 98 10 =
(15)(30) 10 Y X =
450 10 Y
Y = 45
STANDARDS 4 and 5
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