Chapter 8 Similarity
232 GeometryBEHS
Mrs. Prescott
Ratios• Ratio – a comparison of 2 numbers in the
same unit of measure• Example: 2 females to 3 males– 2 to 3– 2:3–
2
3
2
Simplifying Ratios• Reduce using common factors as you would
simplify any fraction• Pages 461-462
12.
16.
22.
26.
26
11
52
22
feet
feet
5
8
20
32
20
2
oz
oz
oz
lb
2
1
2
1
2
12
ft
ft
feet
in
2
1
24
12
2
12
in
in
feet
inor
17
7
CF
BD
Proportions
• Proportion – a statement equating 2 ratios• Example:
• Also written - –3:6 = 5:10
10
5
6
3
extremes
means
Proportions• The product of the means is equal to the
product of the extremes–3:6 = 5:10
• The cross products in a proportion are always equal which is why we cross multiply to solve a proportion
10
5
6
3
310 = 65 30 = 30
31065
Solve the Proportion
6
xx
4
3
5
)3(45 xx
1245 xx12x
-4x-4x
Properties of Proportions
1. Cross Product Property – The Product of the extremes equals the
product of the means.–
• Reciprocal Property– If two ratios are equal, then their
reciprocals are also equal. –
., bcadthend
c
b
aIf
.,c
d
a
bthen
d
c
b
aIf
Page 463#56.
7
5
XV
WX
7
5
2
2
k
k
)2(527 kk
kk 10147 -7k -7k
k314 3 3
k3
14324kor
Extended Ratios• Simplify the extended ratio of the 4 sides of
the quadrilateral.• 20:16:40:36 =• 5:4:10:9 20cm
16cm
36cm
40cm
Examples:1. The perimeter of a parallelogram is 42ft, and ratio
of 2 of its unequal sides is 3:4. Find each side length.
4 sides – 3:4: 3:4 or 3x:4x:3x:4x 3x + 4x + 3x + 4x = 42 14x = 42 x = 3
2. If the extended ratio of the angles of a triangle are 5:6:7, find each angle measure.
5x:6x:7x 5x + 6x + 7x = 180
18x = 180 x = 10
Side lengths are 9 ft, 12 ft, 9 ft,
and 12 ft.
Angle measures are 50º, 60º, 70º,
Section 8.2 Problem Solving in Geometry
with Proportions
Page 465232 Geometry
BEHS, Mrs. Prescott
Properties of Proportions from Section 8.1
1.
2.
. then , Ifc
d
a
b
d
c
b
a
Additional Properties of Proportions:
3. If
4. If
Ex. ¾ =9/12, then 3/9=4/12
Ex. ¾ = 9/12, then
(3+4)/4 = (9+12)/127/4 =
21/12
Examples:
True or False.
1.
2.
Complete the statement.
3.
4. .
15
23
9 then ,
15
8
9 If m
True
False
15
m+9
Example – page 469
26. 2
7
5
9 xX + 5
-7x -7x
K J
P
S
Q
Geometry Mean - definition
The geometric mean of 2 positive numbers a and b is the positive number x such that
Geometry Mean-Example
Find the geometric mean of 3 and 48.
√ √
Geometry Mean-Example
Find the geometric mean between 6 and 15.
Section 8.3 Similar Polygons
Page 473232 Geometry
BEHS, Mrs. Prescott
Madeleine Wood
Similar Polygons
• Definition: Similar Polygons – 2 polygons such that:
1. their corresponding angles are congruent2. the lengths of their corresponding sides are
proportional• Symbol:
Similar Polygons Example:1. You can see that the corresponding angles are
congruent.2. Corresponding sides are in proportion means that
the ratios of every 2 pairs of corresponding sides are equal.
m
m
m
m
m
m
m
m
4
2
6
3
2
1
10
5
They all reduce to ½ ,
which is called the
scale factor.
Similarity Statement: ABCD~EFGHThe order of the vertices indicates which angles correspond and which segments correspond in the similarity statement.To write a proportionality statement, write the ratios of all pairs of corresponding sides.
List all pairs of congruent corresponding angles.
Write the proportionality statement.
Examples: Page 475
2.
3.
15
83
10
Rotate to make it easier to match the corresponding sides
No, because 15/10 ≠ 8/3
Yes, all corr. ∡s are ≅ and the ratios of all corr. sides are =.
Examples: Page 475. Given: TUVW~ABCD
4.
5.
∡A≅∡T, ∡B≅∡U, ∡C≅∡V, ∡D≅∡W, andTW
AD
VW
CD
UV
BC
TU
AB
5
3
15
9
70º
A
D
B
T
C
U
VW
D
9
6
23
15
Scale factor of ABCD to TUVW
Scale factor of TUVW to ABCD is 5/3
Examples: Page 475
6.
7. Find all missing segment lengths and angle measures.
5
3
TW
AD
5
36x
10
303
x
x
110TUVm70º
A
D
B
T
C
U
V70º
)70180(TUVm
W
9
6
15
23
6
70º
70º
x =10 10
y
5
3
23
y
8.13
695
y
y
110º
110º 110º
110º
Given that ΔRST~ΔJKL. find the value of x and y.
x
5
7
4
354 x75.8x
2
6
7
4
y
42)2(4 y
5.8y
4284 y
50º
70º
(2w-5)º
70º
70 + 50 + (2w – 5) = 180115 + 2w = 180
2w = 65w = 32.5
344 y
Complete each.1. Find the scale factor of the
triangles.
2. Find the lengths of the missing segment lengths.
3. Find the perimeter of each triangle.
4. Find the ratio of the perimeter of the 2 triangles.
4
1
12
3
20,16 XZYZ
48201612
12543
4
1
20
12
Similar Polygon Perimeter TheoremIf 2 polygons are similar, then the ratio of their
perimeters is equal to the ratios of their corresponding side lengths.
MNONPOMP
ABCBDCAD
MN
AB
ON
CB
PO
DC
MP
AD
thenMPONADCBIf
,~
SECTION 8.4
1. Write the statement of proportionality.
2. m∡RSV=_______, m∡U= _______3. RS = ______ RT=______ SV= ______ RV=______
Given that ΔRSV~ΔRTU, answer each of the following.
R
S
T
V
U70º
80º5
10
8
12
𝑅𝑆𝑅𝑇
=𝑅𝑉𝑅𝑈
=𝑆𝑉𝑇𝑈
70˚ 80˚
5 15
4 4
• If 2 angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar.
•Given that ∡D ≅ ∡A and ∡C ≅ ∡F, then ΔABC ~ ΔDEF.
AA ~ Postulate
A
BCF
D
E
Chapter 8- Section 8.5Proving Triangles are Similar
Page 488232 GeometryMrs. Prescott
BEHS
• If the lengths of the corresponding sides of two triangles are proportional, then the two triangles are similar.
SSS ~ Theorem
A
BCF
D
E
9
12
66
8
48
12
6
9
4
6
.~ then ,Given ABCDEFBC
EF
AC
DF
AB
DE
If an angle of one triangle is congruent to an angle of a 2nd triangle, and the lengths of the sides including these angle are proportional, then the two triangles are similar.
SAS ~ Theorem
A
BC
F
D
E
9
12
6
8
8
12
6
9,38,38 andBmEm
.~ then , andGiven ABCDEFBC
EF
AB
DEBE
38º
38º
1. AA~ Post., ΔABC ~ΔDEF
2. SAS~ Thm., ΔABC ~ΔDFE
3. Both ΔJKL and ΔMNP; SSS~
4. Ratio = 1:6, the triangles are similar by SSS~
Practice Problems: page 492 #2-5
Chapter 8- Section 8.6Proportions and Similar Triangles
Page 498232 GeometryMrs. Prescott
BEHS
A line that intersects 2 sides of a triangle is parallel to the 3rd side if and only if it divides the 2 sides proportionally.
Triangle Proportionality Theorem
R
S
T
V
U
5
10
8
4
.V then , If TUSVU
RV
ST
RS
. then ,V IfVU
RV
ST
RSTUS
and
Examples:
96
2 x
186 x3x
3
3
5
2x
103 x
3
10x
If 3 parallel lines intersect two transversals, then they divide the transversals proportionally.
Theorem:
AB
C
F
D
E
.,CE
BC
DF
ADthenFEDCABIf
Solve for x and y.Example:
84
3 x
84
5 y
6x
10y
• If a ray bisects an angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the other 2 sides.
Theorem:
F
D
E
9
12
3
4G
. DFEbisectsfFE
FD
GE
DGthenFGI
12
9
4
3
Find the value of A.Example:
A
6
8
5
485 A
6.9A