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Transcript
Geometric Similarities
Math 416
Geometric Similarities
Time Frame







1)
2)
3)
4)
5)
6)
7)
Similarity Correspondence
Proportionality (SSS) (Side-side-side)
Proportionality (SAS) (side-angle-side)
Similarity Postulates
Deductions
Dimensions
Three Dimensions
Similarity Correspondence
 Similarity – Two shapes are said to be
similar if they have the same angles
and their sides are proportional
 Note – we see shape by angles & we
see size with side length
Consider
A
8
100
Similar & Why?
D
95
24
X
16
100
15
W
95
10
85
80
B
5
80
32
C
Y
85
20
Z
Proportionality (SSS)
 We say the two shapes are similar
because their angles are the same
and their sides are proportional
 We can note corresponding points
AX
DW
BY
CZ
Angles





We note corresponding angles
< ADC = < XWZ (95°)
< DCB = < WZY (85°)
< CBA = < ZYX (80°)
< BAD = < YXW (100°)
Notes






Hence we would say ADCB
XWZY
Hence we note corresponding angles
< ADC = < XWZ
< DCB = < WZY
< CBA = < ZYX
< BAD = < YXW
Proportionality
 Next is proportionality which we will
state as a fraction
 AD=8 DC=16 CB=32 BA=24
XW 5 WZ 10 ZY 20 YX 15
 What is the proportion (not in a
fraction)?
 8/5 which is reduced to 1.6
Question #1
 Identify the similar figures and state
the similarity relationship, side
proportion and angle equality
A
BIG
T
MED
MED
B
Z
BIG
SMALL
C
SMALL
C
Notes for Solution
 By observing you need to establish
the relationship.
 Look at angles or side lengths
 Important: An important trick
when comparing angles and sides
is that the biggest angles is
always across the biggest side,
the smallest from the smallest
and medium from the medium.
Solution #1
 Triangle ABC ˜ TCZ
 AB = BC
= CA
TC
CZ
ZT
 < ABC = < TCZ
< BCA = <CZT
< CAB = <ZTC
Important Note
 Make sure the middle angle letters
are all different because the middle
letter is the actual angle that you are
looking at.
 AC = CA
 < ACB = < BCA
 Both the above are the same
Question #2
R
K
MED
MED
SMALL
Q
SMALL
T
MED
MED
L
X
With isosceles (or equilateral triangles) you
may get two (or three) different answers).
However, you are only required to provide
one.
Solution a for #2
 The question is to identify similar figures
and state the similarity relationship, side
proportion and graph equality.
QKT ˜
RXL
 QK = KT = QT
RX XL RL
 < QKT = < RXL
< KTQ = < XLR
< TQK = < LRX
Solution b for #2
 You can also have another solution
 Triangle QKT is still congruent to RLX
 QK = KT = QT
RL
LX
RX
 < QKT = < RLX
< KTQ = < LXR
< TQK = XRL
More Notes 
 There are other ways of establishing
similarity in triangles
 At this point we will abandon reality for
simple effective but not accurate drawings
of triangles… (it is not to scale).
 Please complete #1 a – o
 For Question #3, again, state similarity
relationship, side proportion and angle
equality.
Question #3
T
A
18
B
21
35
Q
45
30
R
27
C
If the three sides are proportional to
the corresponding three sides in the
other triangle, the two will be similar.
Solution Notes
 You need to check…
 SMALL with SMALL
 MEDIUM with MEDIUM
BIG with BIG
Solution #3
ABC ˜

QRT
Small
Small
Med
Med Big
18 = 21 = 27
30
0.6
35
Big
45
= 0.6 = 0.6; YES SIMILAR
Proportionality SAS
 We can also show similarity in
triangles if we can find two set
corresponding sides proportional and
the contained angles equal; we can
determine similarity
X
A
18
14°
15
42
14°
Y
B
C
35
Z
Question #4
Show if the triangle is similar
Solution… since
<BAC = <XYZ = 14°
18 = 42
15 35
= 6/5 6/5

BAC ˜
XYZ
 Notice BAC = Small, Angle, Big &
compared to Small, Angle, Big




Triangle Similarity Postulates
 There are three main postulates we use to
state similarity
 SSS all corresponding sides proportional
 SAS two sets of corresponding sides and
the contained angle are equal
 AA two angles (the third is automatically
equal since in a triangle, the interior angle
must add up to 180°) are equal
Example #1
 Why are the following statements
true?

QPT ˜
ZXA
X AA
Q
84°
42°
54°
P 84° 54° T
A
Z
Example #2
 Why are the following statements
true?
M
SAS

KTR ˜
PMN
R
18
27
51°
T
24
K
N
51°
16
P
Solution: since 24/16 = 27/18
Example #3
K
A
16
B
12
24
C
32
T
6
9
Since 16 = 24 = 32
6
9
12
8/3 = 8/3 = 8/3
SSS
P
Parallel Lines
 Facts: If two tranversals intersect
three parallel lines, the segments
between the lines are proportional
a
c
b
d
Therefore, a = c
b
d
Notes
 Also note that…
C
B
A
•
•
BC = 1
•
AC
2
Parallel and the Triangle
 If a parallel line to a side of a triangle
intersects the other two sides it
creates two similar triangles
A
B
C
Therefore,
ABE ˜ ACD
E
D
Question #1
3
4
9
x
3= 9
4
x
3x = 36
Find x
x = 12
Question #2
A
We note BE // CD
Thus,
x
B
40
C
50
ABE ˜
ACD
AB = BE = AE
E
AC
CD
AD
x = 50 = AE
150
D
x+40 150 AD
Question #2 Sol’n Con’t
 We only need x = 50
x+40 150
 150x = 50(x+40)
 150x = 50x + 2000
 100x = 2000
 x = 20
Proportion Ratio
 Consider
1
10
1
10
Dimensions
SIDE
SMALL
BIG
RATIO
SIDE 1D
1
10
1:10 or
1/10
AREA 2D
1
100
1:100 OR
1/100
Dimensions
 In general in 1D if a:b
then in 2D a2 : b2
 Ex. In 1D if 5:3 then in 2D?
 In 2D then 25:9
 You can go backwards by using
square root
 Ex. In 2D if 36:49 then in 1D
 6:7
3D or Volume
 Consider
1
5
1
1
5
5
3D or Volume
Side 1D
Small Big
Ratio
1
5
1:5 or 1/5
125
1:125 or 1/125
Volume 3D 1
3D or Volume
In general in 1D if a:b then in 3D…
Then in 3D a3 : b3
Ex. In 1D if 6:7 then in 3D
In 3d 216:343
You can go backwards by using the
cube root
 Ex. In 3D if 27:8 then in 1D
 In 1D 3:2





Practice
Complete the following
1 D Length
2 D Area 3 D Volume
2:9
4:81
8:729
2:11
4:121
8:1331
5:3
25:9
125:27
3D Question #1
 Two spheres have a volume ratio of
64:125. If the radius of the large one
is 11cm, what is the radius of the
small one?
3D Ratio 64:125
 Big
Small
1D Ratio 4:5
 r 11
x
5x = 44
x = 8.8
Thus 4 = x
5
11
3D Question #2
V=?
V=200m3
A
Base
=
100m2
A
base
= 16 m2
Question #2 Solution
Big Small
Area of Base 100 16
Volume
200
x
Thus 200 = 1000
x
64
1000x = 12800
x = 12.8
1 Ratio 10 / 4
2 Ratio 100/16
3 Ratio 1000/64