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Unformatted text preview: 21. PT SR, so P
R and T
S because they
are alternate interior angles. PQT
SQR
because they are vertical angles. Thus,
PQT
RQS.
PQ
RQ
6x
6x IH
that HG x)
x2
x2
x2
x2
x (6
18
18
18
0
0 x)(3)
3x
3x IH
HG
IH
HG 6
6
QS 3
3
The scale 23. 24. 25. x
0 or 6
x
0 or 3
3
3
factor is TQ
3x
3 0 or 1.
SQ
Triangles ABC and DFE are isosceles triangles.
A
D, and BA CA and FD ED so
BA
CA
DFE by SAS
FD
ED . Thus, ABC
Similarity.
HI JK, so GHI
GJK and GIH
GKJ
because they are corresponding angles. Thus,
GHI
GJK by AA Similarity.
m L m Q m LMQ ¬180
35 85 m LMQ ¬180
m LMQ ¬60
LMQ
NMP, so m NMP 60.
m N m P m NMP 180
m N 40 60 180
m N 80
LMQ is not similar to PMN because the
angles of the triangles are not congruent.
Since AB DE, B
E and A
D because
they are alternate interior angles. By AA
Similarity, ABC
DEC. Using the definition
AC
of similar polygons, BC
DC .
EC
x3
11x 2 6(x
6x IK x
4
5 x x or 2x. IH
2x
x or 2, and HG 30. In order to
IH
that HG
HG 3x
IH
HG IK
22
11 or 2. KL IH
HG . Thus, show that GL HK, we must show
IK
KL . Let HI x. Then IG 3x, so
x or 2x. IK
x
1
2x or 2 , and KL 18
6 or 3. So the sides are not proportional and GL HK.
31. From the Triangle Proportionality Theorem,
BC
ED
AB
AE .
Substitute the known measures.
4
6 ED
9 4(9)
36
6
32. From
BC
AB 6(ED)
6(ED)
ED
the Triangle Proportionality Theorem,
ED
AE . Substitute the known measures.
16 12 5
AE
5
AE 12
4
12 4(AE) 12(5)
4(AE) 60
AE 15
33. Since BE CD, ABE
ACD and AEB
ADC by the corresponding angles postulate.
Then ABE
ACD by the AA Similarity. Using
AD
the definition of similar polygons, CD
BE
AE . 2 Substitute the known measures.
CD
6
CD
6 4 4) (x 2)(x 5)
16 x2 5x 2x 10
16 x2 7x 10
16 x2 x 10
0 x2 x 6
0 (x 3)(x 2)
x 3 0 or x 2 0
x3
x
2
Reject x
2 because otherwise UT
2 2 or 0. So x 3. 28
7
8 or 2 . GL HK. 1
6 3 3x IK
KL 1(11x 2)
11x 2
5x 2
5x
x
26. V
UST and T
T, so by AA Similarity,
RVT
UST. Using the definition of similar
RT
VT
polygons, UT
ST .
4
2
4
2 IK
35
7
10 or 2 , and KL
IK
KL , so GL HK. 29. In order to show that GL HK, we must show
IH
IK
that HG
KL . Let KL x. Then IL 3x and 3)
18
18
20
4 2x
x
2x
x 15
5
9 or 3 . Because the side lengths are not proportional,
GL HK.
28. IL IK KL
36 28 KL
8 KL
In order to show that GL HK, we must show
IH
IK
that HG
KL . PQ 22. IK
KL .
IK
21
3
14 or 2 , and KL IH
HG TQ
SQ
3
3x (6 x)(3
6x 3x
18 3x
18 18 27. In order to show that GL HK, we must show 4(2x
8x
8x 8(CD)
8(CD)
CD x 8 4 8
12
8 6(12)
72
9 2 185 Chapter 6 ...
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This note was uploaded on 12/19/2011 for the course MAC 1140 taught by Professor Dr.zhan during the Fall '10 term at UNF.
 Fall '10
 Dr.Zhan
 Calculus, PreCalculus, Angles

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