Unformatted text preview: 6(x
6x 4)
24
6x
x
21. In order 17. EB DC, so by the Triangle Proportionality
Theorem, ED
AE BC
AB . Substitute the known measures.
2x 3 ¬6
2
2(2x 3) ¬3(6)
4x 6 ¬18
4x ¬24
x ¬6
ED 2x 3
2(6) 3 or 9
3 HG
GF 4)
60
60
3x
x
22. In order to
HG
GF ¬ x
8 10
3 10 ¬ 2y
¬ 2y 3 16y BC
FE PQ
QR
PQ
QR y ¬ 40 y
3
¬ 40 y
3
¬ 8y
3 3 PT
TS 3x and PT 3x x or 2x. 2
PT
TS , QT is not parallel to RS. 25. In order to show QT RS, we must show that
PQ
QR ¬10
y
9
DE 2y 3
2(9) 3 or 15
20. In order to have GJ FK, it must be true that
HG
HJ
GF
JK . PT
TS . Let RQ CD Chapter 6 x. Then SP 2x
x PQ
Because QR 10
1
3 or 13 3 ¬x PT
TS .
¬ 65 22 22
¬ 43
22 Let TS 24
24
9 ¬y
x
10
¬x 10
3 6
12 54.25
54.25
54.25
203
14.5 x)(x 8.5)
148.75
8.5x
26x 148.75
148.75
148.75 24. In order to show QT RS, we must show that 3 3) 8(2y 12x
12x ¬(17.5
¬17.5x
¬ x2
¬ x2
¬26x
¬14x
¬14x
¬x PT
TS .
¬ 30 9 9
9
¬ 21 or 3
7
PT
¬ 18 12 12
TS
¬ 12 or 2
6
PQ
PT
Because QR
TS , QT is not parallel to RS. y ¬ 2y 40
3 (x 3.5)(15.5 x)
x2 54.25 3.5x ¬ 7 x (x 8.5
8.5)
x
¬ 15.5 8.5x PQ
QR
PQ
QR y 8 3.5
(x 3.5)
x 3.5
17.5 x 23. In order to show QT RS, we must show that 8 10
3 x x2 10
3 8 HJ
JK . 15.5x ¬6 x 10
3
¬6x 20
¬20
¬10 8x
8x
2x
x ¬18(x 5)
¬18x 90
¬ 90
¬ 30
¬10
have GJ FK, it must be true that 21 19. BF CE and AC DF, so by the Triangle
FE
Proportionality Theorem, BC
AB
AF and
AF
CD
FE
DE . Substitute the known measures.
x HJ
JK .
x4
¬ x 15 5
18 15(x
15x
3x 18. BC ED, so by the Triangle Proportionality
CD
Theorem, BE
AB
AC . Substitute the known
measures.
20
x5
16 ¬ x 3
20(x 3) ¬16(x 5)
20x 60 ¬16x 80
4x 60 ¬80
4x ¬140
x ¬35
AC x 3
35 3 or 32
CD x 5
35 5 or 40 x
6 ¬12(8)
¬96
¬120
¬20
to have GJ FK, it must be true that PQ
QR
PT
TS x. Then PQ x
2. x
2 ¬ x or 1
2
¬ 12.9 6 8.6
8.
¬ 4.3 or 1
8.6
2
PQ PT
1
Thus, QR
TS
2 . Since the sides have
proportional lengths, QT RS. 8 4 170 ...
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Full Document
 Fall '10
 Dr.Zhan
 Calculus, PreCalculus, Harshad number, Trigraph, QR, QT RS

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