28. Given:
ABC
S
is the midpoint of
AC
.
T
is the midpoint of
BC
.
Prove:
ST
1
2
AB
Midpoint
S
is
b
2
0
,
c
2
0
or
b
2
,
2
c
.
Midpoint
T
is
a
2
b
,
0
2
c
or
a
2
b
,
2
c
.
Proof:
ST
¬
a
2
b
b
2
2
2
c
2
c
2
¬
a
2
2
0
2
¬
a
2
2
or
a
2
AB
¬
(
a
0)
2
(0
0)
2
¬
a
2
0
2
or
a
ST
¬
1
2
AB
29. Given:
ABD
,
FBD
AF
6,
BD
3
Prove:
ABD
FBD
Proof:
BD
¬
BD
by the Reflexive Property.
AD
¬
(3
0)
2
(1
1)
2
9
0 or 3
DF
¬
(6
3)
2
(1
1)
2
9
0 or 3
Since
AD
DF
,
AD
DF
.
AB
(3
0)
2
(4
1)
2
9
9 or 3
2
BF
(6
3)
2
(1
4)
2
9
9 or 3
2
Since
AB
BF
,
AB
BF
.
ABD
FBD
by SSS.
30. Given:
BPR
PR
800,
BR
800
Prove:
BPR
is
an isosceles
right triangle.
Proof:
Since
PR
and
BR
have the same measure,
PR
BR
.
The slope of
PR
8
0
00
0
0
or 0.
The slope of
BR
8
8
0
0
0
0
8
0
00
, which is undefined.
PR
BR
, so
PRB
is a right angle.
BPR
is an
isosceles right triangle.
31. Given:
BPR
,
BAR
PR
800,
BR
800,
RA
800
Prove:
PB
BA
Proof:
PB
(800
0)
2
(800
0)
2
or
1,280,000
BA
(800
1600)
2
(800
0)
2
or
1,280,000
PB
BA
, so
PB
BA
.
32. Given:
JCT
Prove:
JCT
is a right triangle.
Proof:
The slope of
JC
3
5
0
0
0
0
0
0
or
3
5
.
The slope of
TC
5
3
0
0
0
0
0
0
or
5
3
.
The slope of
TC
is the opposite reciprocal of the
slope of
JC. JC
TC
, so
TCJ
is a right angle.
JCT
is a right triangle.
33.
Use the Distance Formula to find the distance
between
J
(
500, 300) and
T
(300, 500).
JT
¬
[300
(
500)]
2
(500
300)
2
¬
680,000
¬
824.6
The distance between Tami and Juan is
680,000
or approximately 824.6 feet.
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 Fall '10
 Dr.Zhan
 Calculus, PreCalculus, Cartesian Coordinate System, Right triangle, Hypotenuse, FBD

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