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G
B
I
A
=
r
2
⋅
π
⋅
α
360
r
=
A
⋅
360
α
⋅
π
α
=
A
⋅
360
r
2
⋅
π
b
=
2
⋅
r
⋅
π
⋅
α
360
r
=
b
⋅
360
α
⋅
π
⋅
2
α
=
b
⋅
360
r
⋅
π
⋅
2
Geometrie-Kreis-Kreissektor (Grad)
A
=
r
2
⋅
π
⋅
α
360
1
2
3
4
5
6
7
8
r
=
A
⋅
360
α
⋅
π
1
2
3
4
5
6
α
=
A
⋅
360
r
2
⋅
π
1
2
3
4
5
6
b
=
2
⋅
r
⋅
π
⋅
α
360
1
2
3
4
r
=
b
⋅
360
α
⋅
π
⋅
2
1
2
3
4
α
=
b
⋅
360
r
⋅
π
⋅
2
1
2
3
4
Beispiel Nr: 01
ä
Gegeben:
Winkel
α
[
∘
]
Kreiszahl
π
[
]
Radius
r
[
m
]
Gesucht:
Fläche
A
[
m
2
]
A
=
r
2
⋅
π
⋅
α
360
Gegeben:
α
=
34
∘
π
=
3
16
113
r
=
52
m
Rechnung:
A
=
r
2
⋅
π
⋅
α
360
α
=
34
∘
π
=
3
16
113
r
=
52
m
A
=
(
52
m
)
2
⋅
3
16
113
⋅
34
∘
360
A
=
802
m
2
a
l
p
h
a
=
34
°
2
,
04
⋅
10
3
'
1
,
22
⋅
10
5
''
37
7
9
g
o
n
0
,
593
r
a
d
r
=
52
m
520
d
m
5
,
2
⋅
10
3
c
m
5
,
2
⋅
10
4
m
m
5
,
2
⋅
10
7
μ
m
A
=
802
m
2
8
,
02
⋅
10
4
d
m
2
8
,
02
⋅
10
6
c
m
2
8
,
02
⋅
10
8
m
m
2
8
,
02
a
0
,
0802
h
a
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