Die Kugel

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O ( r ) = 4π r 2
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4 3
πr
3
V (r ) =
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O ( r ) = 4 ⋅π ⋅ r 2
O (18 ) = 4 ⋅ π ⋅182
= 1296π
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⋅π ⋅ r3
3
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=
⋅ π ⋅183
3
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d2
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Raumdiagonale :
e = a⋅ 3
Radius :
1
r =
⋅ 3⋅a
2
Nach a (= Kantenlänge des Würfels) aufgelöst:
2
a (r ) =
⋅ 3 ⋅r
3
r = 18 eingesetzt :
2
a (18 ) =
⋅ 3 ⋅18 = 12 ⋅ 3
3
Volumen des Würfels:
V (a )
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= a3
V 12 ⋅ 3
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=
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= 8.972,97 [VE ]
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