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d=16cm
r=16cm:2=8cm
P=πr²
P=π(8cm)²
P=64πcm²
Pole małego koła:
d=8cm
r=8cm:2=4cm
P=πr²
P=π(4cm)²
P=16πcm²
Pole zacieniowanej figury:
64πcm²-16πcm²=48πcm²
Odp.: B