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zacienowane jest 50% ojregu czyli ½pola(p)
r=3√2cm
π=3,14
p=2π(3√2)cm²
p=2π√18²
p=2π18cm
p=36πcm²
p=36*3,14
p=113,04
1/2p=56.52cm2
Pole zacieniowanej figury wynosi 56,52cm²