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h=8 cm
r=?
pole koła
P1=1/4 π * 8²
P1=1/4π * 64
P1=16π cm²
pole trójkąta
P2=1/2 * 8 * 8
p2= 32 cm²
pole figury
P3=2*(P1-P2)
P3=2*(16π - 32)
P3= 32π - 64 cm²
Odp.: Pole zacieniowanej figury wynosi 32π-64cm²