III gimnazjum

Matematyka z plusem 3

a) Bryła ta składa się z dwóch walców. Pole obliczamy ze wzoru: Pc​=Pp​+Pb​   Pp​=πr2 Pb​=2πrH Duży walec: Pp​=π⋅42=16π Pb​=2⋅π⋅4⋅5=40π

Rozwiązanie 1

Pb​=21​πrH⋅2 Pb​=21​π⋅24​⋅5⋅2=10π

a) V=πr2H H_1=20cm=2dm H_2=25cm=2,5dm r=πH1​V​ r1​=3,14⋅21​=0,16dm=1,6cm

x2+x2=(102<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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H400000v40H845.2724
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h=8 2r=10π⇒r=5π

V=πr2⋅H