III gimnazjum

Matematyka z plusem 3

Rozwiązanie 1

r=10cm H=10cm h1​=6cm V1​=πr2⋅H=π⋅102⋅6=600πcm3=1884cm3

Pole powierzchni walca obliczamy ze wzoru: Pc​=2πr2+2πrH a) Pc​=2⋅π⋅42+2⋅π⋅4⋅8

a) r=6 h=6 Obliczam P_b ze wzoru: Pb​=2πrH Pb​=2⋅π⋅6⋅6 Pb=72πcm2 b) Pc​=2πr(r+h) a obliczamy z tw. Pitagorasa: a2+a2=82 2a2=64 a=32<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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c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
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M834 80h400000v40h-400000z'/></svg>​=42<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
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
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M834 80h400000v40h-400000z'/></svg>​cm

V=0.1cm3 d=2⋅r=710cm r=355cm   V=Pp⋅h   V=Πr2⋅h