III gimnazjum

Matematyka z plusem 3

Rozwiązanie 1

Promień walca będzie miał długość 6, natomiast jego wysokość będzie mieć długość 4.

Objętość świeczki pozostaje taka sama, więc:    π⋅(4:2)2⋅10=π⋅(2:2)2⋅h

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/29072/vFWeTfPh4vzym6n0Qil9Xw%3D%3D_05bc8565bcb39fbd7b3def92473b39eb988efbb1356d035cc88057242707d236.jpg"> 2b=10   ⇒   b=10:2=5   ⇒   b3<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>​=53<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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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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Vwalca​=πr2⋅h

31​⋅π⋅r2⋅10=120   ∣:10

Tworząca stożka ma długość 4. Obliczamy pole boczne na 2 sposoby (jako  43​

Przed zmianą:  d1​=38 mm r1​=38 mm:2=19 mm