III gimnazjum

Matematyka wokół nas 3

Moneta jednozłotowa ma kształt walca o średnicy podstawy długości 2,2 cm. Oznacza to, że promień podstawy ma 1,1 cm (promień jest dwa razy krótszy od średnicy). Grubość monety (jej wysokość) wynosi 1,5 mm=0,15 cm. Obliczamy jaką objętość ma ta moneta (przyjmujemy, że π=3,14).  V=π⋅1,12⋅0,15=3,14⋅1,21⋅0,15=0,56991   [cm3]   Moneta ta jest wykonana ze stopu miedzi i niklu o gęstości 8,25 g/cm3, czyli 1 cm3 waży 8,25 g.  Obliczamy ile waży jedna moneta.  0,56991⋅8,25 g=4,7017575 g   Jedna moneta waży 4,7017575 g. Chcemy wybić monety o wartości 1000 zł, czyli potrzebujemy 1000 takich monet jednozłotowych.    Obliczamy ile waży 1000 monet jednozłotowych.  4,7017575 g⋅1000=4701,7575 g=4,7017575 kg≈4,7 kg   1000 monet jednozłowotych waży około 4,7 kg.  Odpowiedź: Potrzeba około 4,7 kg stopu, aby wybić monety o wartości 1000 zł.

Rozwiązanie 1

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M834 80h400000v40h-400000z'/></svg>​=6 r=3cm h=6cm P=2⋅32Π+2⋅3Π⋅6

Objętosˊcˊ walca:  V=πr2⋅h   V=216πcm3

1,25=2Πr r=Π0,625​

V1​=Πr2h

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M834 80h400000v40h-400000z'/></svg>​cm

36Π=Πr2 r2=36 r=6cm Przekrój poprzeczny stożka jest trójkątem równobocznym o boku 12cm. l=12 h=2123<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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<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/25104/RaJfTgnACGB/b02UX4aUQw%3D%3D_4819d4199d737fa41466570f36ab06e2099da8527b037d37acd3231a2ad713dd.jpg" width="640" height="380"> ∣∠CBA∣=30o 2a=c b=2c3<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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M834 80h400000v40h-400000z'/></svg>​​ 23<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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M834 80h400000v40h-400000z'/></svg>​=21​ab 23<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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Rysunek przekroju osiowego stożka:  <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/30609/z7TMEmAcqb4GQs8WDbkBMQ%3D%3D_627985898ffa3539dd45013e630ebd9cb2f88d6f056a135ef5d368fa60980117.png"> Jest to trójkąt równoramienny zatem jego wysokość jest również dwusieczną kąta prostego. Wysokość opada na podstawę pod kątem prostym zatem kąt przy podstawie będzie miał miarę 45o. Stąd wiemy, że: h=r       Pole dużego trójkąta wynosi 32 cm2, zatem wysokość dzieli ten trójkąt na dwa trójkąty o równych polach. Stąd : P=16   21​⋅h⋅r=16   h⋅r=32

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