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).&nbsp; V=π⋅1,12⋅0,15=3,14⋅1,21⋅0,15=0,56991   [cm3]

36<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>​=6 r=3cm h=6cm P=2⋅32Π+2⋅3Π⋅6

Objętosˊcˊ walca:  V=πr2⋅h &nbsp; V=216πcm3

1,25=2Πr r=Π0,625​

V1​=Πr2h

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/25095/hpcBS8EC87x/foC9Uj5z4g%3D%3D_ea49d50725a74dc2bedda09fb9f079f222b8a50bcbc6bf1f3fdc00787ae324a8.jpg" width="419" height="480"> ∣∠CBA∣=30o h=21​⋅12=6cm r=122−62<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.28em' viewBox='0 0 400000 1296' preserveAspectRatio='xMinYMin slice'><path d='M263,681c0.7,0,18,39.7,52,119
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M834 80h400000v40h-400000z'/></svg>​=108<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>​=63<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>​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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M834 80h400000v40h-400000z'/></svg>​​=63<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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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​cm

<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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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​=21​⋅21​c⋅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>​​∣:3<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: &nbsp;<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 &nbsp; &nbsp; &nbsp; Pole dużego trójkąta wynosi 32 cm2, zatem wysokość dzieli ten trójkąt na dwa trójkąty o równych polach. Stąd : P=16 &nbsp; 21​⋅h⋅r=16 &nbsp; h⋅r=32

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