III gimnazjum

Matematyka wokół nas 3

Rozwiązanie 1

Koło wpisane w sześciokąt foremny ma promień równy wysokości trójkąta równobocznego, z których składa się ten sześciokąt. Mamy zatem (a - bok sześciokąta foremnego): 9=2a3<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>​​ a3<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>​=18 a=18⋅33<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

Krótsza przekątna tworzy razem z dłuższym ramieniem i podstawą dolną trójkąt prostokątny. Mamy zatem: 8 cm - przekątna trapezu 16 cm - podstawa dolna 8√3 cm - dłuższe ramię   Kąt między dłuższym ramieniem i dolną podstawą wynosi 30°. Zatem z własności trapezu kąt między tym ramieniem i górną podstawą wynosi 180 - 30 = 150°. Z tego wynika, że kąt między krótszą przekątną i górną podstawą wynosi 150 - 90 = 60°. Krótsza przekątna z górną podstawą i krótszym ramieniem(a także wysokością trapezu) tworzy zatem trójkąt prostokątny o kątach 30° i 60°. Mamy więc: 8 cm - przekątna trapezu 4 cm - górna podstawa 4√3 cm - krótsze ramię   Obwód trapezu wynosi: O=4+43<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>​+83<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>​+16=20+123<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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M834 80h400000v40h-400000z'/></svg>​ cm=4(5+33<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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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​) cm

6 cm - długosˊcˊ jednego boku

2,4x - długość jednej przyprostokątnej x - długość drugiej przyprostokątnej   Z tw. Pitagorasa: x2+(2,4x)2=262 x2+5,76x2=676

Sprawdźmy, czy zachodzi tw. Pitagorasa:   A. 1,52+62=2,25+36=49=72   B. (0,82<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>​)2+(1,52<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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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​)2=0,64⋅2+2,25⋅2

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