III gimnazjum

Policzmy to razem 3

Wyznaczamy wzór na długość przekątnej sześcianu. <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31516/htIemczwz1rZu3ospUBgyA%3D%3D_34bb786cc4aa0111d9e40ec279347ec5cbba82a2b7c869ca9454f55422d47b08.png"> Z twierdzenia Pitagorasa: a2+(2<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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c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
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M834 80h400000v40h-400000z'/></svg>​a)2=x2 a2+2a2=x2 3a2=x2           /<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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c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​ x=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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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>​a     Przyrównujemy znaną przekątną sześcianu do wyprowadzonego wzoru na tą przekątną: 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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M834 80h400000v40h-400000z'/></svg>​a=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
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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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>​         /: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
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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H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​ a=8 Obliczamy objętość tego sześcianu: V=a3 V=83=512​​ <section class="answer"><h4 class="answer-heading">Odpowiedź:</h4> Odpowiedź C.</section>

Rozwiązanie 1

Z treści zadania: krawędź boczna ma długość x cm, a krawędź podstawy jest o 2 cm dłuższa, zatem ma długość x+2. Graniastosłup prawidłowy czworokątny ma 4 krawędzie boczne i 8 krawędzi podstawy. Sumę tych krawędzi możemy zatem zapisać jako:

Jeśli wszystkie krawędzie czworościanu foremnego mają długość 14, to ostrosłup ten ma cztery ściany w kształcie trójkąta równobocznego. Aby obliczyć pole całkowite obliczamy pole jednej ściany i mnożymy ją razy 4 razy. Ze wzoru na pole trójkąta równobocznego: P=4a23<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>​​=4143<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>​​=41963<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>​​=493<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>​

Oznaczamy sobie promień stożka jako r a jego wysokość jako H1. Wtedy wzór na jego objętość ma postać: V=31​πr2⋅H1​ Wiemy, że promień walcy jest dwa razy większy: 2⋅r=2r

Wzory na pole powierzchni i objętość kuli: Pc=4πr2 V=34​πr2

a) Ze względu na to, że ściany boczne każdego graniastosłupa są w kształcie prostokątów(a kwadrat to również prostokąt), to każda ze ścian bocznych tego graniastosłupa będzie pokazanym na fragmencie jego siatki kwadratem, a podstawą będzie trójkąt równoboczny.  <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31532/syRBuXHo9xAtVJ76fi74TQ%3D%3D_3561bb2e02b02b9341ee2dd2106ef9046fd9bd23ece5319c7c6d44954db6a79a.png" width="372" height="461"> Pp​=4a23<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>​​=41223<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>​​=41443<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>​​=363<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>​ V=Pp​⋅H=363<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>​⋅12=4323<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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