III gimnazjum

Matematyka wokół nas 3

Pp - pole podstawy h - wysokość V - objętość   Mamy: V=31​⋅Pp​⋅h

Rozwiązanie 1

a) ∣AC∣=∣AB∣2+∣BC∣2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.88em' viewBox='0 0 400000 1944' preserveAspectRatio='xMinYMin slice'><path d='M983 90
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M834 80h400000v40h-400000z'/></svg>​=2500<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>​=50

Pole powierzchni całkowitej składa się z podstawy i czterech ścian bocznych. Mamy: Pc​=Pp​+4⋅Pb​ 384=64+4⋅Pb​ 4Pb​=320

Połowa przekątnej podstawy tworzy razem z krawędzią boczną i wysokością ostrosłupa trójkąt prostokątny o kątach 30°  i 60°. Wysokość ostrosłupa (H) wynosi zatem ½ ∙ 10 = 5 cm, a połowa przekątnej podstawy - 5√3 cm. Krawędź podstawy jest zatem równa: a=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
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
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
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
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>​2⋅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
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l0 -0
c5.3,-9.3,12,-14,20,-14
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>​​=102<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>​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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l0 -0
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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>​​=56<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
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>​ cm   Policzmy jeszcze wysokość ściany bocznej (h). Wraz z połową krawędzi podstawy i krawędzią boczną tworzy trójkąt prostokątny. Z tw. Pitagorasa: h=102−(21​a)2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='3.08em' viewBox='0 0 400000 3240' preserveAspectRatio='xMinYMin slice'><path d='M473,2793
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Krawędź boczna ostrosłupa wynosi połowę długości krawędzi sześcianu, czyli ½ ∙ 12 = 6 cm.

3x - wysokosˊcˊ ostrosłupa 2x - krawędzˊ podstawy

Objętość sześcianu wynosi: V1​=33=27 cm3   Objętość ostrosłupa (o wysokości 3 cm) jest równa: V2​=31​⋅32⋅3=9 cm3