II gimnazjum

Matematyka wokół nas 2

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31309/MRqd9iGhseSlR9i%2Bscg7XQ%3D%3D_23837bf906c75ffbf1c5958b833e8a69a531bc92b08c992e8ef93c33f1508c58.jpg" width="522" height="477"> Obliczmy długość wysokości ostrosłupa korzystając z twierdzenia Pitagorasa dla zamalowanego trójkąta:&nbsp; 102+H2=202 &nbsp; 100+H2=400   ∣−100

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31310/QskVHyXnQlPewX0SsUpoHQ%3D%3D_7eb7bb334b8b4e8b40aee71b22eb9ea1cb8572d21eb45a3936084caadfa6bdca.jpg"> Ten trójkąt jest prostokątny równoramienny, więc możemy długości ramion obliczyć z twierdzenia Pitagorasa.&nbsp; x2+x2=122 &nbsp; 2x2=12⋅12   ∣:2 &nbsp; x2=612

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31312/oQ3RdUuTMwoZbsfbbe2i9A%3D%3D_b0c0015eaac0f566b19b4ba29c2b030b3a5de367d5a40856abb9bad149f87857.jpg"> Przypomnijmy sobie, jakie długosˊci bokoˊw ma troˊjkąt o kątach 90o,60o,30o. <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/22341/oYpmszt038tlbWOAz7Plg%3D%3D_032fc89220bce6c83d37bbabab70a48243742857de1e66e57d578c102bdbba6f.jpg"> Zastosujmy to dla troˊjkąta ODS: ∣DS∣=2x=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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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>​ &nbsp; ∣OD∣=x=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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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=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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l0 -0
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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>​ &nbsp; ∣OS∣=x3<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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s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
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H400000v40H845.2724
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M834 80h400000v40h-400000z'/></svg>​=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
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
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>​⋅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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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>​=4⋅3=12 &nbsp; &nbsp; Podstawą ostrosłupa prawidłowego troˊjkątnego jest troˊjkąt roˊwnoboczny.

Nie, ponieważ nie wszystkie romby o boku 6 cm mają równe pola. Pole rombu obliczamy jako połowę iloczynu długości przekątnych.&nbsp; <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31315/84NopaEzR9ccijB3RY35Qw%3D%3D_3469c9e4970ffc11957141f904e71b6bcb0efa7adc4bd9b21b56bc3d4d3d2185.jpg" width="541" height="262"> Odcinki x i y mogą mieć różne długości.&nbsp; Pokażmy dwa przykłady rombów o bokach 6 cm, które mają różne pola.&nbsp; &nbsp; &nbsp; pierwszy &nbsp; x=2 &nbsp; Obliczamy y z twierdzenia Pitagorasa 22+y2=62 &nbsp; 4+y2=36   ∣−4 &nbsp; y2=32

a) Oznaczmy P1​ - pole podstawy graniastosłupa, P2​ - pole podstawy ostrosłupa Wiemy, z˙e objętosˊci obu brył są roˊwne, więc moz˙emy zapisacˊ: P1​⋅H=31​⋅P2​⋅H   ∣:H

Czworosˊcian foremny ma 4 sˊciany, kaz˙da z nich jest takim samym troˊjkątem roˊwnobocznym. &nbsp; Policzmy pole jednej sˊciany. 643<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>​ cm2:4=163<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
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>​ cm2 &nbsp; &nbsp; Oznaczmy długosˊcˊ boku tego troˊjkąta jako a. Znamy wzoˊr na pole troˊjkąta roˊwnobocznego o boku a, więc moz˙emy zapisacˊ:

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31319/43mmnLXjqm1I42Su7K3tmA%3D%3D_8c059621850aa22a6606b5fdcc2b2bade77e2de1a53275b100cdfbfec863d72d.jpg"> x to 32​ wysokosˊci podstawy, czyli wysokosˊci troˊjkąta roˊwnobocznego o boku a. x=32​h=32​⋅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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H400000v40H845.2724
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M834 80h400000v40h-400000z'/></svg>​​=3a3<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>​​

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