III gimnazjum

Matematyka z plusem 3

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/29054/tb0Xt3Jgi57T5Y0bYE2SZg%3D%3D_f130d7c9a8d35b0902a3444de7e7e458da0d65956b57c5b6908894555eb94c36.jpg"> 6=2b   ⇒   b=3   ⇒   b3<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>​=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
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>​ 6=2b   ⇒   b=3   ⇒   b3<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>​=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
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
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M834 80h400000v40h-400000z'/></svg>​ Promień stożka ma długość 3, a wysokość stożka ma długość 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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Rozwiązanie 1

α   - kąt nachylenia tworzącej do podstawy β   - kąt rozwarcia stożka   a) α=70o

Rysunek pomocniczy:  <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/29059/yWWwovWy3H96BpaWh7/GPA%3D%3D_aa1be4b7d77801b678e0c847ef6ca95e6ba9b1f257d49f1e1556e125661881b5.jpg"> Trzeci kąt w żółtym trójkącie ma także miarę 45°, ponieważ 180°-90°-45°=45°, w trójkącie o takich kątach długości boków można wyrazić tak jak pokazano na rysunku. l=12,     r=122<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>​

a) promień tego koła także wynosi 10 cm, czyli pole koła wynosi: PO​=π⋅(10 cm)2=100π cm2

Jeśli obracamy prostokąt wokół dłuższego boku, to otrzymany walec będzie miał promień 3 cm i wysokość 8 cm.

średnica wewnętrzna:  8,6 cm−3 mm−3 mm=8,6 cm−0,3 cm−0,3 cm=8 cm=0,8 dm

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