III gimnazjum

Matematyka wokół nas 3

Matematyka

h=l2−r2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.28em' viewBox='0 0 400000 1296' preserveAspectRatio='xMinYMin slice'><path d='M263,681c0.7,0,18,39.7,52,119
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<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/25099/aL73JBrH0xlZJKGv9qHOig%3D%3D_a2bce972ac76cd0a5f979af9489250fec9ed3b063f331aeb9d0d341aac65b4a9.jpg" width="408" height="341"> 246​=8r​ 24r=48

Przekrój osiowy stożka jest trójkątem równobocznym. <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/25103/3QZq39UG5f%2B/mIRJlwNbBg%3D%3D_f2f261a63bb0cb42b72f2077eda59b5021588be5d2bdfb1991a7192c9cda2125.jpg" width="408" height="341"> 2r=l r+l=15 r+2r=15 3r=15 r=5cm l=10 cm h=2103<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>​​=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
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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s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
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M834 80h400000v40h-400000z'/></svg>​cm

Przekrój osiowy stożka o trójkąt, którego podstawa jest zarazem średnicą stożka, wysokość pokrywa się z wysokością stożka, a ramiona to tworzące stożka.&nbsp; W trójkącie równobocznym boki mają jednakowe długości (oznaczyliśmy jako a), mamy także wzór na wysokość w trójkącie równobocznym, co zapisano na rysunku.&nbsp; <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/36027/AFD3S8LWs7mDqhkoYiWywg%3D%3D_79c68c40ec2907dfaa2108b553d81f6a270d9a993e05896554e485a2941818e9.jpg"> &nbsp; Wiemy, ile wynosi objętość stożka.&nbsp; V=9π3<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>​ dm3 &nbsp; Wiemy też, że objętość stożka oblicza się ze wzoru:&nbsp; V=31​πr2⋅h &nbsp; Ale u nas: r=21​a h=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>​​

Obliczamy pole powierzchni bocznej, czyli pole półkola o promieniu 12 cm:&nbsp; Pb​=21​⋅π⋅122=21​⋅144π=72π cm2 &nbsp; Promień tego półkola, czyli promień powierzchni bocznej, to zarazem tworząca stożka.&nbsp; l=12 cm &nbsp; Wiemy, że pole powierzchni bocznej możemy także obliczyć ze wzoru:&nbsp;&nbsp; Pb​=πrl

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/36028/WIOG0LYZKpyU51pwj3TG6A%3D%3D_452246df5357b571e1999e76c0c98bc081890af0ea603be6262e4a559ad2ad1e.jpg" />
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