III gimnazjum

Matematyka wokół nas 3

Przekrój osiowy tego stożka jest trójkątem równobocznym o boku 10 cm. P=41023<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>​​=41003<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>​​=253<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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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

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/24173/xd3AdkBgJfjWNhTGC/F8mA%3D%3D_4332e0a9bda59b0482ecaff2dc2f8094270055a97f56d649780eb04385452b25.jpg" width="640" height="380"> ∣∠CBA∣=30o h=6 r=122−62<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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c340,-704.7,510.7,-1060.3,512,-1067
l0 -0
c4.7,-7.3,11,-11,19,-11
H40000v40H1012.3
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c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1
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c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z
M1001 80h400000v40h-400000z'/></svg>​=144−36<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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M834 80h400000v40h-400000z'/></svg>​=108<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>​=63<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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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

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/24176/ak7PoaKP34Ho5zW9OC9o/w%3D%3D_7aa0228c8546d9debcbc4d11deff5ced18fccfb8ed2c2e6dfc578f33c0bce7c4.jpg" width="518" height="432"> r12​=l2−h2 r22​=(21​l)2−(21​h)2

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/25092/NHKPMiJqPTlT0s7xBrmWpw%3D%3D_376e13651b7119736f08f492a45cfad938dbd4fb8af3eeff437fe6ba75157f07.jpg" width="131" height="480"> P=21​⋅2⋅2+21​⋅2⋅3=5[cm2]

2Π⋅5cm=10Πcm=31cm

Pb​=31​⋅92Π=27Πcm2

Narysujmy najpierw, jak wygląda wycinek kołowy: <img class="img-responsive img-thumbnail dropzone-img" src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/160663/gCHPDCbfypdP2YZVLPc%2BtQ%3D%3D_5e9a4fa721b0162d4597f28e13f89192a4e2455f99a9d80676dfe50a4c507169.jpg" width="464" height="264"> Wiemy, że łuk jest obwodem okręgu w podstawie stożka i jest równy&nbsp; 2πr. &nbsp; Wzór na długość łuku okręgu o danym promieniu R &nbsp;to: l=360∘α⋅2πR​

Przypomnijmy, że wzór na długość łuku wycinka kołowego obliczamy ze wzoru: l=360∘α⋅2πR​, &nbsp;&nbsp;gdzie&nbsp; α &nbsp;to kąt środkowy wycinka, a R &nbsp;to promień wycinka. Przekształćmy wzór i wyliczmy z niego wzór na&nbsp; α, &nbsp;bo to w zadaniu będziemy chcieli obliczyć. l=360∘α⋅2πR​ /⋅2πR360∘​ &nbsp; α=2πRl⋅360∘​ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Wiemy, że długość łuku jest równa obwodowi podstawy stożka, czyli wynosi 2πr, &nbsp; gdzie&nbsp; r

Rysunek pomocniczy: <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/66709/QsHLTaalLLBGW%2BZk3l7YjQ%3D%3D_311d89a3c6c75e30ac9263a259b8ea20ffbd1ee265dec49f8c2bbb3230486ce5.jpg"> Będziemy podstawiać do wzoru na&nbsp; α &nbsp;wyznaczonego w poprzednim zadaniu. Przekrój osiowy jest trójkątem równobocznym, zatem: R=a &nbsp; r=2a​

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