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>​cm2 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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M834 80h400000v40h-400000z'/></svg>​cm2 Odp. Pole przekroju wynosi 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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M834 80h400000v40h-400000z'/></svg>​cm2 .

Rozwiązanie 1

<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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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
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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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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  2πr.   Wzór na długość łuku okręgu o danym promieniu R  to: l=360∘α⋅2πR​

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

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