III gimnazjum

Matematyka z plusem 3

Rozwiązanie 1

Promień okręgu opisanego na trójkącie prostokątnym to połowa długości przeciwprostokątnej w tym trójkącie.  Dla przypomnienia - długości boków w trójkątach prostokątnych o kątach 90°, 60°, 30° oraz 90° 45°, 45°:   <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/23434/ITQ753RZp6YkJ5kQnuel4A%3D%3D_4f4b1642c86f5f780a039da5c1c51f977f4f4be540c875f0471f70f002d78a7f.jpg" width="699" height="201">   a) 9=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>​9​=393<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
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M834 80h400000v40h-400000z'/></svg>​       2b=2⋅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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H400000v40H845.2724
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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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<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/28990/b0n9K1TKFWo4wDlQYr9Drw%3D%3D_f19fad84588c6a7b5eb485713d9dd784f555b3606b5e8082da97beee9e4baf14.jpg" width="260" height="482"> Średnica płyty to zarazem bok kwadraowego pudełka. Obliczam pole kwadratu: P□​=12⋅12=144 cm2     Sześciokąt foremny składa się z 6 jednakowych trójkątów równobocznych. Połowa średnicy płyty to wysokośc trójkąta równobocznego 21​⋅12=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>​​   12=a3<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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Obracamy cegłę tak, aby przecisnąć ją na długość, czyli aby prostokąt o wymiatach 12 cm x 6,5 cm zmieścił się w kole o średnicy 13 cm.  <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/28991/hWorikZ1iQxSSXy7sWOAYw%3D%3D_bb14cc092c58d16896f68dece3a84c584fc4c8d44670094e2bd432c1f2ddd855.jpg"> Musimy sprawdzić, czy przekątna prostokąta (czyli przeciwprostokątna w trójkącie prostokątnym) ma długość mniejszą od średnicy koła, jeśli tak, to cegła się zmieści.  Obliczamy (z twierdzenia Pitagorasa) długość przekątnej prostokąta:  122+6,52=x2

Musimy obliczyć pole koła wpisanego w sześciokąt foremny.Sześciokąt foremny o boku długości 2 m składa się z 6 trójkątów równobocznych o boku długości 2 m.  Promień koła wpisanego w sześciokąt foremny to wysokość trójkąta równobocznego.  r=223<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>​​=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
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a) 10-kąt foremny α=360o:10=36o     b) 7-kąt foremny 360o:7=(350o+10o):7=50o+710​ o=50o+173​ o=5173​ o

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