III gimnazjum

Matematyka z plusem 3

Sprawdzimy, korzystając z twierdzenia odwrotnego do twierdzenia Pitagorasa - jeśli kwadrat długości najdłuższego boku trójkąta równy jest sumie kwadratów długości pozostałych boków, to trójkąt jest prostokątny   a) 132=?112+52            169=?121+25            169=121+25  , więc ten trójkąt nie jest prostokątny     b) (32<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>​)2+(23<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>​)2=?(42<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>​)2

Dla przypomnienia - długości boków w szczególnych trójkątach prostokątnych:    <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/23434/ITQ753RZp6YkJ5kQnuel4A%3D%3D_4f4b1642c86f5f780a039da5c1c51f977f4f4be540c875f0471f70f002d78a7f.jpg" width="720" height="207">   3=2b   ⇒   b=23​   ⇒   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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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​=233<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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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​​

Pole kwadratu o boku a jest równe: a⋅a=a2     PA​=(a+1)⋅(a−1)=a(a−1)+1⋅(a−1)=a2−a+a−1=a2−1=a2

A. α=180o−50o=130o

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