2 szkoły ponadpodstawowej

II liceum

II technikum

Matematyka 1. Zakres podstawowy i rozszerzony. Po gimnazjum

Rozwiązanie 1

Oznaczmy długości przyprostokątnych tego trójkąta przez 5x i 12x, korzystając z twierdzenia Pitagorasa wyrazimy długość przeciwprostokątnej y za pomocą x. (5x)2+(12x)2=y2 25x2+144x2=y2

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/32486/v17kYG1w7x8R5JqyqInalQ%3D%3D_4164bb6267aefd2602f53de12043e68adb4ccf76ab15cf6acd2f61110594f7b4.jpg" width="294" height="278"> Zamalowany trójkąt jest równoramienny - jego ramiona to promienie okręgu.  Szukana odległość środka okręgu od cięciwy to zarazem wysokość zamalowanego trójkąta.

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/32487/pY00Q6IF%2BTjdKygABGNURw%3D%3D_3d3f687c7404dd34cd5cbd537124aa7683100b28b48899d91f637e4e5fa308ce.jpg" width="229" height="231"> a2+a2=d2 a2⋅2=d2 d=a2⋅2<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<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/32488/ixUjdDUMuzPRnWxsGfhN5g%3D%3D_034ffe20d680a81a9bf8a166e8c55a617ab3c85252d960de6fdb578471dc7684.jpg" width="375" height="216"> (x−1)2+112=x2 x2−2x+1+121=x2   ∣−x2

Będziemy korzystać z twierdzenia 3.    a) (2,5)2+62=6,25+36=42,25=6,52   ⇒   prostokątny

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