1 szkoły ponadpodstawowej

I liceum

I technikum

MATeMAtyka 1. Zakres podstawowy. Po gimnazjum

Rozwiązanie 1

a) 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/22340/f6gfU1KyPAE0gXAZKqnvuw%3D%3D_b83fbd8cdf0f4eafda0aefebce679a097ea552834fc1724fc7984f6d2ec61058.jpg"> a)   b+b+b2<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>​)   b(2+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>​)=6(1+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+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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H400000v40H845.2724
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
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Oznaczmy miary kątów ostrych tego trójkąta przez a i 2a: a+2a+90o=180o   ∣−90o 3a=90o   ∣:3 a=30o 2a=2⋅30o=60o

Trójkąt ABC to trójkąt o kątach 90°, 60°, 30°.  Oznaczmy długości jego boków:

W trójkącie prostokątnym o kątach 90°, 60°, 30° boki mają długość:  21​a,   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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Trójkąty ABE oraz ACD to prostokątne trójkąty równoramienne.  ∣AB∣=∣BE∣=2,5

Odcięte trójkąty są prostokątne, ponieważ wszystkie kąty kwadratu są proste.  Wiemy, że powstał ośmiokąt foremny, więc z każdej strony odcięto po równo, czyli odcięte trójkąty są także równoramienne.

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