1 szkoły ponadpodstawowej

I liceum

I technikum

MATeMAtyka 1. Zakres podstawowy. Po gimnazjum

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Rozwiązanie 1

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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
M834 80h400000v40h-400000z'/></svg>​)

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
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
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
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​​,   a

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.