II gimnazjum

Matematyka wokół nas 2

Rozwiązanie 1

Wybieramy punkt S na środku boku (dlatego mamy 3 rozwiązania).

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31175/4PtU8AYP2BMVeqiRvnrPvw%3D%3D_83dab04f3cea2f2283da74da7cd74aeb9f3501fb758f07b750e076978bacf248.jpg" width="838" height="471"> Czworokąt ABA'B' jest rombem (ma równe boki, ponieważ trójkąty CA'B, CA'B, CBA, CAB' są przystające na mocy cechy bok-kąt-bok). Obliczamy jego pole (połowa iloczynu długości przekątnych): PABA′B′​=(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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a) FAŁSZ - w równoległoboku punkt przecięcia przekątnych jest środkiem symetrii, ale przekątne równoległoboku nie są jego osiami symetrii

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31179/JapGfgs73ajmcOZeQ/sb/w%3D%3D_7575fc5bfcfda0c1c8e089a1a1d7db476ceeefdd49328f18577ebd28b7a4fd29.jpg">
Skoro punkt D jest symetryczny do punktu G względem punktu C, to odcinki DC i CG są równe.