Witam mam kłopot z powtórzeniowym zadaniem z zestawu wakacyjnego.
średnica okręgu wynosi 2cm. Oblicz ple zamalowanego obszaru.
wygląda to tak:
pole zamalowanego obszaru
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silicium2002
- Użytkownik
- Posty: 786
- Rejestracja: 9 lip 2009, o 15:47
- Płeć: Mężczyzna
- Lokalizacja: Wrocław
- Podziękował: 2 razy
- Pomógł: 114 razy
pole zamalowanego obszaru
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}}\)
W okrąg wpisujemy sześciokąt foremny. Trzy jego punktu już mamy są to przeciącia tych "kawałków" okręgów z twojego rysunku. Bierzemy pod uwagę. Fragment wskazany strzałką.
Mamy tam trójkąt równoboczny o boku 1. Awięc jego pole to \(\displaystyle{ \frac{ \sqrt{3} }{4}}\) A także wycinek koła o promieniu 1 (jeśli dodamy również że tak powiem łuczek.) A jego pole to \(\displaystyle{ \frac{\pi r^{2}}{6}}\) Teraz zauważamy że różnica między wycinkiem koła a trójkątem (w sensie pól) to połowa jedenego z trzech zaznaczonych przez ciebie na różowo obszarów. Tak więc szukane pole to
\(\displaystyle{ P = 6 (\frac{\pi r^{2}}{6} - \frac{ \sqrt{3} }{4} = \pi r^{2} - 1,5 { \sqrt{3} }{4}}\)
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W okrąg wpisujemy sześciokąt foremny. Trzy jego punktu już mamy są to przeciącia tych "kawałków" okręgów z twojego rysunku. Bierzemy pod uwagę. Fragment wskazany strzałką.
Mamy tam trójkąt równoboczny o boku 1. Awięc jego pole to \(\displaystyle{ \frac{ \sqrt{3} }{4}}\) A także wycinek koła o promieniu 1 (jeśli dodamy również że tak powiem łuczek.) A jego pole to \(\displaystyle{ \frac{\pi r^{2}}{6}}\) Teraz zauważamy że różnica między wycinkiem koła a trójkątem (w sensie pól) to połowa jedenego z trzech zaznaczonych przez ciebie na różowo obszarów. Tak więc szukane pole to
\(\displaystyle{ P = 6 (\frac{\pi r^{2}}{6} - \frac{ \sqrt{3} }{4} = \pi r^{2} - 1,5 { \sqrt{3} }{4}}\)