III gimnazjum

Matematyka wokół nas 3

Okrąg wpisany w trójkąt równoboczny ma promień równy ¹/₃ wysokości tego trójkąta, czyli: 31​⋅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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Rozwiązanie 1

Szesˊciokąt foremny wpisany w okrąg ma bok roˊwny promieniowi okręgu, czyli a. Jego pole wynosi: P=6⋅4a23<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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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
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Rysunek pomocniczy: <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/23413/E1SCDTuD%2BwJqMEBIrKJDZw%3D%3D_55b4cf0e98b7d11973fe71d6170869f059e5c4cc9aa2728810aa6bfa7ba66aa8.png" width="238" height="237"> Szukamy boku a wpisanego trójkąta. Ponieważ trójkąt prostokątny o bokach a, x i 6-x ma kąty 30° i 60°, to możemy zapisać równanie: 6−x=2x 3x=6

a)  O=2Π⋅4=8Πcm b)  O=2Π⋅5=10Πdm

a) d=3,1412,56​=4 cm

Na stoˊł w kształcie koła o sˊrednicy 1 m potrzeba:

a) nie O=π⋅2,15≈3,14⋅2,15=6,751 cm&gt;6,5 cm

Obwoˊd koła wynosi:

a)   2⋅π⋅1=2π cm

Promienˊ zewnętrznego obwodu wynosi:

a)   P=π⋅32=9π cm2 b)