Oblicz długości łuk i ple powierzchni wycinka,gdy dane są: a). jedna ósma koła o promieniu 20dm. B).Jedna szesnasta koła o średnicy 50m.
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a)
L = (1/8)*2 pi* r = (1/4) pi* 20 dm = 5 pi dm
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P = (1/8) pi*r^2 = (1/8) pi * [ 20 dm]^2 = (1/8) *400 pi dm^2 = 50 pi dm^2
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b)
d = 2 r = 50 m
r = 25 m
L = (1/16)*2 pi*r = (1/8) pi * 50 m = 6,25 pi m
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P = (1/16) pi*r^2 = (1/16) pi* ( 50 m)^2 = (1/16)*2 500 pi m^2 = 156,25 pi m^2
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zad 2 1/8 * πr^2
pole wycinka 1/8* (20)^2= 400/8=50π [dm]^2 1/8 to tak jak 45/360 45 stopni kat wycinka na 8 czesci
luk dlugosc luku to 1/8 2πr = 40/8=5π[dm] dlugość < ósma cześć calego luku
zad 3
pole wycinek o kacie α a lukiem = 22,5 stopnia
1/16 *π r^2=2500/16 =156,25*π[m]^2 ≈490,6m^2 powierzchni
dlugosc luku = 2πr=2*50m*π*1/16=100/16=6,25mπ ≈ 19,6m dlugosci luku