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l=α/360°*2πr π≈3 /kreska ułamkowa
l=60°/360°*2*3*15
l=6°*90
l=540 - długość łuku
P=α/360°*πr²
P=60°/360°*3*15²
P=6°*3*225
P=4050 pole wycinka koła
b)
l=α/360°*2πr
l=135°/360°*2*3*20
l=135°/360°*120
l=135°/3°
l=45
P=α/360°*πr²
P=135°/360°*3*20²
P=135°/360°*3*400
P=135°/360°*1200
P=162000/360
P=450
c)
l=α/360°*2πr
l=200°/360°*2*3*15
l=200°/360°*90
l=200°/4°
l=50
P=α/360°*πr²
P=200°/360°*3*15²
P=200°/360°*3*225
P=200°/360°*675
P=135000/360°
P=375