Odpowiedzi

2009-12-20T13:03:14+01:00
2*π*r=4*a
a=2*π*r/4=π*r/2
pole koła
s1=π*r²
pole kwadratu
s2=a²=(π*r/2)²=π²*r²/4
s1/s2=π*r²/(π²*r²/4)=4/π
2009-12-20T13:15:54+01:00
2πr=4a => a = ½πr
pole kwadratu
P₁= (½πr)² = ¹/₄π²r²
pole koła
P₂ = πr²

stosunek

P₁/P₂ = (¹/₄π²r²)/πr² = ¹/₄π = π/4