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12dm : 4 = 3dm
P = 3 * 3 = 9dm ²
Koło
12dm = 2πr / 2π
6/π = r
π to ok. 3
6/3 = r
r = 2
P = πr²
P = 4π
p = 4 * 3 = 12 dm ²
Koło jest większe
r=6πdm
P(koła)=πr²= 6²=36πdm²
O(kwadratu)=4a
4a=12dm/:4
a=3dm
P(kwadratu)=a²
P=9dm²
odp:wieksze pole ma koło