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b) trójkąt równoramienny
DE=6
DF=EF
DF≈6.7
O=6+2*6.7=19.4
c)
GI≈3.9
GH≈8.3
HI≈5.4
O=3.9+8.3+5.4=17.6
|EF|=[(1-4)^2+(4+2)^2]^0,5=[9+36]^0,5=45^0,5=3*5^0,5
|DF|=[(3)^2+(6)^2]^0,5=45^0,5=3*5^0,5
OB=6 + 3*5^0,5+3*5^0,5=6+6*5^0,5
b) |GH|=[(4+4)^2+(-1+1)^2]^0,5=[64]^0,5=8
|HI|=[(-1-4)^2+(3+1)^2]^0,5=[25+16]^0,5=41^0,5
|GI|=[(-1+4)^2+(3+1)^2]^0,5=[9+16]^0,5=5
OB= 8+ 41^0,5+5=13+41^0,5
x^2 tzn x do potegi 2
x^0,5 tzn pierwiastek z x