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bok=6
podstawa=8
bok=10
rozwiazanie:
0b = 24
(x+2) + x + (x-2) = 24
3x = 24
x=8
stąd:
1bok = 8-2 = 6
3bok = 8
2bok = 8+2= 10
odp. boki tego trojkata maja dlugosc 6,8,10.
x-2cm drugi bok
x trzeci bok
24:3=8
8+2=10cm pierwszy bok
8-2=6cm drugi bok
8cm trzeci bok