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y - drugi bok
x+4 - dlugosc przekatnej
z Twierdzenia Pitagorasa
(x+4)^2= X^2 + y^2
x^2 + 8x + 16 = x^2 + y^2
8x + 16 = y^2
8x= (y^2 - 16) |:8
x = (y^2 - 16)/8
Ob = 17
Ob = (x+y)*2
2([(y^2 - 16)/8] + y ) = 17 | *4
y^2 + 8y - 84 = 0
delta = 400
pierwiastek z delty = 20
y1 = - 14 nie nalezy do dziedziny
y2 = 6 nalezy do dziedziny
x = (y^2 - 16)/8
x= (36 - 16)/8
x= 2,5
P=x*y
P=2,5 *6
P=15