Bonsoir
Aire IJKL = 100
AireIJKL = [ (x-4)(y-2) ] / 2
[(x-4)(y-2)] = 100
(x-4)(y-2) = 200
(y-2) = 200 /(x-4)
y = (200/x-4) +2
y = [2(x-4)+200] / (x-4)
y = (2x+192) / (x-4)
comme L(x) = DB alors L(x) = y - 2
L(x) = [(2x+192)/(x-4) ) - 2
L(x) = [2x+192-2(x-4)]/(x-4)
L(x) = 200/(x-4)
3)
On veut que DB = 2 * AC
comme DB = 200/(x-4) et AC = x - 4
200/(x-4) = 2(x-4)
2(x-4)(x-4) = 200
2(x²-4x-4x+16) = 200
2x²-16x-168 = 0
delta = 1600 alors Vdelta = 40
deux solutions mais une seule est positive donc
x = (16+40)/4 = 14