in ABC isoscel ducem inaltimea AD⊥BC si facem notatiile:
AD = x
BC = y
xy = 50√3, aria tr ABC
4x^2+y^2=400, pitagora in ADC
y=50√3 / x
4x^2 + 50^2 x 3/x^2 = 400
x^4 - 100x^2 + 1875 = 0
x^4 - 25x^2 - 75x^2 + 1875 = 0
x^2(x^2 - 25) -75(x^2 - 25) = 0
(x^2 - 25)(x^ - 75) = 0
(x-5)(x+5)(x-√75)(x+√75) = 0
ne intereseaza numai solutiile pozitive
x1 = 5 , AD = 5
x2 = √75 = 5√3, AD = 5√3
y1=10√3, BC = 10√3
y2 = 10, BC = 10
rezulta 2 triunghiuri:
AB=10, AC=10, BC=10√3, AD = 5 triunghi isoscel
AB=10, AC=10, BC = 10, AD = 5√3 triunghi echilateral
in ambele cazuri aria tr ABC este:
10 x 5√3 / 2 = 10√3 x 5/2 = 25√3
