Dane:
x - wysokość w trójkącie ABC opuszczona na podstawę AB, zał. x > 0
I AB I = x + 3 - długość postawy na którą opuszczona jest wysokość x
PΔ ABC = 35 cm²
[tex]P\Delta _{ABC} =\dfrac{1}{2} \cdot x \cdot (x+3)\\\\P\Delta _{ABC} =\dfrac{1}{2}x^{2} +\dfrac{3}{2} x~~\land~~P\Delta _{ABC} =35~cm^{2} \\\\\dfrac{1}{2}x^{2} +\dfrac{3}{2} x=35~~\mid ~~\cdot 2\\\\x^{2} +3x=70\\\\x^{2} +3x-70=0\\\\a=1,~~b=3,~~c=-70\\\\\Delta=b^{2} -4ac\\\\\Delta=3^{2} -4\cdot 1\cdot (-70)\\\\\Delta =9+280\\\\\Delta =289\\\\\sqrt{\Delta} =\sqrt{289} =\sqrt{17^{2} } =17\\\\x_{1} =\dfrac{-b-\sqrt{\Delta} }{2a} ~~\lor~~x_{2} =\dfrac{-b+\sqrt{\Delta} }{2a}\\\\\\[/tex]
[tex]x_{1} =\dfrac{-3-17}{2\cdot 1} ~~\lor~~x_{2} =\dfrac{-3+17}{2\cdot 1} \\\\x_{1} =\dfrac{-20}{2} ~~\lor~~x_{2} =\dfrac{14}{2} \\\\(~~x_{1} =-10~~\lor~~x_{2} =7~~)~~\land~~x > 0~~\Rightarrow~~x=7~cm\\[/tex]
[tex]\mid AB \mid = x + 3~cm~~\land~~x=7~cm~~\Rightarrow ~~\mid AB \mid = 10~cm[/tex]
Odp:
Długość podstawy AB wynosi 10 cm.
