Răspuns: [tex]\bf A_{\Delta ABC}=2\sqrt{6}[/tex]
Explicație pas cu pas:
(●'◡'●) Salutare!
Folosim formula lui Heron fiindcă stim toate laturile
[tex]\bf Aria_{\Delta ABC} =\sqrt{p\cdot(p-a)\cdot(p-b)\cdot(p-c)}~, unde\\\\a,b,c~sunt~cele~3~laturi,~iar~p~este~semiperimetrul\\\\p=\dfrac{a+b+c}{2}[/tex]
[tex]\bf p= \dfrac{2\sqrt{3} +2\sqrt{2} +2\sqrt{5} }{2} =\dfrac{2(\sqrt{3} +\sqrt{2} +\sqrt{5})}{2} \implies p = \sqrt{3} +\sqrt{2} +\sqrt{5}[/tex]
[tex]A=\sqrt{(\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{3}+\sqrt{5}-2\sqrt{2})(\sqrt{2}+\sqrt{3}+\sqrt{5}-2\sqrt{3})(\sqrt{2}+\sqrt{3}+\sqrt{5}-2\sqrt{5})}[/tex][tex]A=\sqrt{(\sqrt{2}+\sqrt{3}+\sqrt{5})(-\sqrt{2}+\sqrt{3}+\sqrt{5})(\sqrt{2}+\sqrt{5}-\sqrt{3})(-\sqrt{5} +\sqrt{3} +\sqrt{2})}[/tex]
[tex]A=\sqrt{[(\sqrt{3}+\sqrt{5})^{2}-2](\sqrt{2} +\sqrt{5}-\sqrt{3})(-\sqrt{5} +\sqrt{3} +\sqrt{2})}[/tex]
[tex]\bf A=\sqrt{(3+2\sqrt{15}+5-2)\cdot [2-(\sqrt{5}-\sqrt{3})^{2}]}[/tex]
[tex]\bf A=\sqrt{(6+2\sqrt{15})\cdot [2-(8-2\sqrt{15})]}[/tex]
[tex]\bf A=\sqrt{(6+2\sqrt{15})\cdot (2-8+2\sqrt{15})}[/tex]
[tex]\bf A=\sqrt{(6+2\sqrt{15})\cdot (-6+2\sqrt{15})}[/tex]
[tex]\bf A=\sqrt{(2\sqrt{15})^{2}-6^{2}}[/tex]
[tex]\bf A=\sqrt{4\cdot15-36}[/tex]
[tex]\bf A=\sqrt{60-36}[/tex]
[tex]\bf A=\sqrt{24}[/tex]
[tex]\boxed{\bf A=2\sqrt{6}}[/tex]
Formule de calcul prescurta folosite:
(a + b)(a – b) = a² – b²
sau
a² – b² = (a + b)(a – b)
==pav38==
