Question

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Answer 1

Răspuns: $\bf A_{\Delta ABC}=2\sqrt{6}$

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(●'◡'●) Salutare!

Folosim formula lui Heron fiindcă stim toate laturile

$\bf Aria_{\Delta ABC} =\sqrt{p\cdot(p-a)\cdot(p-b)\cdot(p-c)}~, unde\\\\a,b,c~sunt~cele~3~laturi,~iar~p~este~semiperimetrul\\\\p=\dfrac{a+b+c}{2}$

$\bf p= \dfrac{2\sqrt{3} +2\sqrt{2} +2\sqrt{5} }{2} =\dfrac{2(\sqrt{3} +\sqrt{2} +\sqrt{5})}{2} \implies p = \sqrt{3} +\sqrt{2} +\sqrt{5}$

$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})}$$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})}$

$A=\sqrt{[(\sqrt{3}+\sqrt{5})^{2}-2](\sqrt{2} +\sqrt{5}-\sqrt{3})(-\sqrt{5} +\sqrt{3} +\sqrt{2})}$

$\bf A=\sqrt{(3+2\sqrt{15}+5-2)\cdot [2-(\sqrt{5}-\sqrt{3})^{2}]}$

$\bf A=\sqrt{(6+2\sqrt{15})\cdot [2-(8-2\sqrt{15})]}$

$\bf A=\sqrt{(6+2\sqrt{15})\cdot (2-8+2\sqrt{15})}$

$\bf A=\sqrt{(6+2\sqrt{15})\cdot (-6+2\sqrt{15})}$

$\bf A=\sqrt{(2\sqrt{15})^{2}-6^{2}}$

$\bf A=\sqrt{4\cdot15-36}$

$\bf A=\sqrt{60-36}$

$\bf A=\sqrt{24}$

$\boxed{\bf A=2\sqrt{6}}$

Formule de calcul prescurta folosite:

(a + b)(a – b) = a² – b²  

            sau  

a² – b² = (a + b)(a – b)

==pav38==

Answer 2

Răspuns:

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