[tex]\it Fie \ \ a=2\sqrt5,\ \ b=2\sqrt3,\ \ c=2\sqrt2\ lungimile\ laturilor\ triunghiului.\\ \\ Folosim\ formula\ lui\ Heron:\ \ S=\sqrt{p(p-a)(p-b)(p-c)}\\ \\ p=\dfrac{a+b+c}{2}=\dfrac{2\sqrt5+2\sqrt3+2\sqrt2}{2}=\sqrt5+\sqrt3+\sqrt2=(\sqrt3+\sqrt2)+\sqrt5\\ \\ \\ p-a=\sqrt5+\sqrt3+\sqrt2-2\sqrt5=(\sqrt3+\sqrt2)-\sqrt5\\ \\ \\ p-b=\sqrt5+\sqrt3+\sqrt2-2\sqrt3=\sqrt5-\sqrt3+\sqrt2=\sqrt5-(\sqrt3-\sqrt2)\\ \\ \\ p-c=\sqrt5+\sqrt3+\sqrt2-2\sqrt2=\sqrt5+(\sqrt3-\sqrt2)[/tex]
[tex]\it p(p-a)= (\sqrt3+\sqrt2)^2-5;\ \ \ (p-b)(p-c)=5-(\sqrt3-\sqrt2)^2\\ \\ \\ S^2=p(p-a)(p-b)(p-c)=[(\sqrt3+\sqrt2)^2-5][5-(\sqrt3-\sqrt2)^2]=\\ \\ \\ =5(\sqrt3+\sqrt2)^2-[(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)]^2-25+5(\sqrt3-\sqrt2)^2=\\ \\ \\ 5(3-2\sqrt6+2+3-2\sqrt6+2)-25-1=5\cdot10-26=24 \Rightarrow \\ \\ \\ \Rightarrow S=\sqrt{24}=\sqrt{4\cdot6}=2\sqrt6[/tex]
Remarcă:
E de preferat să verificăm, la început, dacă triunghiul e dreptunghic
[tex]\it b^2=(2\sqrt2)^2=4\cdot2=8\\ \\ c^2=(2\sqrt3)^2=4\cdot3=12\\ \\ a^2=(2\sqrt5)^2=4\cdot5=20\\ \rule{150}{0.8}\\ \\ b^2+c^2=8+12=20=a^2\ \stackrel{RTP}{\Longrightarrow }\ \Delta\ dreptunghic, cu\ ipotenuza\ 2\sqrt5\\ \\ \mathcal{A}=\dfrac{c_1\cdot c_2}{2}=\dfrac{b\cdot c}{2}=\dfrac{2\sqrt3\cdot2\sqrt2}{2}=\dfrac{4\sqrt6}{2}=2\sqrt6\ .[/tex]
E ceva mai rapid, dar nu atât de spectaculos (!)
