[tex]\it a)\ \ \mathcal{A}=\dfrac{c_1\cdot c_2}{2}=\dfrac{12\cdot3\sqrt3}{2}=18\sqrt3\ cm^2\\ \\ \\ c)\ \mathcal{A}=\dfrac{\ell^2\sqrt3}{4}=\dfrac{53^2\sqrt3}{4}=\dfrac{2809\sqrt3}{4}cm^2[/tex]
[tex]\it b)\ \ Vom\ nota\ triunghiul\ ABC,\ cu \ AB=AC=7cm,\ BC=6cm\\ \\ \mathcal{A}=\sqrt{p(p-a)(p-b)(p-c)};\ \ p=\dfrac{a+b+c}{2}=\dfrac{6+7+7}{2}=10\\ \\ \\ p-a=10-6=4;\ \ p-b=p-c=10-7=3\\ \\ \\ \mathcal{A}=\sqrt{10\cdot4\cdot3\cdot3}=\sqrt{10\cdot4\cdot9}=2\cdot3\cdot\sqrt{10}=6\sqrt{10}\ cm^2[/tex]
