[tex]\textbf{Formula lui Heron:}\\ \\ A_{\triangle} = \sqrt{p(p-a)(p-b)(p-c)}\\ \\p= \dfrac{a+b+c}{2}\\ \\\\ \textbf{Formula generala pentu aria unui triunghi:}\\ \\A_{\triangle} = \dfrac{\beta\cdot h}{2}[/tex]
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[tex]\textbf{Rezolvare:}\\ \\ a =BC = 13 \text{ cm},\quad b = AC = 5 \text{ cm},\quad c = AB = 12 \text{ cm}\\ \\ p = \dfrac{13+5+12}{2} = \dfrac{30}{2} = 15,\quad B = 13\\ \\\sqrt{p(p-a)(p-b)(p-c)} = \dfrac{\beta\cdot h}{2}[/tex]
[tex]\Leftrightarrow \, \sqrt{15(15-13)(15-5)(15-12)} = \dfrac{13\cdot h}{2}\\ \\ \Leftrightarrow \, \sqrt{15\cdot 2\cdot 10\cdot 3} = \dfrac{13\cdot h}{2}\\ \\ \Leftrightarrow \,\sqrt{3\cdot 5\cdot 2\cdot 2\cdot 5\cdot 3} = \dfrac{13\cdot h}{2}\\ \\ \Leftrightarrow\, \sqrt{2^2\cdot 3^2\cdot 5^2} = \dfrac{13\cdot h}{2}\\ \\ \Leftrightarrow \,2\cdot 3\cdot 5 = \dfrac{13\cdot h}{2}\\ \\ \Leftrightarrow\, 13\cdot h = 30\cdot 2\\ \\ \Rightarrow \, \boxed{h = \dfrac{60}{13}\text{ cm}}[/tex]
