[tex] h_{b}= \frac{8 \sqrt{3} }{2}* \frac{3}{4 \sqrt{6} }= \frac{3 \sqrt{2} }{2} [/tex]aplicam teorema cosinusului
[tex]AC^2=AB^2+BC^2-2AB*BC*cosB \\ (4 \sqrt{2})^2=2AB^2-2AB^2*(- \frac{1}{2}) \\ 32=3AB^2 \\ AB^2= \frac{32}{3} \\ AB=BC= \frac{4 \sqrt{6} }{3} [/tex]
Aflam inaltimea din virful B
BH=[tex] \sqrt{AB^2-AH^2}= \sqrt{ \frac{32}{3}-(2 \sqrt{2})^2 } = \sqrt{ \frac{32}{3}-8 }= \sqrt{ \frac{8}{3} }= \frac{2 \sqrt{6} }{3} [/tex]
Aflam celelalte inaltimi care sint congruente,utilizind formula pentru calcularea ariei unui triunghi
[tex]A= \frac{1}{2}a* h_{a} \\ A= \frac{1}{2}*4 \sqrt{2}* \frac{2 \sqrt{6} }{3}= \frac{4 \sqrt{12} }{3}= \frac{8 \sqrt{3} }{3}cm^2 \\ A= \frac{1}{2}b* h_{b} \\ \frac{8 \sqrt{3} }{2}= \frac{4 \sqrt{6} }{3}* h_{b} [/tex]
