Question

ofer coroanaa va rog ajutor! acum......

Answer 1

   
[tex]\displaystyle 7a) \\ \text{Se da: } \\ AB=5~cm \\ BC = 6 cm \\ \ \textless \ B = 30^o \\ \text{Se cere: } \\ A_{\Delta ABC} =? \\ \\ \text{Rezolvare: } \\ \\ A_{\Delta ABC} = \frac{AB\times AC\times \sin B}{2}=\frac{5\times 6\times \frac{1}{2} }{2}= \frac{30}{4}= \boxed{7,5~cm^2} [/tex]


[tex]\displaystyle 7b) \\ \text{Se da: } \\ AB=AC=BC=12~dm \\ \text{Se cere: } \\ A_{\Delta ABC} =? \\ \\ \text{Rezolvare: } \\ \\ A_{\Delta ABC} = \frac{AB^2\times \sqrt{3} }{2}= \frac{12^2\times \sqrt{3} }{2}= \frac{144 \sqrt{3} }{2}= \boxed{72\sqrt{3} ~dm^2} [/tex]


[tex]\displaystyle 7c) \\ \text{Se da: } \\ AB=4 \sqrt{6} ~m \\ AC=BC \\ \ \textless \ C = 90^o \\ \text{Se cere: } \\ A_{\Delta ABC} =? \\ \\ \text{Rezolvare: } \\ \\ A_{\Delta ABC} = \frac{AB^2}{4}=\frac{(4 \sqrt{6} )^2}{4}= \frac{96}{4}= \boxed{24~m^2}[/tex]


[tex]\displaystyle 7d) \\ \text{Se da: } \\ AB=6~cm \\ AC = 9m \\ BC = 11 m \\ \text{Se cere: } \\ A_{\Delta ABC} =? \\ \\ \text{Rezolvare: } \\ \\ P = AB+AC+BC=6+9+11 = 26~m\\ \\ p = \frac{P}{2} = \frac{26}{2} = 13\\ \\ A_{\Delta ABC} = \sqrt{p(p-AB)(p-AC)(p-BC)} = \\ =\sqrt{13(13-6)(13-9)(13-11)}= \\ =\sqrt{13 \times 7 \times 4 \times 2}= \\ 2\sqrt{13 \times 7 \times 2}=2\sqrt{13 \times 7 \times 2}=2\sqrt{13 \times 14} = \boxed{2\sqrt{182}~m^2} [/tex]