Răspuns:
Explicație pas cu pas:
1)
P = 6 ⇔ 3l = 6 → l = 6 : 3 = 2 cm
[tex] \mathcal A = \frac{2 {}^{2} \sqrt{3} }{4} = \frac{ \not4 \sqrt{3} }{ \not4} = \boxed{ \sqrt{3} cm {}^{2} }\\ [/tex]
_________________
2)
[tex]h = 3 \sqrt{3} \Leftrightarrow \frac{l \sqrt{3} }{2} = 3 \sqrt{3} \Rightarrow l = \frac{2}{ \sqrt{\not3} } \times 3 \sqrt{\not3} = 2 \times 3 = 6cm \\ [/tex]
[tex] \mathcal A = \frac{6 {}^{2} \sqrt{3} }{4} = \frac{ \not36 \sqrt{3} }{ \not4} = \boxed{9 \sqrt{3} cm {}^{2} } \\ [/tex]
