[tex]\it \dfrac{\mathcal{P}_1}{\mathcal{P}_2}=0,(3)=\dfrac{\ \ 3^{(3}}{9}=\dfrac{1}{3}= k\ (raport\ de\ asem\breve anare)\\ \\ \\ \dfrac{\mathcal{P}_1}{\mathcal{P}_2}=\dfrac{1}{3} \Rightarrow \mathcal{P}_1 < \mathcal{P}_2 \Rightarrow \mathcal{A}_1 < \mathcal{A}_2\\ \\ \\ \dfrac{\mathcal{A}_1}{\mathcal{A}_2}= k^2 \Rightarrow \dfrac{5}{\mathcal{A}_2}= \dfrac{1}{9} \Rightarrow \mathcal{A}_2=5\cdot9=45\ cm^2[/tex]