Răspuns: [tex]\color{red} \Large \boxed{\bf \dfrac{P_{1}}{P_{2}}= \dfrac{3}{4}}[/tex]
Explicație pas cu pas:
[tex]\color{blueviolet}\Large\boxed{\bf Perimetrul = 2\cdot( L+\ell )}[/tex]
[tex]\large \bf \dfrac{L_{1}}{L_{2}} =\dfrac{3}{4}\implies \boxed{\bf L_{1} = \dfrac{3\cdot L_{2}}{4} }[/tex]
[tex]\large \bf \dfrac{\ell_{1}}{\ell_{2}} =\dfrac{3}{4} \implies \boxed{\bf \ell_{1} = \dfrac{3\cdot \ell_{2}}{4} }[/tex]
[tex]\large \bf \dfrac{P_{1}}{P_{2}} =\dfrac{2\cdot (L{1}+\ell_{1})}{2\cdot (L{2}+\ell_{2})} =\dfrac{L{1}+\ell_{1}}{L{2}+\ell_{2} }=[/tex]
[tex]\large \bf \dfrac{\dfrac{3\cdot L_{2}}{4}+\dfrac{3\cdot \ell_{2}}{4}}{L{2}+\ell_{2}} =\dfrac{\dfrac{3\cdot (L_{2}+ \ell_{2})}{4}}{(L{2}+\ell_{2})}= \dfrac{3\cdot (L_{2}+ \ell_{2})}{4} : \dfrac{(L_{2}+ \ell_{2})}{1}=[/tex]
[tex]\large \bf \dfrac{3\cdot (L_{2}+ \ell_{2})}{4} \cdot \dfrac{1}{(L_{2}+ \ell_{2})}=\dfrac{3\cdot \not(L_{2}+ \ell_{2})}{4} \cdot \dfrac{1}{\not(L_{2}+ \ell_{2})}\implies[/tex]
[tex]\color{orangered}\Large\boxed{\bf \dfrac{P_{1}}{P_{2}}= \dfrac{3}{4}}[/tex]
Bafta multa!
P.S.: Te rog sa glisezi spre stânga pentru a vedea toata rezolvarea daca esti pe telefon.
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