a)
[tex]\it DC||AB \stackrel{T.f.a.}{\Longrightarrow} \Delta MAB\sim \Delta MDC\ \Rightarrow\ k=\dfrac{AB}{DC}=\dfrac{25}{10}=2,5\ (raport\ de\ asem\breve anare)\\ \\ \\ Fie\ MD=x,\ \ MC=y\\ \\ \\ \dfrac{\mathcal{P}_{MAB}}{\mathcal{P}_{MDC}}=k \Rightarrow \dfrac{9+x+25+12+y}{x+y+10}=2,5 \Rightarrow\dfrac{x+y+10+36}{x+y+10}=2,5 \Rightarrow\\ \\ \\ \Rightarrow 1+\dfrac{36}{x+y+10}=2,5\Big|_{-1} \Rightarrow \dfrac{36}{\mathcal{P}_{MCD}}=1,5 \Rightarrow \mathcal{P}_{MDC}=\dfrac{^{2)}36}{\ 1,5}=\dfrac{72}{3}=24cm[/tex]
b)
[tex]\it \dfrac{\mathcal{A}_{MAB}}{\mathcal{A}_{MDC}}=k^2=2,5^2=6,25\ \ \ \ (*)\\ \\ \\ \dfrac{p}{100}\cdor\mathcal{A}_{MAB}=\mathcal{A}_{MDC}\ \Big|_{:\mathcal{A}_{MDC}} \Rightarrow \dfrac{p}{100}\cdot\dfrac{\mathcal{A}_{MAB}}{\mathcal{A}_{MDC}}=1\ \stackrel{(*)}{\Longrightarrow}\dfrac{p}{100}\cdot6,25=1 \Rightarrow\\ \\ \\ \Rightarrow p\cdot\dfrac{6,25}{100}=1 \Rightarrow p=1\cdot\dfrac{^{4)}100}{\ 6,25}=\dfrac{400}{25}=16[/tex]
