Explicație pas cu pas:
a) notăm AC ∩ BD = {E}
ducem înălțimea: CF ⊥ AB, F ∈ AB, CF = 30 cm
ABCD trapez isoscel => ΔAEB isoscel
=>AE ≡ BE
T.P.: AE² + BE² = AB² <=> 2BE² = AB²
[tex]\bf AB = BE \sqrt{2} \: cm \\ [/tex]
în ΔABC: BE×AC = CF×AB
[tex]BE \cdot AC = CF \cdot AB \iff BE \cdot AC = 30 \cdot BE \sqrt{2} \\ \implies \bf AC = 30 \sqrt{2} \: cm[/tex]
b)
[tex]\frac{DC}{AB} = \frac{2}{3} \implies \frac{Aria_{\triangle MDC}}{Aria_{\triangle MAB}} = {\Big( \frac{2}{3} \Big)}^{2} = \frac{4}{9} \\ \frac{Aria_{\triangle MDC}}{Aria_{\triangle MAB} - Aria_{\triangle MDC}} = \frac{4}{9 - 4} \\ \implies \bf \frac{Aria_{\triangle MDC}}{Aria_{\triangle MAB} - Aria_{ABCD}} = \frac{4}{5} \\ [/tex]
