[tex]Aplicam~teorema~lui~Pitagora~in~triunghiul~\Delta ABD-~dreputnghic \\ in~D: \\ \\ AD^2+BD^2=AB^2 \Rightarrow AD= \sqrt{AB^2-BD^2}= \sqrt{25^2-20^2}= \\ \\ =\sqrt{625-400}= \sqrt{225}=15~(cm). \\ \\ d(D,AB)= \frac{AD \cdot BD}{AB}= \frac{20 \cdot 15}{25}=12~( cm). \\ \\ A_{ABCD}= AB \cdot d(D,AB)=25 \cdot 12=300~(cm^2)[/tex]
