imc

$△ADC$ can be split into a $45-45-90$ right triangle and a $30-60-90$ right triangle by dropping a perpendicular from $D$ to side $AC$. Let $F$ be where that perpendicular intersects $AC$. Because the side lengths of a $45-45-90$ right triangle are in ratio $a:a:a\sqrt{2}$, $DF=AF$. Because the side lengths of a $30-60-90$ right triangle are in ratio $a:a\sqrt{3}:2a$ and $AF+FC=5$, $DF=\frac{5-AF}{\sqrt{3}}$. Setting the two equations for $DF$ equal to each other, $AF=\frac{5-AF}{\sqrt{3}}$. Solving gives $AF=DF=\frac{5\sqrt{3}-5}{2}$. The area of $△ADC=\frac{1}{2}\cdot DF\cdot AC=\frac{25\sqrt{3}-25}{4}$. $△ADC$ is congruent to $△BEC$, so their areas are equal. A triangle's area can be written as the sum of the figures that make it up, so $[ABC]=[ADC]+[BEC]+[CDE]$. $\frac{25}{2}=\frac{25\sqrt{3}-25}{4}+\frac{25\sqrt{3}-25}{4}+[CDE]$. Solving gives $[CDE]=\frac{50-25\sqrt{3}}{2}$, so the answer is $\boxed{\frac{50-25\sqrt{3}}{2}}$