INMO Problem!

Let $P$ be the foot of perpendicular from $C$ to $MN.$
Since $BM$ is a tangent to the circle $,$
$\angle DNC=\angle NMC$
$\Rightarrow sin\angle DNC=sin\angle NMC$
$\Rightarrow \frac{DC}{CN}=\frac{CP}{CM}$
$\Rightarrow \frac{DC}{CP}=\frac{CN}{CM}$
Similarly $\frac{CB}{CP}=\frac{CM}{CN}$
Multiplying these two relations,we get $DC\times CB=CP^2=5^2=25;i.e.,$ the area of the rectangle is $\boxed{25 sq. units}.$

Let $O$ be the center of circle $(CMN)$ and $Q$ the midpoint of $MN$ . Since the circle is tangent to $AB$ and $AD$ at $M$ and $N$ respectively, we have $\angle MON=90^\circ \Rightarrow \angle MCN=45^\circ \Rightarrow \angle QCM=22.5^\circ$ and $\angle MCB=22.5^\circ$ . Since $\angle MQC=\angle MBC=90^\circ$ , it follows that triangles $MQC$ and $MBC$ are congruent. Hence, $BC=5$ , Area $= 5^{2} = 25 units^{2}$