Do you need a hint in an area problem?

In rectangle $ABCD$ , quadrilateral $CBMH$ and quadrilateral $CDNH$ are cyclic quadrilateral because, $\angle CHM=\angle CBM=\angle CDN =90$

In quadrilateral $CBMH$ , $\angle CMB=\angle CHB$ and $\angle CBH=\angle CMH$

As $ND$ is a tangent to the circle, $\angle DNC=\angle CMN=\angle CMH=\angle CBH$

In quadrilateral $CDNH$ , $\angle DNC=\angle DHC$ and $\angle CNH=\angle CDH$

As $MB$ is tangent to the circle, $\angle CMB=\angle CNM=\angle CNH=\angle CDH$

From the above relations we get that, $angle CDH=\angle CHB$ and $\angle CHD=\angle CBH$

Therefore $\vartriangle CHD \sim \vartriangle CBH$

From this we get that, $\frac {CH}{CB}=\frac {CD}{CH}$

Therefore, ${CH}^{2}=CD.CB$

$CH=5$ is given.

Therefore, $CD.CB={5}^{2}=25$

Therefore, The area of the rectangle $ABCD$ = $25$

There are infinite number of rectangles with this condition. Since no more data is given, and if no more data is needed, we can safely assume that all have the same area.
$\therefore \text{Let us assume that MC is the diameter of the circle.}\\ \therefore~ CN \perp MN.~~~~~\implies~MN=5. \\ \text{Rectangle ABCD is made up of two equal squares and 5units is the diagonal of both.}\\ \therefore ~~ The ~area ~ABCD=2*\{\dfrac 5{\sqrt2}* \dfrac 5{\sqrt2} \}=~~~~~~~~~\Large \color{#D61F06}{25}$