Information Insufficient?

Let AB and AD be along x-axis and y-axis, let the coordinates of P and Q be $(h,0)$ and $(0,h)$ therefore coordinates of centre will be $(h,h$ ). (radius of circle will also be h)

Let coordinates of vertex C be (a,b) and is distance from chord PQ will be

$\displaystyle \left| \dfrac{a + b - h}{\sqrt{2}} \right| = 5$

Also

$\displaystyle h = \sqrt{ (a-h)^2 + (b -h)^2}$

Simplifing above equation to get

$\displaystyle a^2 + b^2 - 2ah - 2bh + h^2 =0$ .......... (1)

Also

$\displaystyle |a + b - h| = 5\sqrt{2}$

Squaring to get

$\displaystyle a^2 + b^2 + h^2 + 2ab - 2ah - 2bh = 50$ ............ (2)

Subtracting (2) from (1) to get

$2ab = 50 \implies \boxed{ ab = 25}$

Note : a and b are sides of rectangle

Since no other data is supplied, we can solve this question only by assuming ABCD is a SQUARE. Let O be the center of the circle radius R. M is the midpoint of the chord PQ. COMA is the diagonal. APOQ is the square with sides R and diagonals PQ and AO. It is easy to see that $MO=\dfrac{R}{\sqrt 2}, MC=R+\dfrac{R}{\sqrt 2}=R*(1+\dfrac{1}{\sqrt 2})=R*(\dfrac{\sqrt 2 +1}{ \sqrt 2} )=5..(1) \\Diagonal~~ AC=CO+OA=R+ \sqrt 2R=R(1+\sqrt 2).~~\\\therefore~AB=\dfrac{R(1+\sqrt 2)}{\sqrt 2}=R*\dfrac{1+\sqrt 2}{\sqrt 2} =5~~by (1)\\ABCD=5^2=\color{#D61F06}{\boxed {25} }\\$

So I think the wording should be ABCD is a SQUARE. Otherwise the problem has insufficient data.