my bad, i was too focus on circle radius. I neglect crucial detail

Hi There!

Let me try helping

So, there are two things that one should know while solving questions which says a rectangle is inscribed in a circle:
1. The rectangle is also a square and thus the above said rectangle can be a square
2. It can a rectangle

In the above case it cannot be a square, because if we assume the radius to be an integer, the sides are not an integer. In that case the figure is a rectangle.

Now, as we know the figure is a rectangle, it will for a right triangle as well, indicating that we can use the pythagoras theorem as well.
The smallest triplet that I could think of os of 3-4-5 but in this case, 5 is the diagonal making 2.5 as the radius which is not an integer.
Next we can take 6-8-10 (Multiplying the above triplet by 2)

Using the above, we can get the radius and the sides as integer to be integer. Therefore 6 x 8= 48

IMO D

Note here knowing some basic concept and triplets comes in handy.

Hope this helps!

Remember: Whenever a rectangle is inscribed in a circle, the diameter and diagonal are of equal length

Diagonal = Diameter
\(\sqrt{l^2 + b^2} = 2r\)
\(l^2 + b^2 = 4r^2\)

We have been given that \(r, l\), and \(b\) are all integers.
Also, \(l^2, b^2\) and \(r^2\) should be perfect squares which can take values: 1, 4, 9, 16, 25, 36, 49, ...

When, \(r = 1\)
\(l^2 + b^2 = 4\)
No possible case

When, \(r = 2\)
\(l^2 + b^2 = 16\)
No possible case

When, \(r = 3\)
\(l^2 + b^2 = 36\)
No possible case

When, \(r = 4\)
\(l^2 + b^2 = 64\)
No possible case

When, \(r = 5\)
\(l^2 + b^2 = 100\)
\(6^2 + 8^2 = 100\) or \(8^2 + 6^2 = 100\)

Area of Rectangle = (8)(6) = 48

Hence, option D