A symmetric lamina of mass `M` consists of a square shape with a semicircular section over each of the edge of the square as in fig. The side of the square is `2 a`.
The moment of inertia of the lamina about an axis through its centre of mass and perpendicular to the plane is `1.6 Ma^(2)`. The moment of inertia of the lamina about the tangent `AB` in the plane of lamina is.
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Explanation

`(4.8 Ma^(2))`
Assuming symmetric lamina to be in xy plane, we will have `I_(X)= I_(Y)`. (Since the mass distribution is the same about the x-axis and y-axis).

`I_(X) = I_(Y) = I_(Z)//2` (perpendicular-axis theorem)
It is given that
`I_(L) = 1.6 Ma^(2)`
Hence, `I_(X) = I_(Y) =I_(Z)/2 = 0.8 Ma^(2)`
Now, according to the parallel-axis theorem, we get
`I_(AB) = I_(XT) + M(2a)^(2) =0.8 Ma^(2) + 4Ma^(2) = 4.8 Ma^(2)`