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

Assuming symmetric lamina to be in `xy` plane we will have

`I_(x)=I_(y)`(since the mass distribution is the mass about the `x`-axis and `y`-axis)
`I_(x)=I_(y)=I_(z)` (perpendiclar axis theorem)
it is given that
`I_(z)=1.6Ma^(2)`
Hence `I_(x)=I_(y)=(I_(z))/2=0.8Ma^(2)`
Now according to the paralel axis theorem, we get
`I_(AB)=I_(XT)M(2a)^(2)`
`=0.8Ma^(2)+4Ma^(2)=4.8Ma^(2)`