Moment of inertia of a squARE lamina

ZxcvbnM2000

a square lamina is made of 4 uniform thin rods each of which was a moment of inertia Ml^2 /12 about an axis perpendicular to their length and passing through their centres.

My way of understanding it is the following : the two rods to whom the axis of rotation is perpendicular and passes through their centres have a total moment of inertia:

2*(1/12)*M*(L/2)^2 so ML^2 / 24 .

The rest two thin rods who are parallel to the axis of rotation are again rotating "forming a cylinder" and are L/2 away from the axis of rotation ( perpendicular distance ).So

the total is : 2 * ML^2/24 = ML^2 / 12 which is the moment of inertia of each rod .

Is my thinking correct ? If not could you please help me understand it ?

Thank you very much :)

Bob S

As you say, the moment of inertia I (see http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html) of the two rods where the axis of rotation passes through their center is I = M·L²/12, so for two of them it is M·L²/6.
For each rod parallel to the axis of rotation, the moment of inertia is M·(L/2)² = M·L²/4 , so for two it is M·L²/2.