Moment of inertia for thin uniform rectangle/hexagon

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Homework Statement


the first problem is i have a rectangle split into four triangles the long side labeled a the short side labeled b and the inner angles of the triangles are [tex]\theta[/tex] [tex]_{1}[/tex] and [tex]\theta[/tex] [tex]_{2}[/tex]
(since the traingles are made from a rectangle their are 2 sets of triangles with different [tex]\theta[/tex])
also not [tex]\theta[/tex] is the angle towards the center of the mass not an outer ridge angle(all together 4 [tex]\theta[/tex] representing all 4 triangles)



Homework Equations




im given the eqaution
I=moment of inertia= [tex](1/12)[/tex]M(a[tex]^{2}[/tex] + b[tex]^{2}[/tex])
and also earlier in class worked out for isosceles trianlges
I=([tex]1/2)[/tex]M[1+([tex]1/3[/tex])tan[tex]^{2}[/tex]((1/2)[tex]\theta[/tex])]h[tex]^{2}[/tex]

The Attempt at a Solution


given those 2 equations i have to prove that I=[tex](1/12)[/tex]M(a[tex]^{2}[/tex]b[tex]^{2}[/tex]) really gives the moment of inertia for full mass.
I started working backwards on the problems replacing M with [tex]\sigma h^{2} tan((1/2)\theta[/tex])
being that [tex]\sigma = M/A[/tex] and A = h[tex]^{2}[/tex][tex]\ tan((1/2)\theta[/tex])
btw M=total mass of object, A=total area
(sorry trying my best to not confuse on problem)
since the height of one triangle is 1/2b or 1/2a depending on which triangle picked
i also replaced the terms for a[tex]^{2}[/tex] and b[tex]^{2}[/tex]

but I am at a lost at this point and don't know where to go past this
i figured if i reverse engineered the problem i may be able to figure the 4 triangles relationship to the total moment of inertia so then i could return the sumed up eqaution into the original eqaution for an isosceles triangle so i can understand how to use the triangles for a hexagon and octagon.

please help if possible with this problem.
 
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When you specify a moment of inertia for an object, you need to also specify the axis about which the moment is calculated.

You basically need to write down h and [itex]\theta[/itex] in terms of the sides a,b, and use the fact that m(triangle) = (M/ab)*(area of triangle). If necessary you may need to use the Parallel Axis theorem to have all moments about the same point. Then it's just plugging and adding up the 4 moments to get the total.