Moment of Inertia of an L-shaped object A uniform thin bar formed into a L-shaped object of mass $m=2.5kg$ with a longer side of length $l=0.8m$ and a shorter side of length $l/2$. Initially the object is positioned with one end at the origin and the longer side along the $x$ axis. The centre of mass of the object has coordinates $r_{cm}=\frac{2l}{3} \hat i -\frac{l}{12} \hat j$ . I should also add that the object is held in position by a massless wire that makes an angle $\phi=50  ^{\circ}  $ with the longer side of the object.
The object is attached to the wall by a pivot (at the origin). Compute the moment of inertia of the object about an axis through the pivot perpendicular to the plane of the object. 
I know that moment of inertia is equal to $I=r^2 m $. I broke down the moment of inertia into two components, one calculating $I_1$ over the longer side of L (length of $l$), and the other calculating $I_2$ over the shorter side of L ( length of $ \frac{l}{2} $ ). However, the correct answer provided clearly states that $I_1=\frac{2}{9}ml^2$ and $I_2= \frac{13}{12}\frac{1}{3} ml^2$ (using the parallel axis theorem). I don't know how to achieve these results, my reasoning behind calculating $I_1$ is:
Since $I=mr^2$,
$I_1=\frac{2}{3}m l^2$   (since the longer side is twice as long as the shorter side)
which clearly gives me an incorrect result. 
If someone could explain the logic behind calculating the total moment of inertia of this type of object that would be great! 
 A: Without doing integration you need to have a plan:
1) Draw good diagram with a well-defined coordinate system.
2) Consider the object to be composed of two slender rods of uniform density. Based on their lengths, you can assign masses to each rod. Also find the coordinates of the CM of each rod.
3) Calculate the MOI of each rod about its own CM. (Easily found in any physics text)
4) Use the parallel axis theorem and calculate the MOI of each rod about the pivot point.
5) Add these final MOIs together.
A: Consider the general case:

Consider a cross section, with density $\rho$ and thickness $w$. The domain is split into two rectangular areas:


*

*The horizontal leg with:


*

*Mass $m_1 =\rho w t_1 (\ell_1-t_2)$

*Centroid $(x_1,y_1) = \left( \frac{\ell_1+t_2}{2}, \frac{t_1}{2} \right)$

*MMOI about centroid $I_1 = \frac{m_1}{12} \left(t_1^2+(\ell_1-t_2)^2\right)$


*The vertical leg with:


*

*Mass $m_2 =\rho w t_2 \ell_2$

*Centroid $(x_2,y_2) = \left( \frac{t_2}{2}, \frac{\ell_2}{2} \right)$

*MMOI about centroid $I_2 = \frac{m_2}{12} \left(t_2^2+\ell_2^2\right)$


*Combined properties


*

*Mass $m = m_1+m_2 = \rho w \left( \ell_1 t_1 + t_2(\ell_2 - t_1) \right)$

*Centroid $ (x_c,y_c) = \left( \frac{m_1 x_1+m_2 x_2}{m_1+m_2}, \frac{m_1 y_1+m_2 y_2}{m_1+m_2} \right)$

*MMOI about centroid $I_c =\sum \limits_{i=1}^2 \left[I_i + m_i \left( (x_c-x_i)^2 + (y_c-y_i)^2 \right) \right]$


A: A diagram of the situation:

The moment of inertia of the entire system can be written as the sum of the moment of inertia of each element about its center of mass, plus a component due to the fact that they are rotating about a point that is not its center of mass.
This gives us for $I_1$:
$$\begin{align}I_1 &= \frac{1}{12} \frac{2m}{3} L^2 + \frac{2m}{3} \left(\frac{L}{2}\right)^2\\
&= \frac29 m L^2\end{align}$$
as given in your solution.
A similar approach can be used for $I_2$
