Multi-Segment Moment

I initially got this question wrong, but afterwards I obtained a solution. Here it is:

First we denote the point $(0,0)$ to be $A$ , $(1,2)$ to be $B$ , $(3,1)$ to be $C$ and $(4,5)$ to be $D$ . Then, the midpoint of $AB$ is given by $(0.5,1)$ with a distance of $\sqrt{5}$ from $A$ . The midpoint of $BC$ is $(2,1.5)$ , with a distance of $\sqrt{6.25}$ from the origin. The midpoint of $CD$ is $(3.5,3)$ , with a distance of $\sqrt{21.25}$ from the origin.

Now, we observe that the mass of segment $AB$ , $BC$ and $CD$ are given by $\frac{|AB|}{|AB|+|BC|+|CD|}M\\\frac{|BC|}{|AB|+|BC|+|CD|}M\\\frac{|CD|}{|AB|+|BC|+|CD|}M$

We know that the moment of inertia relative to an axis perpendicular to the $xy$ plane and passing through $A$ can be expressed as $\sum mr^2$ , where $m$ is the mass of each point and $r$ is the distance of that point to $A$ . Hence, we have

$\begin{aligned}I&=m_{ab}|AB|^2+m_{bc}|BC|^2+m_{cd}|CD|^2\\&=5\frac{|AB|}{|AB|+|BC|+|CD|}M+6.25\frac{|BC|}{|AB|+|BC|+|CD|}M+21.25\frac{|CD|}{|AB|+|BC|+|CD|}M\\&\approx13.12M \end{aligned}$

So the answer is 13.