Answer
$I=209in^4$
Work Step by Step
We can find the required moment of inertia as follows:
The area of the first rectangle is given as
$A_1=\frac{1}{2}hb$
$\implies A_1=\frac{1}{2}(3)(3)=4.5in^2$
The area of the second rectangle is
$A_2=\frac{1}{2}hb$
$A_2=\frac{1}{2}(6)(6)=18in^2$
The area of the third rectangle is
$A_3=hb$
$\implies A_3=9in^2$
Now $I=\Sigma (I+Ady^2)$
We plug in the known values to obtian:
$I=\frac{3\cdot 3^3}{36}+4.5(4)^2+(\frac{6\times 6^3}{36}+18\times (2)^2)+(\frac{3\times(3)^3}{12}+9(15)^2)$
This simplifies to:
$I=209in^4$
