Answer
$1971in^4$
Work Step by Step
We can find the required moment of inertia as follows:
We know that
$I_{triangle}=\frac{bh^3}{36}$
and $I_{rectangle}=\frac{bh^3}{12}$
We find the area for each part in the given figure
$A_1=\frac{hb}{2}$
$A_1=\frac{(3)(6)}{2}=9in^2$
$A_2=hb$
$\implies A_2=(6)(6)=36in^2$
and $A_3=\frac{hb}{2}=\frac{(9)(6)}{2}=27in^2$
Now $I_{sum}=\Sigma (I+Ad^2_x)$
We plug in the known values to obtain:
$I_{sum}=\frac{bh^3}{36}+9(7)^2+(\frac{bh^3}{12}+36(3)^2)+(\frac{bh^3}{36}+27(6)^2)$
This simplifies to:
$I_{sum}=1971in^4$
