Answer
$2376~in^4$
Work Step by Step
We can determine the required moment of inertia as follows:
$I_{triangle}=\frac{bh^3}{36}$
and $I_{rectangle}=\frac{bh^3}{12}$
Now the area for each part in the given figure can be calculated as
$A_1=\frac{bh}{2}=\frac{3(6)}{2}=9in^2$
$A_2=hb$
$A_2=(6)(6)=36in^2$
and $A_3=\frac{hb}{2}$
$A_3=\frac{(9)(6)}{2}=27in^2$
The required moment of inertia is given as
$I_{sum}=\Sigma (I+Ad^2_x)$
We plug in the known values to obtian:
$I_{sum}=(\frac{bh^3}{36}+(9)(7))+(\frac{bh^3}{12}+36(3)^2)+(\frac{bh^3}{36}+27(6)^2)$
This simplifies to:
$I=2376in^4$
