P$\mathrm{.}5.(10$ points$)$ Determine the following:

$\frac{2}{6}$ a$.$ the centroid $(\stackrel{}{x}\frac{,}{b}ar(y))$ of the area shown

b$.$ the location of the neutral axis parallel to the $x-$ axis $($Draw on image and indicate dimensions$)$

c$.$ the moment of inertia $I_{x,NA}$ with respect to the neutral axis.

Extra Credit: the moment of inertia $I_x$ with respect to the indicated $x-$axis.

$A_1=120*100=12m$

$A_2=\frac{1}{2}(60)(120)=3\mathrm{.}6m$

$I_{yy}=Ix+A*d^2$

$I_{yy}=Iy+A*d^2$

$=I_x=\frac{1}{12}b{n}^3=\frac{1}{12}(100)(1200{)}^3=1400m$

$I_y=\frac{1}{12}h{b}^3=\frac{1}{12}(120)(100{)}^3=10000m$

$=I_x=\frac{1}{36}b{h}^3=\frac{1}{36}(60)(120{)}^3=2880m$

$I_y=\frac{1}{36}h{b}^3=\frac{1}{36}(120)(60{)}^3=7200m$

$I_{}=(14400)+(12)({)}^2+$

$(2880)+(3\mathrm{.}6)({)}^2$

$I_{yy}=(10,000)+(12x,{)}^2+$

$(720)+(3\mathrm{.}1)12$

Centroid of the composite body

$\stackrel{}{x}=\frac{,}{b}ar(y)=$

Moment of Inertia

$\frac{?}{\text{b}\text{a}\text{r}}(I{)}_{x,NA}=\frac{,}{b}ar(I{)}_x=$