Problem #$9$: For the cross section shown in the figure below, calculate the following section properties:

Distance to centroid $\frac{?}{b}=ar(Y)=\frac{(A_{y_S})}{A_T}=,y=\frac{11,006}{550}=20mm$

A$)$ Area $=$

$m{m}^2$

B$)$ Distance from the bottom to the horizontal, $x-x,$ section centroidal axis $=$ mm

C$)$ Distance from the left$-$most edge to the vertical, $y-y,$ section centroidal axis $=$ mm

D$)$ Moment of Inertia, $I_{x-x}=m{m}^4$

E$)$ Section modulus about the vertical, $y-y,$ section centroidal axis, $S_{y-y},m{m}^3$

F$)$ Section modulus about the vertical, $x-x,$ section centroidal axis, $S_{x-x}=$ Please explain and show steps