P$\mathrm{.}4.(10$ points$)$ Locate the centroid $(\stackrel{}{x}\frac{,}{b}ar(y))$ of the composite shape relative to the origin. Note: Set a $=4in.$

$A_1=B*H=4*4=16in$

$A_2=\frac{1}{2}*B*H=\frac{1}{2}(4)(3)=6in$

$A_3=\frac{1}{2}*B*H=\frac{1}{2}(7)(4)=14in$

$A_T=16it6it14in$

$=36in$

Al square

$\stackrel{}{x}=\frac{1}{12}b*h^3=\frac{1}{12}(4)(4{)}^3=213in\frac{?}{b}$

$ar(y)=\frac{1}{12}h*b^3=\frac{1}{12}(4)(4{)}^3=21\mathrm{.}3in$

Aa Trimule

$\stackrel{}{x}=\frac{1}{36}*b*h^3=\frac{1}{36}(4)(3{)}^3=3in\frac{?}{b}$

$ar(y)=\frac{1}{36}*h*b^3=\frac{1}{36}(3)(4{)}^3=5\mathrm{.}33in$

As Tryingle

Tryingle

$\stackrel{}{x}=\frac{1}{36}*b*h^3=\frac{1}{36}(4)(7{)}^3=38\mathrm{.}11in\frac{?}{b}$

$ar(y)=\frac{1}{36}*h*b^3=\frac{1}{36}(7)(4{)}^3=12\mathrm{.}4in$

Centroid of the composite body

$\stackrel{}{x}=62\mathrm{.}41in\frac{,}{b}ar(y)=39\mathrm{.}03in$