Step 1

We are required to locate the centroid $\left( {\overline x ,\overline y } \right)$ of the shaded area.

 
Step 2

The diagram can be represented as below:

Here, ${a_1}$ represents the side of the square 1, ${b_2}{\rm{ \;and\; }}{h_2}$ represents the base and height of the triangle 2, ${b_3}{\rm{ \;and\; }}{h_3}$ represents the base and height of the triangle 3.

 
Step 3

To locate the centroid $\overline x $ we will use the relation,

\[\overline x = \frac{{{A_1}{x_1} + {A_2}{x_2} + {A_3}{x_3}}}{{{A_1} + {A_2} + {A_3}}}\] ….. (1)

Here, ${A_1}$ represents the area of the region 1, ${A_2}$ represents the area of the region 2, ${A_3}$ represents the area of the region 3, ${x_1}$ represents the distance of center of gravity of region 1 from the origin, ${x_2}$ represents the distance of center of gravity of region 2 from the origin and ${x_3}$ represents the distance of center of gravity of region3 from the origin.

 
Step 4

To find the center of gravity in horizontal axis of region 1 which is a square, we will use the relation,

\[\begin{array}{c} {x_1} = - \frac{{{a_1}}}{2}\\ = - \frac{{6\;{\rm{in}}{\rm{.}}}}{2}\\ = - {\rm{3 in}}{\rm{.}} \end{array}\]
 
Step 5

To find the center of gravity in horizontal axis of region 2 which is a triangle, we will use the relation,

\[\begin{array}{c} {x_2} = - \frac{{{b_2}}}{3}\\ = - \frac{{6{\rm{ }}{\rm{in}}{\rm{.}}}}{3}\\ = - 2{\rm{ in}}{\rm{.}} \end{array}\]
 
Step 6

To find the center of gravity in horizontal axis of region 3 which is a triangle, we will use the relation,

\[\begin{array}{c} {x_3} = \frac{{{h_3}}}{3}\\ = \frac{{6{\rm{ in}}{\rm{.}}}}{3}\\ = 2{\rm{ in}}{\rm{.}} \end{array}\]
 
Step 7

To locate the centroid $\overline y $ we will use the relation,

\[\overline y = \frac{{{A_1}{y_1} + {A_2}{y_2} + {A_3}{y_3}}}{{{A_1} + {A_2} + {A_3}}}\] ….. (2)

Here, ${A_1}$ represents the area of the rectangle 1, ${A_2}$ represents the area of the rectangle 2, ${A_3}$ represents the area of the rectangle 3, ${y_1}$ represents the distance of center of gravity of rectangle 1 from the top, ${y_2}$ represents the distance of center of gravity of rectangle 2 from the bottom and ${y_3}$ represents the distance of center of gravity of rectangle 3 from the bottom.

 
Step 8

To find the center of gravity in vertical axis of region 1 which is a square, we will use the relation,

\[\begin{array}{c} {y_1} = \frac{{{a_1}}}{2}\\ = \frac{{6\;{\rm{in}}{\rm{.}}}}{2}\\ = {\rm{3 in}}{\rm{.}} \end{array}\]
 
Step 9

To find the center of gravity in vertical axis of region 2 which is a triangle, we will use the relation,

\[\begin{array}{c} {y_2} = {a_1} + \frac{{{h_2}}}{3}\\ = 6{\rm{ in}}{\rm{.}} + \frac{{3{\rm{ }}{\rm{in}}{\rm{.}}}}{3}\\ = 6{\rm{ in}}{\rm{.}} + 1{\rm{ in}}{\rm{.}}\\ = 7{\rm{ in}}{\rm{.}} \end{array}\]
 
Step 10

To find the center of gravity in vertical axis of region 3 which is a triangle, we will use the relation,

\[\begin{array}{c} {y_3} = 2 \times \frac{{{b_3}}}{3}\\ = 2 \times \frac{{9{\rm{ in}}{\rm{.}}}}{3}\\ = 6{\rm{ in}}{\rm{.}} \end{array}\]
 
Step 11

To find the area of region 1 which is a square, we will use the relation.

\[{A_1} = {a_1} \times {a_1}\]

Here, ${a_1}$ represents the sides of the square 1.

 
Step 12

On plugging the values in the above relation, we get,

\[\begin{array}{c} {A_1} = 6{\rm{ in}}{\rm{.}} \times 6{\rm{ in}}{\rm{.}}\\ = {\rm{36}}\;{\rm{in}}{{\rm{.}}^2} \end{array}\]
 
Step 13

To find the area of region 2 which is a triangle, we will use the relation.

\[{A_2} = \frac{1}{2} \times {b_2} \times {h_2}\]

Here, ${h_2}$ and ${b_2}$ represents the length and breadth of the triangle 2.

 
Step 14

On plugging the values in the above relation, we get,

\[\begin{array}{c} {A_2} = \frac{1}{2} \times 6\;{\rm{in}}{\rm{.}} \times {\rm{3}}\;{\rm{in}}{\rm{.}}\\ = 9\;{\rm{in}}{{\rm{.}}^2} \end{array}\]
 
Step 15

To find the area of region 3 which is a triangle, we will use the relation.

\[{A_3} = \frac{1}{2} \times {b_3} \times {h_3}\]

Here, ${h_3}$ and ${b_3}$ represents the length and breadth of the triangle 3.

 
Step 16

On plugging the values in the above relation, we get,

\[\begin{array}{c} {A_3} = \frac{1}{2} \times \left( {6{\rm{ in}}{\rm{.}} + 3{\rm{ in}}{\rm{.}}} \right) \times 6{\rm{ }}{\rm{in}}{\rm{.}}\\ = \frac{1}{2} \times 9{\rm{ }}{\rm{in}}{\rm{.}} \times 6{\rm{ }}{\rm{in}}{\rm{.}}\\ = {\rm{27}}\;{\rm{in}}{{\rm{.}}^2} \end{array}\]
 
Step 17

On plugging the values in the equation (1), we get,

\[\begin{array}{c} \overline x = \frac{{\left( {36{\rm{ in}}{{\rm{.}}^2} \times \left( { - 3{\rm{ in}}{\rm{.}}} \right)} \right) + \left( {9{\rm{ in}}{{\rm{.}}^2} \times \left( { - 2{\rm{ in}}{\rm{.}}} \right)} \right) + \left( {27{\rm{ in}}{{\rm{.}}^2} \times 2{\rm{ in}}{\rm{.}}} \right)}}{{36{\rm{ in}}{{\rm{.}}^2} + 9{\rm{ in}}{{\rm{.}}^2} + 27{\rm{ in}}{{\rm{.}}^2}}}\\ = \frac{{ - 108{\rm{ in}}{{\rm{.}}^3} - 18{\rm{ in}}{{\rm{.}}^3} + 54{\rm{ in}}{{\rm{.}}^3}}}{{72{\rm{ in}}{{\rm{.}}^2}}}\\ = \frac{{ - 72{\rm{ in}}{{\rm{.}}^3}}}{{72{\rm{ in}}{{\rm{.}}^2}}}\\ = - 1{\rm{ in}}{\rm{.}} \end{array}\]
 
Step 18

On plugging the values in the equation (2), we get,

\[\begin{array}{c} \overline y = \frac{{\left( {36{\rm{ in}}{{\rm{.}}^2} \times 3{\rm{ in}}{\rm{.}}} \right) + \left( {9{\rm{ in}}{{\rm{.}}^2} \times 7{\rm{ in}}{\rm{.}}} \right) + \left( {27{\rm{ in}}{{\rm{.}}^2} \times 6{\rm{ in}}{\rm{.}}} \right)}}{{36{\rm{ in}}{{\rm{.}}^2} + 9{\rm{ in}}{{\rm{.}}^2} + 27{\rm{ in}}{{\rm{.}}^2}}}\\ = \frac{{108{\rm{ in}}{{\rm{.}}^3} + 63{\rm{ in}}{{\rm{.}}^3} + 162{\rm{ in}}{{\rm{.}}^3}}}{{72\,{\rm{in}}{{\rm{.}}^2}}}\\ = \frac{{333{\rm{ in}}{{\rm{.}}^3}}}{{72{\rm{ in}}{{\rm{.}}^2}}}\\ = 4.625{\rm{ in}}{\rm{.}} \end{array}\]