From the given figure, we can divide it into three separate shapes: a semicircle, a rectangle and a right triangle.
The centroid of the body is the arithmetic mean of the centroid of the individual shapes.
Centre of the diameter of the semicircle is located at [tex]\left( \frac{1.25}{2} ,\ \frac{1.25}{2}\right)=(0.625,\ 0.625)[/tex] and the centroid of the semicircle is located at [tex]\left( \frac{0.625}{2} ,\ 0.625\right)=(0.3125,\ 0.625)[/tex].
The length of the base of the right triangle is obtained as follows:
[tex] \sqrt{(1.95)^2-(1.25)^2} = \sqrt{3.8025-1.5625} = \sqrt{2.24} =1.4967ft[/tex]
The length of the base of the rectangle is given by
[tex]3.17-1.4967-0.625=1.0483ft[/tex]
The centroid of the rectangle is given by
[tex]\left( \frac{0.625+1.0483}{2} ,\ 0.625\right)=\left( \frac{1.6733}{2} ,\ 0.625\right)=(0.8367,\ 0.625)[/tex]
The centroid of the rectangle with the hole is given by
[tex]\left( \frac{0.8367+1.05}{2} ,\ 0.625\right)=\left( \frac{1.8867}{2} ,\ 0.625\right)=(0.9434,\ 0.625)[/tex]
The midpoint of side c of the triangle is given by
[tex]\left( \frac{0.625+1.0483+3.17}{2},\ 0.625\right)=\left( \frac{4.8433}{2},\ 0.625\right)=(2.4217,\ 0.625)[/tex]
The midpoint of the altitude of the right triangle is given by (0.625 + 1.0483, 0.625) = (1.6733, 0.625)
The equation of the line joining points (1.6733, 0) and (2.4217, 0.625) is given by y = 0.8351x - 1.3974
The equation of the line joining points (3.17, 0) and (1.6733, 0.625) is given by y = -0.4176x + 1.3238
The x-value of the point of intersection of line y = 0.8351x - 1.3974 and y = -0.4176x + 1.3238 is given by
0.8351x - 1.3974 = -0.4176x + 1.3238
1.2527x = 2.7212
x = 2.1723
The y-value of the point of intersection of line y = 0.8351x - 1.3974 and y = -0.4176x + 1.3238 is given by
y = 0.8351(2.1723) - 1.3974 = 1.8141 - 1.3974 = 0.4167
Thus, the centroid of the right triangle is given by (2.1723, 0.4167).
Therefore, the centroid of the object is given by:
[tex]\left( \frac{0.3125+0.8367+2.1723}{3} ,\ \frac{0.625+0.625+0.4167}{3} \right) \\ \\ = \left(\frac{3.3215}{3} ,\ \frac{1.6667}{3}\right)=(1.1072,\ 0.5556)[/tex]
