724 Rectangle, semicircle, quarter-circle, and triangle | Centroid of Composite Area

$a = \dfrac{4(4)}{3\pi} = 1.698''$

$b = \dfrac{4(6)}{3\pi} = 2.546''$

$c = \frac{1}{3}(6) = 2''$
 

$A_1 = 18(12) = 216 \, \text{ in.}^2$

$x_1 = \frac{1}{2}(18) = 9''$

$y_1 = \frac{1}{2}(12) = 6''$
 

$A_2 = \frac{1}{2}\pi (4^2) = 25.133 \, \text{ in.}^2$

$x_2 = 4''$

$y_2 = 12 - a = 12 - 1.698 = 10.302''$
 

$A_3 = \frac{1}{4}\pi (6^2) = 28.274 \, \text{ in.}^2$

$x_3 = 18 - b = 18 - 2.546 = 15.454''$

$y_3 = 12 - b = 12 - 2.546 = 9.454''$
 

$A_4 = \frac{1}{2}(6)(6) = 18 \, \text{ in.}^2$

$x_4 = 18 - c = 18 - 2 = 16''$

$y_4 = c = 2''$
 

$A = A_1 - A_2 - A_3 - A_4$

$A = 216 - 25.133 - 28.274 - 18$

$A = 144.593 \, \text{ in.}^2$
 

$A\bar{x} = \Sigma ax$

$144.593\bar{x} = 216(9) - 25.133(4)- 28.274(15.454)- 18(16)$

$\bar{x} = 7.736''$           answer
 

$A\bar{y} = \Sigma ay$

$144.593\bar{y} = 216(6) - 25.133(10.302) - 28.274(9.454) - 18(2)$

$\bar{y} = 5.075''$           answer