$\mathrm {The \ centroid\ always\ lies \ on\ the \ axis\ of \ symmetry \ \overline{x}=120mm}$

$\mathrm {To\ find \ \overline{y} :}$

$\mathrm {A_1 = Area\ of\ rectangle=120\times240 =28800\ mm^2}$

${y_1 = \bigg(\dfrac{120}{2}\bigg) = 60mm}$

$A_2 = Area\ of \ semicircle = \dfrac{\pi (90)^2}{2}=-1272304mm^2$

$y_2=120-\dfrac{4(90)}{3\pi}=81.8mm$

$\overline{y} = \dfrac{A_1y_1+A_2y_2}{A_1+A_2}=\dfrac{28800\times60-112723.4\times81.8}{28800-12723.4} =42.747mm$

$\\ …