Finding the center of mass of a composite lamina

Question

The following lamina is given and I have to find the distance from center of mass from AB and AC

I started by splitting the lamina into two shapes, one rectangle of width $4a$ height $5a$ and one of height $3a$ and width $3a$ which is a square.

$$\because \bar{x}=\frac{\sum_{i=0}^n \widehat{x}_i A_i}{\sum_{i=0}^n A_i}$$
$$\therefore \bar{x}=\frac{(\frac{4a}{2}((4a)(5a))+(\frac{3a}{2}(3a)^2)}{(3a)^2+(4a)(5a)}=\frac{107a}{58}$$

And similarly

$$\bar{y}=\frac{(\frac{5a}{2}((4a)(5a))+(\frac{3a}{2}(3a)^2)}{(3a)^2+(4a)(5a)} =\frac{127a}{58}$$ Hence the coordinates of the center of mass are $$<\frac{107a}{58}, \frac{127a}{58}>$$

However the actual answer is

$\frac{179a}{58}$ from AB and $\frac{127a}{58}$ from AC

I am entirely baffled on how to get this answer and have little to no clue what Ive done wrong in solving this.

Answer 1

Try

$$\therefore \bar{x}=\frac{(\frac{4a}{2}((4a)(5a))+((\frac{3a}{2}+4a)(3a)^2)}{(3a)^2+(4a)(5a)}= ? $$

Where $4a$ is a shift from the left side. The right rectangle does not start at the origin so you must take this shift into account

Answer 2

Your mistake is the $x$ cordinate of the square that you have written. If you are taking the origin at A, so the $x$ coordinate would be $4a+\frac{3a}{2}=\frac{11a}{2}$. Now if you do the calculation, you will get the right answer.