Q:

Asymmetrical Center of Gravity Problems

Calculating center of gravity asymmetrical objects
find the position of the center of gravity of relative to the edges AB and AC 
https://s.yimg.com/hd/answers/i/904d22de93984f86a082e538d2ca941a_A.png?a=answers&mr=0&x=1432762599&s=f70331198adfb371d2b6073408571fcd
The answers are 32.5mm from AB and 7.5 from BC 
I can't seem to solve this problem for the side where the answer 7.5 mm is obtained 
I know that the object consists of three smaller squares so I found the area of each square and made my reference lines and solved each axis for the perpendicular midpoints of each rectangle 
I found each area : A1=400 ,A2=500,A3=300 , total area=1200mm^2 
then I input 
20x400+35x500+45x300/1200=32,5mm 
when I input the formula for the other side I obtain 5x400+35x500+65x300/1200 i obtain 32.5 ,which is the same answer as before
I would appreciate any help solving this problem and some advice with calculating the center of gravity as I have spent several hours unsuccessfully solving it.I am sure I am making some mistake somewhere. Thank you 

A:

Draw a vertical line $\ell$ through $A$. We use the same decomposition as you did, and find that the distance of the centre of mass from this line is
$$\frac{(20)(400)+(35)(500)+(45)(300)}{1200}.$$
This is $32.5$, so our centre of mass is $32.5$ to the right of $\ell$. It is therefore $7.5$ to the left of $BC$.
Or else we could take distances to the left of $BC$ as negative, and distances to the right as positive. So we are taking the line through $B$ and $C$ as the $y$-axis. Then the centre of mass has $x$-coordinate
$$\frac{(-20)(400)+(-5)(500)+(5)(300)}{1200}.$$
This is $-7.5$.
Remark: Your calculation was exactly right, except that the question asked one to  describe  the location of the centre of mass in terms of distance from $BC$, not in terms of distance from $\ell$.  

