Centre of gravity question. Should an equilateral triangle be removed and if so why an equilateral? I am really struggling with the last part of this question and would really appreciate some help. The question, regarding centres of gravity, is from an old O level text and reads as follows:
$ABCD$ is a uniform square plate, of side $2a$, $M$ and $N$ are the midpoints of $AB$, $CD$. If $E$ is a point on $MN$ such that $NE = $$\frac{1}{2}a$ and the triangle AEB is cut away, find the distance from $N$ of the centre of gravity, $G$, of the remainder.
Find the length of $NE\:$ if $G$ is coincident with $E$.
Solutions give $\frac{7}{10}a$ for part one and $(\sqrt3 -1)a$ for part two.
I have been successful in gaining the correct solution for part one but am not at all sure about part two. I am not even sure how to set up the question to obtain the correct answer. My only method so far has been to consider that an equilateral triangle is removed. The reason for this choice of triangle was the $\sqrt3$ in the answer and the altitude of an equilateral triangle of base $2a$ is $a\sqrt3$. However, try as I might I am unable to obtain correct answer. Also, if an equilateral triangle is the right choice, then I don't know why and so can only attempt this second part by checking solution and by using trial and error - clearly an unsatisfactory approach.
I should add that I have used the $\bar{x}$$=\frac{\Sigma wx}{\Sigma w}$ and $\bar{y}$$=\frac{\Sigma wy}{\Sigma w}$ approach with origin at $D$.$\:$ Also, as weight is proportional to area, I have let unit weight equal unit area.
Many thanks in advance for any advice offered as I have been stuck on this for some time.
 A: Let $A = (0, 0), B=(2a, 0), C(2a, 2a), D= (0, 2a)$
Then $M = (a, 0 ), N = (a, 2a ) $
Point $E$ is on $MN$, if the distance $NE = y$ then $E = (a, 2a - y) $
Hence,  the centroid of $\triangle AEB = ( a , \dfrac{y}{3} )$
And therefore, after removing $\triangle AEB$, the centroid of the remainder is given by
$G = \dfrac { a^2 (a, a ) -  a y (a, \dfrac{y}{3} ) } { 4a^2 - a y } $
Simplifying,
$G = (a, \dfrac{ 12 a^2 - y^2 }{12 a - 3 y} )$
If the y-coordinate of $G$ is equal to that of $E$ then we have
$ y = \dfrac{ 12 a^2 - y^2 }{12 a - 3 y}$
Therefore,
$ y (12 a -3 y) = 12 a^2 - y^2 $
Re-arranging,
$ 2 y^2 - 12 a y + 12 a^2 = 0 $
Hence, $ y = \dfrac{1}{4} ( 12 a - \sqrt{ 144 a^2 - 96 a^2 } ) =\dfrac{1}{4} ( 12 a - 4 \sqrt{3} a ) = (3 - \sqrt{3} )a $
From which $NE = 2a - y = 2a - (3 - \sqrt{3}) a = a (\sqrt{3} - 1) $
A: 
In figure $a_2$ (i.e A'B'DC )is equivalent to $a_1$ and $b_2$(i.e MA'CDB')  is equivalent to $b_1$.
For figure a we have:
$CB'^2=(\frac{3a}2)^2+(2a)^2=\frac{25a^2}4\rightarrow CF=\frac52 a$
$CG=\frac12 CF=\frac 54 a$
$NG=\sqrt{(\frac54a)^2-a^2}=\frac 34 a$
For figure b we have $DB'=x$:
$G_1$from A'B'=$\frac13(2a-x)$
Area of MA'B'=$2a(2a-x)\frac12$
Area of A'B'DC=$\frac 12x(2a\times x)$
so we must have:
$[\frac13(2a-x)][2a(2a-x)\frac12]=\frac 12x(2a\times x)$
which gives:
$x^2+2ax-2a^2=0\rightarrow x=a(\sqrt3-1)$
