Find the area of the shaded region. (Apparently this is a Chinese primary school math question!) 
Rectangle $ABCD$ has an area of $1$. $AE = ED$. $3BF = AB$. What is the shaded area?
 A: The simplest way, in my opinion, is to write the equations of lines $BE$, $EC$, $FC$, $FD$.  Because scaling doesn't change the ratios of areas, we can assume that we have a square of unit side length, where
$ B = (0, 0)$
$ E = (\frac{1}{2}, 1)$
$ C = (1, 0) $
$ F = (0, \frac{1}{3} )$
$ D = (1,1) $
It is straight forward to derive the following equations
$BE: y = 2 x $
$EC: y = -2 (x - 1) $
$FC: y = -\frac{1}{3} ( x- 1 ) $
$FD: y = \frac{1}{3} + \frac{2}{3} x $
Now, we proceed to find the intersection points
$H: BE \& FC : (\frac{1}{7}, \frac{2}{7} ) $
$G: FD \& BE :  (\frac{1}{4}, \frac{1}{2} ) $
$I: EC \& FD : (\frac{5}{8} , \frac{3}{4} ) $
Finally, applying the so-called shoe-lace formula, we get
$[CIGH] = \frac{1}{2} \left( C_x I_y - C_y I_x + I_x G_y - I_y G_x + G_x H_y - G_y H_x + H_x C_y - H_y C_x \right) $
And this is
$[CIGH] = \frac{1}{2} \left( (1)(\frac{3}{4}) - 0 + (\frac{5}{8})(\frac{1}{2}) - (\frac{3}{4})(\frac{1}{4}) + (\frac{1}{4})(\frac{2}{7}) - (\frac{1}{2})(\frac{1}{7}) + 0 - (\frac{2}{7})(1) \right)$
which reduces to
$[CIGH] = \frac{1}{2} \left( \frac{3}{4} + \frac{5}{16} - \frac{3}{16} + \frac{1}{14} - \frac{1}{14} - \frac{2}{7} \right) $
And finally this comes to
$[CIGH] = \frac{1}{2} \left( \frac{ 84 + 35 - 21  - 32 }{(16)(7)} \right) = \boxed{\frac{33}{112}} $
This is assuming we have a square of side $1$. For a rectangle of area $A$, we get
$[CIGH] =\boxed{ \frac{33}{112} A } $
A: We start by writing $$A_{ABCD}=A_{GHCI}+A_{HBC}+A_{BGA}+A_{AGD}+A_{DIC}$$
Let's call $BC=w$ and $AB=h$. We want to find the coordinates of points $G,H,I$ in terms of $w$ and $h$. Assume that origin is at $B$. Draw perpendiculars from $H$ and $E$ to $AB$. You will form similar triangles, where you can write:
$$\frac{y_H}{h}=\frac{x_H}{w/2}$$and$$\frac{y_H}{h/3}=\frac{w-x_H}w$$
From these two equations, you get that $$x_H=\frac17w\\y_H=\frac27h$$
Then $$A_{HBC}=\frac12y_Hw=\frac17hw$$
You just need to proceed in the same fashion for the other triangles.
A: As we are on methods rather far from primary school, here is one based on special coordinates.
Let us first take cartesian coordinates centered in $A$  with horizontal axis directed by $\vec{AD}$ and vertical axis directed by the opposite of $\vec{AB}$.

Any point in $\Delta:=GHCI$ is the intersection of $2$ unique lines, one passing through $E$, the other through $F$, with resp.  equations:
$$\begin{cases}py&=&x-\frac12\\y&=&qx-\frac23\end{cases} \ \ \iff \ \ \begin{cases}x&=&(3-4p)/(6(1-pq))\\y&=&(3q-4)/(6(1-pq))\end{cases}\tag{1}$$
with $p \in [-1/2,1/2]$ and $q \in [-1/3,2/3]$. Please note that $q$ is a slope, whereas $p$ is the inverse of a slope.
The unicity of $(p,q)$ allows to consider them as a new system of coordinates adapted to region $\Delta$.
The inverse correspondance given by (1) allows to compute the coordinates of points $G,H,I,C$:
$$\begin{cases}
G:& (p_1,q_1)&=&(1/2,2/3) &\iff &(x_1,y_1)&=&(1/4,-1/2)\\
H:& (p_2,q_2)&=&(1/2,-2/3) &\iff &(x_2,y_2)&=&(1/7,-5/7)\\
I:& (p_3,q_3)&=&(-1/2,2/3) &\iff &(x_3,y_3)&=&(5/8,-1/4)\\
C:& (p_4,q_4)&=&(-1/2,-1/3) &\iff &(x_4,y_4)&=&(1,-1)
\end{cases}$$
The area of $\Delta$ is the sum of the areas of triangles $CIG$ and $GHC$, i.e., using a classical determinant formula:
$$\frac12 \begin{vmatrix}x_4&x_3&x_1\\
y_4&y_3&y_1\\
1&1&1\end{vmatrix}+\frac12 \begin{vmatrix}x_1&x_2&x_4\\
y_1&y_2&y_4\\
1&1&1\end{vmatrix}=\frac{33}{112}$$

An alternative method : let us use formulas (1) in a double integral calculation:
$$A:=\iint_{\Delta} dxdy = \int_{p=-1/2}^{1/2} \int_{q=-1/3}^{2/3} \underbrace{\frac{\partial(x,y)}{\partial(p,q)}}_Jdp dq\tag{2}$$
with jacobian (absolute value of the determinant of partial derivatives) $$J=\frac{(3-4p)(4-3q)}{36(1-pq)^3}$$
Numerical computation of (1) gives
$$A=0.294642857142857...$$
from which we can intuite the exact value (we know that is is necessarily a rational number) by remarking the presence of the characteristic period $142857$ of the decimal expansion of $1/7$, giving :
$$A=\frac{1}{10000}(2945+10/7)=\frac{1}{10000}\frac{20625}{7}=\frac{33}{16 \times 7}=\frac{33}{112}$$
