What is the $kite $ area ? (with respect to figure) 
IN the square $ABCD$ we have 
$$MC=MD, AB=BC=CD=AD=1$$ 
I am looking for a simple Idea to find area of $kite$ with orange color .
  I solve it by analytic geometry ,but I must solve it for $k-10$ students , they don't learn integral or analytic geometry .
 A: The area of a kite is half the product of the diagonals. In this figure it is easy to see that the diagonals are $1/2$ and $1/3$.
A: 
\begin{align} 
[MNOP]&=[COD]-2[CNM]=\tfrac14-2[CNM]
,\\
[BMC]&=\tfrac14=2[CNM]+[BNM_3]
,\\
[BNM_3]&=[CM_3N]=[CNM]=\tfrac13[BMC]=\tfrac1{12}
,\\
[MNOP]&=\tfrac14-2\cdot\tfrac1{12}=\tfrac1{12}
.
\end{align}  
A: 
$$|\square OPMQ| = 2\cdot|\triangle OPM| = 2\cdot\frac{1}{3} \cdot|\triangle OBM| = 2\cdot\frac{1}{3}\cdot\frac{1}{8}\cdot|\square ABCD| = \frac{1}{12}\cdot |\square ABCD|$$
A: $AC$ and $BM$ are medians of triange $BCD$ so you can easily calculate the lengths of two sides of the kite: $1/6$ of $AC$ and $1/3$ of $BM$. The perpendicular through $M$ cuts the kite with length $1/2$ and divides it into two congruent triangles. 
Finish with Heron's formula.
A: So one diagonal is $e={1\over 2}$ (vertical). You have to find $f$ (horizontal). Say $AC$ and $MB$ meet at $P$. Then $MCP$ is similar to $BAP$ and so $MP:PB = 1:2$. Thus $f= 1-2\cdot {2\over 3}\cdot {1\over 2}={1\over 3}$. So $S= {e\cdot f\over 2} ={1\over 12}$
A: My trial is :

$$\frac{x}{\frac 12}+y=1\\y=x \\\to 3x=1 \\x=\frac 13 $$ so 
Area of kite $$=\frac{1}2\times \frac 13\times\frac{1}{2}=\frac1{12}$$
A: 
From the picture on the right (visual proof), we see that we need pieces from $3$ kites to fill $\frac 14$ of the square. 
Thus the area of the kite is $\frac 1{12}$ of the square.

The actual proof is not too complicated. Let's replicate the construction for the kite in the square $ABCD$ into the translated square $EFGH$ where $F$ is the intersection point of the diagonals.
By construction we already have many isometric triangles: $OFJ\sim QDJ\sim MFI\sim PDI$.
As already noticed by other posters, the median properties leads to  a simple length for the small side of the kite: $FK=\frac 13 FD$
And since $DL$ verifies the same property, then the unknown $KL=DL=FK$.
Now let's trace the lines passing through $L$ and $K$ and parallel to the sides of square, then $FKM\sim KLN$ ($FLMN$ is twisted parallelogram since $K$ is the middle of $[FL]$)
And it's now straightforward to complete all symmetries and colourize the isometric triangles.
A: Here is my second and geometrical solution:

A: We can easily divide this square into twelve equal kites, therefore the area of a kite is equal to $\frac1{12}$

