The area of the trapezium LBMN? The figure on the right shows a square whose sides are 8cm long.
L and M are the mid-points of AD and BC respectively;
N is the mid -point of LD. The area of the trapezium LBMN=?
I got stuck at this question. The answer is 24cm^2
I could only calculate 16.
My best attempt
ABCD=8*8= 64
ABL=8*4/2=16 
NMDC= 8+8/2 *4=32
So, 64-32=16cm^2
Thanks,
CJ
The image:
 A: The area of a trapezoid is the product of its height times midsegment.
In this problem, the trapezoid height is $8\,\mbox{cm}$, and the midsegment length is the arithmetic mean of $BM$ and $LN$, i.e. ${1\over2}(2+4)=3\,\mbox{cm}$, because $BM=4\,\mbox{cm}$ and $LN=2\,\mbox{cm}$. 
Therefore
$$
\mbox{Area}(LBMN) = \mbox{height}\times\mbox{midsegment} = 8\cdot3 = 24\,\mbox{cm}^2.
$$
Self-check: To check the answer, we note that the area of the $8\times8$ square $ABCD$ is the sum of the areas of triangle $ABL$, which is $16\,\mbox{cm}^2$, plus two trapezoids $LBMN$ and $MNDC$ of equal area $24\,\mbox{cm}^2$ each; thus 
$$
\mbox{Area}(ABCD) = 16 + 24 + 24 = 64\,\mbox{cm}^2.
$$
So our result $\mbox{Area}(LBMN)=24\,\mbox{cm}^2$ is correct.
A: The area of a trapezium is $\frac12(u+v)h$ where $u$ and $v$ are its parallel sides, and $h$ is its height (the distance between the lines defined by its parallel sides). Here $h=8$ (side-length of square) and
$u$ and $v$ are $4$ and $2$ ($BM$ is half the side-length of square
and $LD$ is a quarter of the square's side-length). So the area is
$$\frac12(2+4)\times 8=24.$$
My calculation of area of $NMDC$ is virtually identical to the above,
getting 24. How did you get 32?
