how to calculate area of the image attached I need to solve the equation. I need to findout the area. what will be the formula to get the area?

regarding this image link is https://dl.dropbox.com/u/106401419/test%20area.png
 A: 
Subtract the area of gpx from fphi. 
A: While still tedious, the calculations are easier than I thought. Let the upper right hand corner of the big rectangle be the origin, the positive $x$-axis points to the left and the positive $y$-axis points downward. Then the longer slanted edge is the line segment joining $(B-C,\,0)$ and $(0,\,A-D)$ and the shorter slanted edge is the line segment joining $(K,\,F+y)$ and $(K+x,\,F)$, where $x$ and $y$ are the width and height of the complementary triangle at the upper right corner of the red rectangle. Since these two line segments are parallel, we have
\begin{equation}
\frac yx = \frac{A-D}{B-C} = -m\ \textrm{(say)},\tag{1}
\end{equation}
where $m$ is the common slope of the two line segments. Recall that the distance from the origin to a line joining passing through a point $(x_0,y_0)$ with slope $m$ is given by
$$
\frac{|y_0 - mx_0|}{\sqrt{m^2+1}}.
$$
Now the distance between these two line segments is $E$. Therefore
$$
\frac{|F+y-mK|-|A-D|}{\sqrt{m^2+1}} = E.
$$
Hence
$$
y = mK-F \pm \left[(A-D)+E\sqrt{m^2+1}\right].
$$
Since $y$ is assumed to be positive, we take the positive sign in the "$\pm$" above. Hence, by $(1)$, the area of the complementary triangle is given by $\frac12xy = \frac{y^2}{-2m}$. Subtract this from $(A-F-H)(B-G-K)$, we get the required area.
A: A bit hard to communicate, but here is what I get:  
Triangle with top left equal to $E$ has right top $E$ as well (it has angles 90, 45, 45), so right top line has length $2E$.  
Then the triangle with right top $2E$ has as other sides $\sqrt{E}$, and area $\frac{1}{2} E$.  
It is similar (right expression?) to the triangle in the top right of the sought area. This area is shrunk from the larger area enclosing it by a factor of. 
$\alpha := \frac{ (A - F - H) (B - K - G) }{ AB}$,  
so its area is $\frac{\alpha}{2} E =: T$, with sides shrunk to $\beta := \alpha \sqrt{E}$. Hence:  
Area = $(A - F - H) (B - K- G - \beta) + (A - F - H - \beta) \beta + T$.
A: Assuming naively for a moment that the thing in red is rectangular (or possibly even square) with a missing triangular corner, you would use the formula for a rectangle and the area of a triangle. First find the area of the rectangle, then subtract the area of the missing triangle corner of the rectangle.
Without more information your diagram is impossibly vague.
A: The area for the new geometric figure is $H*G-\frac{(G-K)(H-F)}{2}$. Im assuming the poligon has three right angles.
