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
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
Subtract the area of gpx from fphi.
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 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 = $(A - F - H) (B - K- G - \beta) + (A - F - H - \beta) \beta + T$.
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.
The area for the new geometric figure is $H*G-\frac{(G-K)(H-F)}{2}$. Im assuming the poligon has three right angles.