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I need to solve the equation. I need to findout the area. what will be the formula to get the area?

enter image description here

regarding this image link is https://dl.dropbox.com/u/106401419/test%20area.png

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3  
Note: the file for the image was named "test: area". I downloaded and cropped the image to omit the bright, huge "PLEASE HELP" from the image (see link to see what I mean.) –  amWhy Dec 26 '12 at 14:52
    
@Ashraful The diagram suggests (but nobody can tell for sure) that some of the lengths are equal in the picture (like F,G,K, and H). Is this the case? –  rschwieb Dec 26 '12 at 14:59
    
Probably the lengths are not equal. However the picture also suggests that certain angles are right angles. Without assuming that, there is no way to compute this area. –  GEdgar Dec 26 '12 at 15:01
    
Are the upper-right diagonal segments supposed to be at 45 degrees to the right and top edges? If not, we will need to use trigonometry to do the problem... –  GEdgar Dec 26 '12 at 15:05
    
@rschwieb no, all length are uniq. –  Ashraful Haque Tushar Dec 26 '12 at 15:49

5 Answers 5

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.

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enter image description here

Subtract the area of gpx from fphi.

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Good idea, we need to get lengths xp and gp, for example... –  GEdgar Dec 26 '12 at 15:11
    
If you assume that segments are parallel, then, you can divide in trapezoids and take the length of the heights, to calculate the respective areas. Finally, substract them. –  Mario De León Urbina Dec 26 '12 at 15:19

The area for the new geometric figure is $H*G-\frac{(G-K)(H-F)}{2}$. Im assuming the poligon has three right angles.

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yes u r right, above formulla is goot, but K and F is unknown –  Ashraful Haque Tushar Dec 26 '12 at 15:53
    
what lengths or angles do you know? @AshrafulHaqueTushar –  Jorge Fernández Dec 26 '12 at 16:37
    
thanks I know a,b,c,d,e,f,g,h,k and three angles right angles. other two r not 45. Main polygon is abcde. inner polygon fgxhi r drown from abcde polygon by offset. offsets r E,F,G,H,K. Now I need the area of polygon fgxhi. –  Ashraful Haque Tushar Dec 27 '12 at 0:39
    
What is line E? –  Jorge Fernández Dec 27 '12 at 1:43
    
thanks, E is parallel distance between those inclined lines. –  Ashraful Haque Tushar Dec 28 '12 at 12:23

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 = $(A - F - H) (B - K- G - \beta) + (A - F - H - \beta) \beta + T$.

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hi , 03 angles are 90, but others are not 45 degree. that is the problem. if it was 45, it was easy. AS this is not 45, I failed to solve it. thanks –  Ashraful Haque Tushar Dec 26 '12 at 15:59
    
If they are not 45, there is no solution w/o an unknown angle parameter. To get a parametrized solution, in my solution you need to back those into $T$ etc, using; hypothenuse = $\frac{E}{\sin{\alpha}} := P$ (not same $\alpha$ as above, angle, which is an unknown parameter), so bottom half top diagonal is $\cos{\alpha} P$, and a similar formula for top half of the top right diagonal using the other angle of $E$ touching the interior area. Plug those into the formula. –  gnometorule Dec 26 '12 at 16:15
    
No solution == no solution using just the values given, as it will depend on the angle of the diagonal. –  gnometorule Dec 26 '12 at 16:22
    
Given A,B,C,D we can find the angle of the diagonal. –  GEdgar Dec 26 '12 at 17:25
    
@GEdgar: agreed now! Doing my thinking on a porch at around 30 while watching it snowing is nice, but eventually leads to brain freeze, –  gnometorule Dec 26 '12 at 17:31

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.

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I'm later than you, but mine agrees with yours. –  GEdgar Dec 26 '12 at 17:22
    
No need to rush during Christmas time :-D –  user1551 Dec 26 '12 at 18:02

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