http://i54.tinypic.com/2gsrfw9.jpg

a || b

Construct a 4 angled figure ( could be Square, trapezoid, deltoid or whatever ) whose Surface will be equal to the surface of the pentagon.

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- Mar 27th 2011, 07:21 AMsturmgewehrPentagon.
http://i54.tinypic.com/2gsrfw9.jpg

a || b

Construct a 4 angled figure ( could be Square, trapezoid, deltoid or whatever ) whose Surface will be equal to the surface of the pentagon. - Mar 27th 2011, 07:33 AMscounged
Isn't this a pentagon?

I guess you can set the angle E or C to be 180 degrees, and you will have a quadrilateral with the same surface as the figure. - Mar 27th 2011, 07:38 AMsturmgewehr
Crap yeah, Pentagon. it is a pentagon.

I edited it.

can u copy my figure and show how is it done cuz I tried as you said and it looks like a pentagon again. - Mar 27th 2011, 08:00 AMscounged
Ok. If we make the angle E 180 degrees, we get this:

Attachment 21279

And If we make the angle C 180 degrees, we get this:

Attachment 21280

Both of these are quadrilaterals, as in the first case E went from being a corner to becoming a point on the side AD, and in the second case the same thing happened to C. - Mar 27th 2011, 08:13 AMearboth
I'll show you how to get whatever (Rofl)

Use shearing to get a quadrilateral:

$\displaystyle \Delta(ECD) \cong \Delta(ECD')~\implies~(ABCD')$ is a quadrilateral.

Use shearing again to get a triangle:

$\displaystyle (\overline{D'B}) \parallel (\overline{CC'})$ therefore: $\displaystyle \Delta(D'BC) \cong \Delta(D'BC')$

The area of the pentagon didn't change therefore the area of $\displaystyle \Delta(AC'D') = (ABCDE)$ - Mar 27th 2011, 08:18 AMsturmgewehr

Smart move.

But I how do we prove that the surface of rectangle EQE1 = the Surface of the Rectangle QDC

http://i56.tinypic.com/ek3st3.jpg

Ok maybe I am too stupid to understand this but as far as I know we need 3 elements in common for 2 Triangles to be equal as u stated ( Triangle ECD = Triangle ECD` ), as far as I can see those 2 triangles have only one thing in common and that is EC base. - Mar 27th 2011, 08:28 AMWilmerCode:
`E`

E F

D A D A

C B C B

triangle on top of a square, then made into a rectangle.

EXAMPLE:

Square ABCD sides = 6, AE = DE = 5 ; so area ABCDE = 36 + 12 = 48.

Rectangle becomes a 8 by 6 (ADEF being 6 by 2), so area 48. - Mar 27th 2011, 09:27 AMscounged
I don't think we have to do that.

Instead, we can do this:

http://i1125.photobucket.com/albums/...f?t=1301246367

$\displaystyle \mbox{Here the height, h, is the same for both}~\triangle {ECE_{1}}~\mbox{and}~\triangle {ECD}.$

As both triangles have the same base and the same height, they will also have the same area of

$\displaystyle \displaystyle{\frac{h(\mid{EC}\mid)}{2}}$ - Mar 27th 2011, 09:55 AMsturmgewehr
- Mar 27th 2011, 10:05 AMscounged
No problem.