# Pentagon.

• Mar 27th 2011, 07:21 AM
sturmgewehr
Pentagon.
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 AM
scounged
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 AM
sturmgewehr
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 AM
scounged
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 AM
earboth
Quote:

Originally Posted by sturmgewehr

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

I'll show you how to get whatever (Rofl)

Use shearing to get a quadrilateral:

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

Use shearing again to get a triangle:

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

The area of the pentagon didn't change therefore the area of $\Delta(AC'D') = (ABCDE)$
• Mar 27th 2011, 08:18 AM
sturmgewehr
Quote:

Originally Posted by scounged
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.

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

Quote:

Originally Posted by earboth
I'll show you how to get whatever (Rofl)

Use shearing to get a quadrilateral:

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

Use shearing again to get a triangle:

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

The area of the pentagon didn't change therefore the area of $\Delta(AC'D') = (ABCDE)$

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 AM
Wilmer
Code:

        E                             E                F D              A            D                A C              B            C                B`
The way question is worded does not prevent using an isosceles
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 AM
scounged
Quote:

Originally Posted by sturmgewehr
Smart move.

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

I don't think we have to do that.

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

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

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

$\displaystyle{\frac{h(\mid{EC}\mid)}{2}}$
• Mar 27th 2011, 09:55 AM
sturmgewehr
Quote:

Originally Posted by scounged
I don't think we have to do that.

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

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

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

$\displaystyle{\frac{h(\mid{EC}\mid)}{2}}$

this is very good answer, simple and precise.

thank you very much :)
• Mar 27th 2011, 10:05 AM
scounged
No problem.