1. ## Pentagon.

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.

2. 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.

3. 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.

4. Ok. If we make the angle E 180 degrees, we get this:

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

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.

5. 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

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 $\Delta(AC'D') = (ABCDE)$

6. Originally Posted by scounged
Ok. If we make the angle E 180 degrees, we get this:

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

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

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

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.

7. 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.

8. 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.

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

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

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

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

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

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

10. No problem.