1. [SOLVED] Paper Folding Questions

Hi, I have a question which has been giving me some grief. Any help would be greatly appreciated.

Question 1
The old-style Kit-Kat bar allowed for a saving of resources and costs in the size of the foil wrapping. The bar was offset on an angle of approx 21 degrees.

A given rectangular bar of chocolate is a cm by b cm, placed in a background rectangle ( c cm by d cm).

a= 12cm, b=8cm

Find the angle of offset, and find the length and width of the rectangular wrapping paper.

EDIT: For the purpose of this question, assume that the chocolate bar is flat.

2. Originally Posted by Dans
The old-style Kit-Kat bar allowed for a saving of resources and costs in the size of the foil wrapping. The bar was offset on an angle of approx 21 degrees.

A given rectangular bar of chocolate is a cm by b cm, placed in a background rectangle ( c cm by d cm).

a= 12cm, b=8cm

Find the angle of offset, and find the length and width of the rectangular wrapping paper.

EDIT: For the purpose of this question, assume that the chocolate bar is flat.
I take it that the picture is meant to look like this:

$\displaystyle \setlength{\unitlength}{4.5mm} \begin{picture}(10,12) \put(0,0){\line(1,0){9}} \put(0,12){\line(1,0){9}} \put(0,0){\line(0,1){12}} \put(9,0){\line(0,1){12}} \put(5,0){\line(-5,2){5}} \put(5,0){\line(2,5){4}} \put(9,10){\line(-5,2){5}} \put(0,2){\line(2,5){4}} \put(7,4){$a$} \put(6,10.3){$b$} \put(2.8,0.1){$\theta$} \end{picture}$

I also assume that "assume that the chocolate bar is flat" means that we disregard its thickness and treat it as two-dimensional. The idea is that if you fold over the four triangular areas then they just cover the inner rectangle. If that is the case, then the area cd of the outer rectangle (width c, height d) must be twice the area ab of the bar of chocolate. If the offset angle is $\displaystyle \theta$, then $\displaystyle c = a\cos\theta + b\sin\theta$ and $\displaystyle d = b\cos\theta + a\sin\theta$. Thus $\displaystyle cd = (a\cos\theta + b\sin\theta)(b\cos\theta + a\sin\theta) = ab + (a^2+b^2)\sin\theta\cos\theta$. If this is equal to 2ab then $\displaystyle \sin(2\theta) = \frac{2ab}{a^2+b^2}$. From that, given a and b you can find $\displaystyle \theta$, and hence c and d.