## Tuesday, July 31, 2012

### More fun with geometry

I ran into a fun problem with parallelograms the other day.

The graphic shows a parallelogram broken up into pieces by its diagonals.

Since the opposing sides of a parallelogram are parallel and equal, it's easy to show that the top and bottom triangles are congruent, just rotated, likewise with the other two.

The problem posed was to express the total area in terms of the obtuse angle at the center, the one which is > 90 degrees. Let's call that angle A, and its supplementary angle (the acute angle at the center) B.

The reason why I've drawn the figure in this way is that, as shown below, the triangles can be rearranged to give a different paralellogram with the same area.

x = length of long diagonal (orange + maroon)
y = length of short diagonal
y/2 = 1/2 length of short diagonal (blue)

Area = x (y/2) sin B

The sine values for supplementary angles are equal.

sin (π/2 - θ) = sin θ (ref)

So:

Area = x (y/2) sin A

Nice! Much better than using the law of sines, and for the vectorists out there, expressed even more compactly as the cross-product.