Question

Two rhombuses have sides with lengths of `2`. If one rhombus has a corner with an angle of `pi/3` and the other has a corner with an angle of `(5pi)/8`, what is the difference between the areas of the rhombuses?

Answer 1

`Difference=|Adiamond_1-Adiamond_2|= 2sqrt(3)-2sqrt(2+sqrt(2))`
`|Adiamond_1-Adiamond_2|= 2(sqrt(3)-sqrt(2+sqrt(2)))`
`|Adiamond_1-Adiamond_2| ~~ 2|1.732-1.848| = .231`

Explanation:

The easiest way to approach is using the following approach to finding the area of a parallelogram:
`Adiamond= vecs_1 xx vecs_2 = |s_1*s_2| sin theta`, where `theta` is the angle between the sides of the rhombuses. Since `s_1=s_2` we write:`Adiamond= |s^2| sin theta` thus,
`Adiamond_1= 2^2 sin (pi/3) = 2sqrt(3)`
`Adiamond_2= 2^2 sin (5pi/8) = 2sqrt(2+sqrt(2))`
`Difference=|Adiamond_1-Adiamond_2|= 2sqrt(3)-2sqrt(2+sqrt(2))`
`|Adiamond_1-Adiamond_2|= 2(sqrt(3)-sqrt(2+sqrt(2)))`
`|Adiamond_1-Adiamond_2| ~~ 2|1.732-1.848| = .231`