Question

Two rhombuses have sides with lengths of `1`. If one rhombus has a corner with an angle of `pi/6` and the other has a corner with an angle of `(3pi)/4`, what is the difference between the areas of the rhombuses?

Answer 1

See explanation.

Explanation:

The area of a rhombus can be calculated using:

`A=a^2sin alpha`

where `a` is the side length and `alpha` is any of the rhombus' angles. Here both rhombuses have `a=1` so their areas are just

`A_1=sin (pi/6)=1/2`

and

`A_2=sin((3pi)/4)`

`A_2=sin((3pi)/4)=sin(pi-pi/4)=sin(pi/4)=sqrt(2)/2`

The difference is then:

`A_2-A_1=sqrt(2)/2-1/2=(sqrt(2)-1)/2`