Question

Two rhombuses have sides with lengths of `8`. If one rhombus has a corner with an angle of `pi/12` and the other has a corner with an angle of `(5pi)/8`, what is the difference between the areas of the rhombuses?

Answer 1

Area of rhombus `= (1/2) * d_1 * d_2 or a * h`
Where `d_1 , d_2` are the diagonals, a is the side and h is the altitude.

In this case we will use the formula Area = a * h.

Rhombus 1
`h = a sin theta = 8 * sin ((pi)/12) = 2.0706`
Area = a * h = 8 * 2.0706 = 16.5648#

Rhombus 2
`h = a sin theta = 8 * sin ((5pi)/8) = 7.391`
Area = a * h = 8 * 7.391 = 59.128#

Hence the difference in areas is `59.128 - 16.5648 = 42.5632`