Question

Two rhombuses have sides with lengths of `8`. If one rhombus has a corner with an angle of `(5pi)/12` and the other has a corner with an angle of `(7pi)/12`, what is the difference between the areas of the rhombuses?

Answer 1

Since both the parallelograms are identical, difference in areas is 0

Explanation:

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 ((5pi)/12 = 7.7274`
Area = a * h = 8 * 7.7274 = 61.8192#

Rhombus 2
`h = a sin theta = 8 * sin ((7pi)/12) = 7.7274`
Area = a * h = 8 * 7.7274 = 61.8192#

If one angle is `(5pi)/12,` the other angle will be `(7pi)/12` as both the angles are supplementary and hence, the areas same.

Since both the parallelograms are identical, difference in areas is `color(red)0`