# 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?

Mar 13, 2016

$D \iff e r e n c e = | A {\diamond}_{1} - A {\diamond}_{2} | = 2 \sqrt{3} - 2 \sqrt{2 + \sqrt{2}}$
$| A {\diamond}_{1} - A {\diamond}_{2} | = 2 \left(\sqrt{3} - \sqrt{2 + \sqrt{2}}\right)$
$| A {\diamond}_{1} - A {\diamond}_{2} | \approx 2 | 1.732 - 1.848 | = .231$

#### Explanation:

The easiest way to approach is using the following approach to finding the area of a parallelogram:
$A \diamond = {\vec{s}}_{1} \times {\vec{s}}_{2} = | {s}_{1} \cdot {s}_{2} | \sin \theta$, where $\theta$ is the angle between the sides of the rhombuses. Since ${s}_{1} = {s}_{2}$ we write:$A \diamond = | {s}^{2} | \sin \theta$ thus,
$A {\diamond}_{1} = {2}^{2} \sin \left(\frac{\pi}{3}\right) = 2 \sqrt{3}$
$A {\diamond}_{2} = {2}^{2} \sin \left(5 \frac{\pi}{8}\right) = 2 \sqrt{2 + \sqrt{2}}$
$D \iff e r e n c e = | A {\diamond}_{1} - A {\diamond}_{2} | = 2 \sqrt{3} - 2 \sqrt{2 + \sqrt{2}}$
$| A {\diamond}_{1} - A {\diamond}_{2} | = 2 \left(\sqrt{3} - \sqrt{2 + \sqrt{2}}\right)$
$| A {\diamond}_{1} - A {\diamond}_{2} | \approx 2 | 1.732 - 1.848 | = .231$