# A parallelogram has sides A, B, C, and D. Sides A and B have a length of 7  and sides C and D have a length of  4 . If the angle between sides A and C is (7 pi)/12 , what is the area of the parallelogram?

Feb 5, 2016

The area is $7 \cdot \left(\sqrt{2} + \sqrt{6}\right) \approx 27.05 {\text{ units}}^{2}$.

#### Explanation:

Let the angle between sides $A$ and $C$ be $\alpha = \frac{7 \pi}{12}$.

The formula to compute the area of the parallelogram is

$\text{Area} = A \cdot C \cdot \sin \left(\alpha\right)$

$= 7 \cdot 4 \cdot \sin \left(\frac{7 \pi}{12}\right)$

$= 28 \sin \left(\frac{7 \pi}{12}\right)$

So, the only thing left to do is compute $\sin \left(\frac{7 \pi}{12}\right)$.

Let me show how to do this without the calculator but with some basic knowledge of $\sin$ and $\cos$ functions:

$\sin \left(\frac{7 \pi}{12}\right) = \sin \left(\frac{\pi}{4} + \frac{\pi}{3}\right)$

... use the formula $\sin \left(x + y\right) = \sin \left(x\right) \cdot \cos \left(y\right) + \cos \left(x\right) \cdot \sin \left(y\right)$...

$= \sin \left(\frac{\pi}{4}\right) \cdot \cos \left(\frac{\pi}{3}\right) + \cos \left(\frac{\pi}{4}\right) \cdot \sin \left(\frac{\pi}{3}\right)$

$= \frac{1}{\sqrt{2}} \cdot \frac{1}{2} + \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}$

$= \frac{1 + \sqrt{3}}{2 \sqrt{2}}$

$= \frac{\sqrt{2} + \sqrt{6}}{4}$

Thus, you have the area of

${\text{Area" = 28 sin((7pi)/12) = 28 * (sqrt(2) + sqrt(6))/4 = 7 * (sqrt(2) + sqrt(6)) ~~ 27.05 " units}}^{2}$