Question

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?

Answer 1

The area is `7 * (sqrt(2) + sqrt(6)) ~~ 27.05 " units"^2`.

Explanation:

Let the angle between sides `A` and `C` be `alpha = (7pi)/12`.

The formula to compute the area of the parallelogram is

`"Area" = A * C * sin(alpha)`

`= 7 * 4 * sin((7pi)/12)`

`= 28 sin((7pi)/12)`

So, the only thing left to do is compute `sin((7pi)/12)`.

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

`sin((7pi)/12) = sin(pi/4 + pi/3)`

... use the formula `sin(x+y) = sin(x)*cos(y) + cos(x)*sin(y)`...

`= sin(pi/4) * cos(pi/3) + cos(pi/4) * sin(pi/3)`

`= 1/sqrt(2) * 1/2 + 1/sqrt(2) * sqrt(3)/2`

`= (1 + sqrt(3))/(2sqrt(2))`

`= (sqrt(2) + sqrt(6))/4`

Thus, you have the area of

`"Area" = 28 sin((7pi)/12) = 28 * (sqrt(2) + sqrt(6))/4 = 7 * (sqrt(2) + sqrt(6)) ~~ 27.05 " units"^2`