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?

1 Answer
Lotusbluete
Feb 5, 2016

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#

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