A parallelogram has sides A, B, C, and D. Sides A and B have a length of #2 # and sides C and D have a length of # 1 #. If the angle between sides A and C is #(3 pi)/8 #, what is the area of the parallelogram?

Question

A parallelogram has sides A, B, C, and D.  Sides A and B have a length of #2  # and sides C and D have a length of # 1  #.  If the angle between sides A and C is #(3  pi)/8    #, what is the area of the parallelogram?

Claire Akin · Accepted Answer

#=1.84# Area of the Parallelogram #=ab sin theta# where#a=2# and #b=1# are the sides and #theta=(3pi)/8# Hence Area #=(2)(1)sin((3pi)/8)=2(0.92)=1.84#

Avery Allen · Answer

undefined The area of the parallelogram can be calculated using the formula: Area = base * height. In this case, the base is the length of side A, which is 2 units, and the height is the perpendicular distance between side A and side C. To find the height, we can use trigonometry.

Since the angle between sides A and C is given as (3π)/8, we can use the sine function to find the height:

sin((3π)/8) = height / 1

Solving for height:

height = sin((3π)/8) * 1

Now, we can calculate the area using the formula:

Area = base * height

Area = 2 * sin((3π)/8) * 1

Area ≈ 1.329 square units.

Violet Aldrich · Answer

undefined The area of the parallelogram is $ A = 2 \sin(\theta) $, where $ \theta $ is the angle between sides A and C. Given $ \theta = \frac{3\pi}{8} $, the area can be calculated as $ A = 2 \sin\left(\frac{3\pi}{8}\right) $.