Two opposite sides of a parallelogram each have a length of #4 #. If one corner of the parallelogram has an angle of #(7 pi)/8 # and the parallelogram's area is #18 #, how long are the other two sides?

Question

Two opposite sides of a parallelogram each have a length of #4    #.  If one corner of the parallelogram has an angle of #(7 pi)/8       # and the parallelogram's area is #18      #, how long are the other two sides?

Sophia Alfred · Accepted Answer

#11.76# each

The area is #A = b xx h#, so select one of the known sides as the base. Then #h = 18/4 = 4.5# The angle between an unknown side and the base (4) is the complement of #7/8pi, or 1/8pi#. The unknown side is thus #x = h/(sin(pi/8))# ; #x = 4.5/0.383 = 11.76# You can check it by using this number, and working it again from the other angle (#pi/8#) to see that you get the length of 4 back for the short sides.

Jayden Jackson · Answer

undefined First, find the length of the diagonal using the area of the parallelogram formula: $ \text{Area} = \text{base} \times \text{height} \times \sin(\theta) $, where $ \theta $ is the angle between the given sides.

Given:
- Length of one side = 4
- Area of parallelogram = 18
- Angle $ \theta = \frac{7\pi}{8} $

Substitute these values into the area formula to get: $ 18 = 4 \times \text{height} \times \sin\left(\frac{7\pi}{8}\right) $

Solve for the height: $ \text{height} = \frac{18}{4 \times \sin\left(\frac{7\pi}{8}\right)} $

Once you have the height, you can use the Pythagorean theorem to find the lengths of the other two sides since the diagonals of a parallelogram bisect each other. This can be expressed as $ \text{Diagonal}^2 = \text{base}^2 + \text{height}^2 $.

Calculate the height and then use it to find the lengths of the other two sides using the Pythagorean theorem.