Two opposite sides of a parallelogram each have a length of #12 #. If one corner of the parallelogram has an angle of #(3pi)/8 # and the parallelogram's area is #24 #, how long are the other two sides?

Question

Two opposite sides of a parallelogram each have a length of #12  #.  If one corner of the parallelogram has an angle of #(3pi)/8   # and the parallelogram's area is #24      #, how long are the other two sides?

Wyatt Anderson · Accepted Answer

Other two sides are #2.165# units long.

Area of a parallelogram whose one pair of sides are #a#, other pair of sides are#b# and included angle is #theta# is given by #axxbxxsintheta# As area of parallelogram is #24#, one pair of sides have length #12# and included angle is #(3pi)/8#, we have #12xxbxxsin((3pi)/8)=24# or #12xxbxx0.9239=24# and hence #b=24/(12xx0.9239)=(2cancel24)/(1cancel12xx0.9239)=2/0.9239=2.165#

Aurora Ayala · Answer

undefined Let $ a $ and $ b $ be the lengths of the other two sides of the parallelogram.

Given that the area of the parallelogram is $ 24 $ and one side has a length of $ 12 $, the area can also be expressed as the product of the base and the height. Since the height is the length of the side perpendicular to the given side, it can be calculated as $ b \sin\left(\frac{3\pi}{8}\right) $.

So, we have:

$$ 24 = 12 \cdot b \cdot \sin\left(\frac{3\pi}{8}\right) $$

Now, solve for $ b $:

$$ b = \frac{24}{12 \cdot \sin\left(\frac{3\pi}{8}\right)} $$

$$ b = \frac{2}{\sin\left(\frac{3\pi}{8}\right)} $$

Now, we can use the fact that opposite sides of a parallelogram are equal in length, so $ a = 12 $ as well.

Thus, the lengths of the other two sides are both $ 12 $ units.