Two opposite sides of a parallelogram have lengths of #3 #. If one corner of the parallelogram has an angle of #pi/8 # and the parallelogram's area is #36 #, how long are the other two sides?

Question

Two opposite sides of a parallelogram have lengths of #3  #.  If one corner of the parallelogram has an angle of #pi/8   # and the parallelogram's area is #36      #, how long are the other two sides?

Wyatt Anderson · Accepted Answer

#=31.36# Area of the parallelogram is #=ab sin theta# where #a=3# is the side and #theta=(pi)/8# and we have to find #b=?# Therefore Area#=36=3b sin((pi)/8)# or #bsin((pi)/8)=36/3# or #b(0.383)=12# or #b=31.36#

Wyatt Anderson · Answer

undefined We can use the formula for the area of a parallelogram, which is given by the product of the length of one side and the perpendicular distance to that side from the opposite side. Let's denote the lengths of the other two sides as $a$ and $b$, and the angle between the known side and one of the unknown sides as $\theta$.

Given that the area of the parallelogram is 36, we have:

$$36 = 3 \times h$$

where $h$ is the perpendicular distance between the two opposite sides of length 3.

To find $h$, we can use the sine of the given angle, $\frac{\pi}{8}$:

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

Now, we can find the lengths of the other two sides using the area formula again:

$$36 = a \times 3 \sin\left(\frac{\pi}{8}\right)$$

$$36 = b \times 3 \sin\left(\frac{\pi}{8}\right)$$

Solving for $a$ and $b$:

$$a = \frac{36}{3 \sin\left(\frac{\pi}{8}\right)}$$

$$b = \frac{36}{3 \sin\left(\frac{\pi}{8}\right)}$$