A parallelogram has sides A, B, C, and D. Sides A and B have a length of #8 # and sides C and D have a length of # 9 #. If the angle between sides A and C is #pi/4 #, 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 #8  # and sides C and D have a length of # 9 #.  If the angle between sides A and C is #pi/4   #, what is the area of the parallelogram?

Sophia Alfred · Accepted Answer

# 36 sqrt2 " units"^2 # Area of a parallelogram is the product of the base and the height. We know that the length of the base is #9#. The height may be calculated by the sine of the angle between the sides. Thus, # h / ("side" A) = sin( pi/4 ) = 1 / sqrt2 # # => h = ( 8 " units" ) / sqrt(2) = 4 sqrt2 " units" # Thus, the required area is # A = 9 " units" * 4 sqrt2 " units" = 36 sqrt2 " units"^2 #.

Sadie Adams · Answer

undefined The area of the parallelogram can be calculated using the formula:

$$ \text{Area} = \text{base} \times \text{height} $$

Given that sides A and C form the base of the parallelogram, and the height is the perpendicular distance between them, the height can be found using trigonometry. Since the angle between sides A and C is $ \frac{\pi}{4} $, the height can be calculated as follows:

$$ \text{Height} = \text{Length of side C} \times \sin\left(\frac{\pi}{4}\right) $$

Then, the area can be calculated as:

$$ \text{Area} = \text{Length of side A} \times \text{Height} $$

Substituting the given values:

$$ \text{Height} = 9 \times \sin\left(\frac{\pi}{4}\right) = 9 \times \frac{\sqrt{2}}{2} = \frac{9\sqrt{2}}{2} $$

$$ \text{Area} = 8 \times \frac{9\sqrt{2}}{2} = 36\sqrt{2} $$

So, the area of the parallelogram is $ 36\sqrt{2} $ square units.