A parallelogram has sides A, B, C, and D. Sides A and B have a length of #6 # and sides C and D have a length of # 2 #. If the angle between sides A and C is #pi/3 #, 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 #6  # and sides C and D have a length of # 2 #.  If the angle between sides A and C is #pi/3   #, what is the area of the parallelogram?

Savannah Aguilar · Accepted Answer

#A_p = color(blue)(10.3923)# sq. units

Area of parallelogram #A_p = A * h = A * C sin A#

Given : #A = 6, C = 2, /_theta = pi / 3#

#A_p = 6 * 2 * sin ((pi)/3) = 6 sqrt3 = color(blue)(10.3923)# sq. units

Allison Allen · Answer

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

$$ \text{Area} = \text{Base} \times \text{Height} $$

In a parallelogram, the base can be taken as the length of either side A or side C, and the height is the perpendicular distance between these two sides. The height can be found using trigonometry.

Given that sides A and B have a length of 6 units, sides C and D have a length of 2 units, and the angle between sides A and C is $ \frac{\pi}{3} $ radians, we can find the height as follows:

$$ \text{Height} = \text{Length of side D} \times \sin(\text{angle between A and C}) $$

$$ \text{Height} = 2 \times \sin\left(\frac{\pi}{3}\right) $$

$$ \text{Height} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} $$

Now, plug in the values for the base and height into the area formula:

$$ \text{Area} = 6 \times \sqrt{3} = 6\sqrt{3} $$

Therefore, the area of the parallelogram is $ 6\sqrt{3} $ square units.