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 #(3 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 #6  # and sides C and D have a length of # 2  #.  If the angle between sides A and C is #(3  pi)/4    #, what is the area of the parallelogram?

Theodore Abad · Accepted Answer

The area of parallelogram is #8.49(2dp) # sq.unit.

The area of parallelogram of parallal sides #A=B=6 # and another parallal sides #C=D=2 # and an angle between #A and C , theta= (3pi)/4 = 3*180/4=135^0# is #A_p=A*C*sin theta = 6*2*sin135=8.49(2dp) #sq.unit [Ans]

Parker Allen · Answer

undefined To find the area of the parallelogram, you can use the formula:

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

For a parallelogram, any side can be considered the base, and the height is the perpendicular distance between the base and its opposite side.

Given that sides $A$ and $C$ form the base and the angle between them is $\frac{3\pi}{4}$, you can use trigonometric principles to find the height.

First, find the height using the sine of the angle between sides $A$ and $C$:

$$ \text{Height} = \text{side C} \times \sin\left(\frac{3\pi}{4}\right) $$

Substitute the values:

$$ \text{Height} = 2 \times \sin\left(\frac{3\pi}{4}\right) $$

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

$$ \text{Height} = \sqrt{2} $$

Now, you have the base ($A$) and the height of the parallelogram. Plug these values into the formula for the area:

$$ \text{Area} = 6 \times \sqrt{2} $$

$$ \text{Area} = 6\sqrt{2} $$

So, the area of the parallelogram is $6\sqrt{2}$.