A parallelogram has sides A, B, C, and D. Sides A and B have a length of #1 # 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 #1  # 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?

Aaliyah Anderson · Accepted Answer

area of given parallelogram is #1.4142#

Area of a parallelogram with sides #a# and #b# and included angle #theta# is given by #axxbxxsintheta#.

Note that in a parallelogram. while one angle is #theta#, other angle is #(180^o-theta)# and as #sin(180^o-theta)=sintheta#, which ever angle we may choose, area is not affected.

As two sides are #1# and #2# and included angle is #(3pi)/4# and

area of given parallelogram is

#1xx2xxsin((3pi)/4)=2xx1/sqrt2=sqrt2=1.4142#

Jayden Jackson · Answer

undefined The area of a parallelogram is given by the formula:

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

In this case, the base of the parallelogram is the length of side $C$, which is 2 units. To find the height, we can use the sine of the angle between sides $A$ and $C$, which is $\frac{3\pi}{4}$ radians. The height is given by:

$$ \text{Height} = \text{side} \times \sin(\text{angle}) $$

Substitute the values:

$$ \text{Height} = 1 \times \sin\left(\frac{3\pi}{4}\right) $$

$$ \text{Height} = 1 \times \frac{\sqrt{2}}{2} $$

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

Now, we can calculate the area:

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

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

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