I kinda struggling with this one. Would really appreciate some help.

**A parallelogram is located in the polar coordinate system with vertices (0,0) , (2, pi/6) , (4,0) , and (a , b). **

**What is the area of the parallelogram?**

I don't understand what first step to take. And what are polar coordinates? The internet isn't helping much. Thanks guys.

Gh0sty15 Sep 12, 2019

#1**+1 **

One way is to convert (2, pi/6) into Cartesian Coordinates.....so we have

To

(2cos (pi/6) , 2sin (pi/6) ) = (2 * √3/2, 2 * 1/2) = (√3/2 , 1 )

One possible point for the remaining vertex we have

(√3/2 + 0, 1 + 0) - (4,0) = (√3/2 - 4, 1)

The area of this parallelogram (ABCD) = base * height = 4 * 1 = 4 units^2

Another possible vertex is given by

(√3/2 + 4, 1 + 0) - (0, 0) = (√3/2 + 4, 1)

The area of this parallelogram (ABCE) is the same as the first one

The last possible vertex is

(4 + 0, 0 + 0) - (√3/2 , 1) = ( 4 - √3/2, -1)

And the area of this paralleogram ( AFBC) = area of triangle ABC + area of triangle ABF = 4 units^2

So all the possible parallelograms have the same areas !!!

See the following image :

CPhill Sep 12, 2019