+590 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.
+129899 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 :
