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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.

 Sep 12, 2019
 #1
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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  :

 

 

cool cool cool

 Sep 12, 2019

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