We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
61
1
avatar+581 

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
avatar+103858 
+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  :

 

 

cool cool cool

 Sep 12, 2019

5 Online Users

avatar