# Finding the are of a Parallelogram with coordinates...? Pre-Calculus

• February 5th 2013, 06:59 PM
(Don't read yet) Finding the are of a Parallelogram with coordinates...? Pre-Calculus
Hello. Pre-cal practice work for test tomorrow.

Problem gives us these coordinates: A(1,2), B(5,2), C(3,6), D(5, -3)

I graphed it. How are we supposed to find the area with these coordinates?

I know the answer is 16, because I checked the back of the book. I would like to know how to get that answer.

Thank you :)

Oh yes, I apologize for the misconveniance.. I will correct this when I get back to my room to access my pre-cal book.
• February 5th 2013, 07:17 PM
Prove It
Re: Finding the are of a Parallelogram with coordinates...? Pre-Calculus
A simple plot of the points shows that this is not a parallelogram...
• February 5th 2013, 09:04 PM
ibdutt
Re: Finding the are of a Parallelogram with coordinates...? Pre-Calculus
It is not a parallelogram. Join any diagonal of the quadrilateral. and use the formula for finding the area of a triangle with given vertices. Attachment 26870
• February 5th 2013, 09:45 PM
Soroban
Re: Finding the are of a Parallelogram with coordinates...? Pre-Calculus

The problem doesn't make sense.

Quote:

Problem gives us these coordinates: A(1,2), B(5,2), C(3,6), D(5, -3) .[1]

I graphed it. How are we supposed to find the area with these coordinates?

I know the answer is 16, because I checked the back of the book. .[2]

I would like to know how to get that answer. . So would I!

[1] The vertices are usually labeled in order around the parallelogram.
. . .The given coordinates are not.
. . .Besides, they do not form a parallelogram.
Code:

```      |       |    C       |    o       |    *:*       |  * :4*       |  *  :  *       |Ao * * * oB       |  * 4  *     - + - - * - *5 - -       |      * *       |        o       |        D       |```
[2] The area is not 16.

Triangle $ABC$ has base 4 and height 4.
. . Its area is $8.$

Triangle $ABD$ has base 4 and height 5.
. . Its area is $10.$

The total area is $18.$
• February 6th 2013, 08:57 AM