(Don't read yet) Finding the are of a Parallelogram with coordinates...? PreCalculus
Hello. Precal 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 precal book.
Re: Finding the are of a Parallelogram with coordinates...? PreCalculus
A simple plot of the points shows that this is not a parallelogram...
1 Attachment(s)
Re: Finding the are of a Parallelogram with coordinates...? PreCalculus
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
Re: Finding the are of a Parallelogram with coordinates...? PreCalculus
Hello, theadage!
Please check for typos.
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 has base 4 and height 4.
. . Its area is
Triangle has base 4 and height 5.
. . Its area is
The total area is
Re: Finding the are of a Parallelogram with coordinates...? PreCalculus
A(1,3) B(5,2) C(3,6) D(7,6)
I know that BXH= Area.
But it's too easy... there must be a trap.
;0