SOLUTION: I've been trying to use the same formula that I use to find the area of a triangle but it is not giving me the correct answer. Find the area of the parallelogram with the follow

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Question 1199552: I've been trying to use the same formula that I use to find the area of a triangle but it is not giving me the correct answer.
Find the area of the parallelogram with the following vertices:

11. (-2, 3), (5, 8), (3, 3), and (0, 8)
A. 16 sq units
B. 20 sq units
C. 34 sq units
D. 18 sq units
E. 25 sq units
F. 36 sq units

12. (-2, 7), (-4, 4), (-11, 4), and (-9, 7)
A. 15 sq units
B. 63 sq units
C. 28 sq units
D. 21 sq units
E. 36 sq units
F. 14 sq units
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Find the area of the parallelogram with the following vertices:

11. A(-2,3), B(5,8), C(3,3), and D(0,8)
 A   D   B   C   A
-2   0   5   3  -2
 3   8   8   3   3

Add the diagonal products starting upper left
-2*8 + 0*8 + 5*3 + 3*3 = -16+15+9 = 8
---
Add the diagonal products starting lower left
3*0 + 8*5 + 8*3 + 3*-2 = 0+40+24-6 = 58
The difference is 50.
The area is 1/2 that, = 25 sq units.
-------------------
This works for all convex polygons, but the points have to be in order around the perimeter. Notice it's A, D, B, C
=========================
12. A(-2,7), B(-4,4), C(-11,4), and D(-9,7)
A. 15 sq units
B. 63 sq units
C. 28 sq units
D. 21 sq units
E. 36 sq units
F. 14 sq units
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

                                Part   (a)


If you draw the given points in a coordinate plane, you will see

(even with unarmed eye)  that one pair of sides is horizontal   (y = 3   and   y = 8)  of the length of  5  units,

while the height of the parallelogram (the distance between its parallel horizontal sides)  is  8 - 3 = 5  units.

It gives the area  5*5 = 25  square units.


In part  (b),  the same reasoning does work,  allowing you to get the answer  MENTALLY.


///////////////////


At the beginning of your post you write

        " I've been trying to use the same formula that I use to find
        the area of a triangle but it is not giving me the correct answer. "


It is not surprising,  because you have no any reason to use
" the same "  formula for totally different problem.

It is like jumping off a tower upside down into a swimming pool with no water.


In opposite,  it would be surprising if this formula would work.



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