SOLUTION: What is the area of the rectangle? Show all work. Coordinates: (-4, 0), (-6, 4), (4, 4), (2, 8) Do not round anything until the very end. Round your final answer to the tent

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Question 1149554: What is the area of the rectangle? Show all work.
Coordinates: (-4, 0), (-6, 4), (4, 4), (2, 8)
Do not round anything until the very end.
Round your final answer to the tenths place.

Found 4 solutions by Alan3354, greenestamps, ikleyn, MathTherapy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
What is the area of the rectangle? Show all work.
Coordinates: A(-4, 0), B(-6, 4), C(4, 4), D(2, 8)
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Here's a method for any polygon, and # of sides:
----

 A    B    C    D    A
-4   -6    4    2   -4
 0    4    4    8    0


Add the diagonal products starting at the upper left:
-4*4 + -6*4 + 4*8 + 2*0 = -16 - 24 + 32 + 0 = 8
---
Add the diagonal products starting at the lower left:
0*-6 + 4*4 + 4*2 + 8*-4 = 16 + 8 -32 = -8
---
The difference is 16
The area is 1/2 that, = 8 sq units.
-----------------------
The points have to be in order around the polygon, the only restriction.
==================
The points are not in order, so the area is not 8 sq units.
You can list them in order, then do the calculations.
---------------
It seems to work for polygons of all shapes, not just convex polygons, but I'm not yet certain of that.
=========================
Update: it works for all "simple polygons." Which means the vertices are listed in order.
---
Ref: CRC Press, 32nd Edition


Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The solution currently posted from tutor @Alan is incorrect.

To use that method, as he states at the end of his post, the coordinates of the vertices of the polygon (rectangle) have to be in order. The way he ordered the coordinates, they are not.

Wait and see if he corrects his response; his method is a good one to know for finding the area of any polygon.

For this example, knowing that the polygon is a rectangle (area length times width), we can use the distance formula to find the lengths of two adjacent sides and multiply those two lengths to get the area.

One side: (-4,0) to (4,4); length
Adjacent side: (-4,0) to (-6,4); length

Area: length * width =

ANSWER: 40 square units


Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

1.  Make a sketch (making sketch is a necessary part of the solution to this problem (!) ).



2.  Conclude your figure into the smallest possible rectangle with horizontal and vertical sides.

    This rectangle has dimensions 10 (=4-(-6)) units horizontally and 8 (=8-0) units vertically, so its area is 80 square units.



3.  Cut (mentally) 4 right angled triangles from this large rectangle to get the smaller rectangle.

    The legs of these triangles are  2 and 4  units  and  8 and 4  units.



4.  So, from the area of the large rectangle you should subtract  the areas of these 4 triangles


         80 -  -  = 80 - 8 - 32 = 40.


Thus you get  the ANSWER  40 square units for the area of your rectangle

without using the distance formula and without calculating radicals.


This simple method often works for more complicated cases of arbitrary polygons,
if the vertices of a polygon lie in integer points of the coordinate grid.

Using this approach, even 3-th or 4-th grade student is able to solve similar problems, even although he (or she)
doesn't know yet the distance formula and radicals.



Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

What is the area of the rectangle? Show all work.
Coordinates: (-4, 0), (-6, 4), (4, 4), (2, 8)
Do not round anything until the very end.
Round your final answer to the tenths place.
Alan's method works, but as he pointed out, the coordinate points have to be in order (CLOCK or COUNTERCLOCKWISE), and one of the ways to
determine the order is to do what @IKLEYN suggests: Sketching the polygon.
Doing so, instead of: A B C D A A B C D A
-4 -6 4 2 -4 -4 -6 2 4 -4
0 4 4 8 0, we get: 0 4 8 4 0
The sum of the "POSITIVE" diagonal-products gives us: (0 * - 6) + (4 * 2) + (8 * 4) + (4 * - 4) = 0 + 8 + 32 - 16 = 24
The sum of the "NEGATIVE" diagonal-products gives us: (- 4 * 4) + (- 6 * 8) + (2 * 4) + (4 * 0) = - 16 - 48 + 8 + 0 = - 56

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