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We are given A = (2,5), B = (-5, 2), C = (-3, -6), and D = (6,-3). Find [ABCD], the area of quadrilateral ABCD.

 Nov 25, 2020
 #1
avatar+110 
+1

Have you tried using the Shoelace Formula?

 Nov 25, 2020
 #2
avatar+116126 
+1

We could figure this using some trig, but there is a "formula"   to figure it faster

 

Listing the points in a counter-clockwise manner  :

 

A =  (x1 , y1 )  = (2,5)

B= ( x2, y2)  = (-5, 2)

C = (x3, y3)  = (-3, -6) 

D = (x4, y4)  = ( 6, -3)

 

Area =  (1/2) {(x1y2 + x2y3 + x3y4 + x4y1)  - (x2y1 + x3y2 + x4y3 + x1y4)}

 

Area  = (1/2)  [ ( 2*2  + (-5)(-6) + (-3)(-3)  + (6)(5) )  -  ( (-5)(5) + (-3)(2) + (6)(-6) + (2)(-3) )  

 

Area  =  (1/2) [  (4 + 30 + 9 + 30)  -   ( -25 - 6 - 36 - 6 ) ]

 

Area = (1/2)  [ 73  -  (- 73) ] 

 

Area =  (1/2)  (146)  =       73 units^2 

 

 

cool cool cool

 Nov 25, 2020
 #3
avatar+116126 
0

THX, OlympusHero !!!.....I didn't realize  that this procedure  had a name

 

Here's a website  that explains it pretty well  :  https://www.onlinemath4all.com/area-of-quadrilateral-when-four-vertices-are-given.html

 

 

cool cool cool

CPhill  Nov 25, 2020

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