We are given A = (2,5), B = (-5, 2), C = (-3, -6), and D = (6,-3). Find [ABCD], the area of quadrilateral ABCD.
+129899 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
