Question

The coordinates of the vertices of a quadrilateral are given below .
(2, 3) (8, 3)
(8, 11)
(11, 7)
What is the area, in square units, of the quadrilateral?

Answer 1

DRAW IT !!!!!

First the triangle on the left
(2,3), (8,3), (8,11)
base = 6, altitude = 8
so area = 3*8 = 24

Now the right side, the hard part
First rectangle
(8,3)(8,11)(11,11)(11,3)
that area = 3*8 = 24
subtract the triangles top and bottom
top is (8,11)(11,7)(11,11)
area top = 2*3 = 6
bottom is (8,3)(11,3)(11,7)
area bottom = 2*3 = 6
so
area right = 24 - 12 = 12
and in the end
24 + 12 = 36

Answer 2

easiest algorithm for area of any polygon:
list the points in a column going counterclockwise, repeat the first point listed:
area = (1/2)(downproducts - upproducts)

8 11
2 3
8 3
11 7
8 11

area = (1/2)(24+6+56+121 - (22+24+33+56))
= 36