Determine the area of the quadrilateral whose vertices are (1,3), (6,5), (9,1) and
(3,-3).
Maybe you can use this method: http://www.mathhelpforum.com/math-he...ngle-help.html
See post #4.
$\displaystyle \dfrac{1}{3} \text{ : } \dfrac{6}{5} \text{ : } \dfrac{9}{1} \text{ : } \dfrac{3}{-3} \text{ : } \dfrac{1}{3} $
(1) sum of all the [ a numerator multiplied by the next denominator]
$\displaystyle 1 \cdot 5 + 6\cdot 1 + 9 \cdot -3 + 3 \cdot 3 = -7 $
(2) sum of all the [ denominator multiplied by the next numerator ]
$\displaystyle 3 \cdot 6 + 5 \cdot 9 + 1 \cdot 3 + -3 \cdot 1 = 63$
subtract the second sum from the first (-7 -63 = -70)
and divide that absolute value by 2 ( 70/2 = 35 )
This method will calculate the area of any polygon, when you know the coordinates of the points around the polygon. If you proceed in a clockwise direction around the perimeter (as was done above) you will have negative quantity; If you proceed in a counterclockwise direction the resulting quantity will be positive.