Determine the area of the quadrilateral whose vertices are (1,3), (6,5), (9,1) and
(3,-3).

2. Originally Posted by sri340
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.

3. Hello sri340
Originally Posted by sri340
Determine the area of the quadrilateral whose vertices are (1,3), (6,5), (9,1) and
(3,-3).
Look at the attached diagram. Calculate the areas of the four right-angled triangles and the area of the rectangle. Add together. Done.

(If my arithmetic is correct, it's 35 sq. units.)

4. Originally Posted by sri340
Determine the area of the quadrilateral whose vertices are (1,3), (6,5), (9,1) and (3,-3).
$\displaystyle \dfrac{1}{3} \text{ : } \dfrac{6}{5} \text{ : } \dfrac{9}{1} \text{ : } \dfrac{3}{-3} \text{ : } \dfrac{1}{3}$

(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.