Discussion
The area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3) is:
Answer: B
We can find the area by dividing the quadrilateral into two triangles, e.g., ABC and ADC, and summing their areas. Let A=(-4,-2), B=(-3,-5), C=(3,-2), D=(2,3).
Area(\(\triangle ABC\)) = \(\frac{1}{2} |-4(-5-(-2)) -3(-2-(-2)) + 3(-2-(-5))| = \frac{1}{2} |-4(-3) -3(0) + 3(3)| = \frac{1}{2}|12+9| = 10.5\).
Area(\(\triangle ADC\)) = \(\frac{1}{2} |-4(-2-3) + 3(3-(-2)) + 2(-2-(-2))| = \frac{1}{2} |-4(-5) + 3(5) + 2(0)| = \frac{1}{2}|20+15| = 17.5\).
Total Area = 10.5 + 17.5 = 28 sq units.
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