Find the area of the quadrilateral with vertices at (a, b), (-b,0), (0, -a), and (a, -b). Given that a b 0. Thank you.

Question

Find  the area of the quadrilateral with vertices at (a, b), (-b,0), (0, -a), and (a, -b). Given that a > b > 0. Thank you.

anonymous · Answer

|dw:1374902626998:dw|

anonymous · Answer

Let's assume that $\bf a =2 ,\ b=1$ which satisfies the inequality $\bf a > b > 0$. So our 4 coordinates become: $\bf A(2,1) ,\ B(-1,0) ,\ C(0,-2) ,\ D(2,-1)$. Let's plot these points: |dw:1374902238505:dw|A you can see, segments AB and AD will always be congruent. Another thing we see is that we have a division of 2 triangles of the original trapezoid.