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.

Answer 1

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: 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.