option 2 ... (refer to the attached diagram)Not gonna lie I don't fully understand what you're saying here
$\dfrac{A}{3} < \dfrac{1}{2} ab \sin{\theta} < \dfrac{A}{2}$
solve the inequality for the length of $\color{red}a$
$A = \text{ area of quadrilateral ABCD}$
$\theta = \angle{DAB}$
$b = AB$
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