2017/18 AHSMC Part 1 Problem 11

Graphing the functions, we conclude that two of the vertices of the quadrilateral are $A = (2,2)$ and $C = (a,0)$. The other two vertices $B = \bigl(\dfrac{a}{2}, \dfrac{a}{2}\bigr)$ and $D = \bigl(\dfrac{4+a}{2}, \dfrac{4-a}{2}\bigr)$ are at the intersection of the lines $y = x$, $y = a - x$ and respectively $y = 4 - x$, $y = x - a$. The area of the quadrilateral $ABCD$ is $$4 - \frac{a^2}{4} - \frac{(4-a)^2}{4} = \frac{4a - a^2}{2}.$$ Solving the equation $\dfrac{4a - a^2}{2} = \dfrac{15}{8}$ one obtains $a = \dfrac{3}{2}$, $a = \dfrac{5}{2}$. The answer is (D).