Area of Rhombus ABCD

Let $A$ be the origin, $D(\vec{d})$ and $B(\vec{b})$ represent the position vectors of point $D$ and $B$ . So, $C(\vec{d}+\vec{b})$ .

Given the area of $\triangle PQR=5$ , therefore,

$\frac{1}{2}\left| \vec{RP} × \vec{RQ}\right|=5 \\ \frac{1}{2}\left|\left(\vec{b}-\frac{\vec{d}}{4}\right)×\left(\frac{2\vec{b}}{3}+\frac{\vec{d}}{4}\right)\right|=5$

Simplifying gives us, $\left|\vec{b}×\vec{d}\right|=24$ .