Let a and $b$ be the longest sides of $T$ and $Q$ respectively. Also let $h$ be the height in $T$ perpendicular to $a$. We need to prove that $h<b$. In fact $3\le b^2\frac{\sqrt{3}}{4}$ and $4< a^2\frac{\sqrt{3}}{4}$. Thus $$h=\frac{2[T]}{a}<\frac{8}{4/\sqrt[4]{3}}=2\sqrt[4]{3}\le b.$$