2. The side length of square $ABCD$ is equal to 7. Points $E$, $F$, $G$, and $H$ are taken on sides $AB$, $BC$, $CD$, and $DA$ respectively. If quadrilateral $EFGH$ is a rhombus with a side length of 5, then the lengths of the diagonals $EG$ and $FH$ of the rhombus are $\qquad$
2. $5 \sqrt{2}, 5 \sqrt{2}$.

As shown in Figure 6, since quadrilateral $E F G H$ is a rhombus, $F H$ and $E G$ are perpendicular to each other. Therefore, it can be proven that $F H=E G$. Hence, quadrilateral $E F G H$ is a square. Thus,
$\triangle A E H \cong \triangle B F E$ $\cong \triangle C G F \cong \triangle D H G$.
Figure 6
Therefore, $A E=B F$.
Let $B E=x$, then $B F=7-x$.
In the right triangle $\triangle B E F$, by $x^{2}+(7-x)^{2}=5^{2}$, solving for $x$ yields $x=3$ or 4. Thus, we calculate $F H=E G=5 \sqrt{2}$.