3. Given points $E$ and $F$ are on the sides $BC$ and $CD$ of square $ABCD$ respectively, $\angle EAF=45^{\circ}$, and $S_{\text {square } ABCD}$ : $S_{\triangle AEF}=5: 2$. Then $AB: EF=(\quad$.
(A) $5: 2$
(B) $25: 4$
(C) $\sqrt{5}: 2$
(D) $5: 4$
3. D.

As shown in Figure 4, if $\triangle A D F$ is rotated $90^{\circ}$ clockwise around point $A$ to get $\triangle A B G$, then points $G$, $B$, and $C$ are collinear, and
$$
\begin{array}{l}
A G=A F, \\
\angle G A B=\angle F A D, \\
\angle G A F=90^{\circ} .
\end{array}
$$

Therefore, $\angle G A E=\angle G A F-\angle E A F=45^{\circ}=\angle E A F$. Since $A E$ is a common side, $\triangle G A E \cong \triangle F A E$. Thus, the corresponding heights $A B=A H$.
From $\frac{S_{\triangle A E F}}{S_{\text {square } A B C D}}=\frac{\frac{1}{2} E F \cdot A H}{A B^{2}}=\frac{E F}{2 A B}=\frac{2}{5}$, we get $\frac{A B}{E F}=\frac{5}{4}$.