1. 如图,在$\triangle ABC$中,$\angle C = 90^{\circ}$,$AC = 4$,$BC = 3$,将$\triangle ABC绕点A逆时针旋转得到\triangle ADE$(点$D与点B$对应),连接$BD$.当点$E落在边AB$上时,线段$BD$的长为______.

$\sqrt{10}$

2. (2024 陕西中考)如图,在$\triangle AOB$中,$\angle AOB = 90^{\circ}$,$OA = OB = 6$,将$\triangle AOB绕点O顺时针旋转45^{\circ}得到\triangle A'OB'$,$A'B'与OB相交于点D$,则$OD$的长为______.

$3\sqrt{2}$ 解: $\because ∠AOB = 90^{\circ}$，$OA = OB = 6$，$\therefore AB = \sqrt{OA^{2} + OB^{2}} = 6\sqrt{2}$。由旋转得$∠A'OB' = ∠AOB = 90^{\circ}$，$∠BOB' = ∠BOA' = 45^{\circ}$，$\because OA' = OB'$，$\therefore D$是$A'B'$的中点，$\therefore OD = \frac{1}{2}A'B' = \frac{1}{2}×6\sqrt{2} = 3\sqrt{2}$。

3. (教材$P_{60}$例题变式)如图,在正方形$ABCD$中,$AB = 4$,$E为AB$的中点,连接$DE$,将$\triangle DAE绕点D逆时针旋转90^{\circ}得到\triangle DCF$,连接$EF$,则$EF$的长为______.

$2\sqrt{10}$

4. (2024 湖北中考)如图,点$A的坐标是(-4,6)$,将线段$OA绕点O顺时针旋转90^{\circ}$,点$A的对应点B$的坐标是()
A. $(4,6)$
B. $(6,4)$
C. $(-6,-4)$
D. $(-4,-6)$

B 解: 分别过点$A$和点$B$作$x$轴的垂线, 垂足分别为$M$和$N$，$\therefore ∠AMO = ∠ONB = 90^{\circ}$。由旋转得$OA = OB$，$∠AOB = 90^{\circ}$，$\therefore ∠AOM + ∠BON = ∠A + ∠AOM = 90^{\circ}$，$\therefore ∠A = ∠BON$。$\therefore △AOM ≌ △OBN$，$\therefore BN = MO$，$ON = AM$。$\because$ 点$A$的坐标为$(-4,6)$，$\therefore BN = MO = 4$，$ON = AM = 6$，$\therefore$ 点$B$的坐标为$(6,4)$。故选 B。

5. 如图,在$\triangle ABC$中,$\angle BAC = 90^{\circ}$,$AB = AC$,$BC = 2$.点$D在BC$上,且$BD:CD = 1:3$,连接$AD$,将线段$AD绕点A顺时针旋转90^{\circ}得到线段AE$,连接$BE$,$DE$.求$\triangle BDE$的面积.

解: 由旋转得$AD = AE$，$∠DAE = ∠BAC = 90^{\circ}$，$\therefore ∠BAE = ∠CAD$，$\because AB = AC$，$\therefore △BAE ≌ △CAD$，$\therefore ∠ABE = ∠ACB = ∠ABC = 45^{\circ}$，$\therefore ∠CBE = 90^{\circ}$，$\therefore S_{△BDE} = \frac{1}{2}BD·BE$ $= \frac{1}{2}BD·CD$。$\because BC = 2$，$BD:CD = 1:3$，$\therefore BD = \frac{1}{2}$，$CD = \frac{3}{2}$，$\therefore S_{△BDE} = \frac{1}{2}×\frac{1}{2}×\frac{3}{2} = \frac{3}{8}$。