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1. 如图，在正方形 $ABCD$ 中，$AB=6$，$E$ 为边 $AD$ 上的动点，连接 $CE$，以 $CE$ 为边作正方形 $ECGF$，连接 $DG$，$EG$，则 $\triangle DEG$ 面积的最大值为
$\frac{9}{2}$
.
答案：
1.$\frac{9}{2}$ 点拨:连接BE,如答图所示.由题意知$\angle CDE = \angle BCD = 90^{\circ}$,$BC = CD$,$CE = CG$,$\angle ECG = 90^{\circ}$,设$DE = x(0\leq x\leq6)$,则$CE^{2} = DE^{2} + CD^{2} = x^{2} + 6^{2} = x^{2} + 36$.
$\because \angle BCE + \angle DCE = \angle DCE + \angle DCG = 90^{\circ}$,$\therefore \angle BCE = \angle DCG$,$\therefore \triangle BCE \cong \triangle DCG$,$\therefore S_{\triangle BCE} = S_{\triangle DCG}$,$\therefore S_{\triangle DEG} = S_{四边形BCGD} - S_{\triangle BCG} = S_{\triangle BCD} + S_{\triangle DCG} - S_{\triangle BCG} = S_{\triangle BCD} + S_{\triangle BCE} - S_{\triangle BCG} = \frac{1}{2}S_{正方形ABCD} + S_{\triangle BCD} - S_{\triangle BCG} = \frac{1}{2} × BC × CD + \frac{1}{2}DE × CD - \frac{1}{2}CE^{2} = \frac{1}{2} × 6 × 6 + \frac{1}{2} × 6x - \frac{1}{2}(x^{2} + 36) = -\frac{1}{2}x^{2} + 3x = -\frac{1}{2}(x - 3)^{2} + \frac{9}{2}$.
$\because -\frac{1}{2} ＜ 0$,$\therefore$当$x = 3$时,$-\frac{1}{2}(x - 3)^{2} + \frac{9}{2}$最大,且最大值为$\frac{9}{2}$,即$\triangle DEG$面积的最大值为$\frac{9}{2}$.
1.$\frac{9}{2}$ 点拨:连接BE,如答图所示.由题意知$\angle CDE = \angle BCD = 90^{\circ}$,$BC = CD$,$CE = CG$,$\angle ECG = 90^{\circ}$,设$DE = x(0\leq x\leq6)$,则$CE^{2} = DE^{2} + CD^{2} = x^{2} + 6^{2} = x^{2} + 36$.
$\because \angle BCE + \angle DCE = \angle DCE + \angle DCG = 90^{\circ}$,$\therefore \angle BCE = \angle DCG$,$\therefore \triangle BCE \cong \triangle DCG$,$\therefore S_{\triangle BCE} = S_{\triangle DCG}$,$\therefore S_{\triangle DEG} = S_{四边形BCGD} - S_{\triangle BCG} = S_{\triangle BCD} + S_{\triangle DCG} - S_{\triangle BCG} = S_{\triangle BCD} + S_{\triangle BCE} - S_{\triangle BCG} = \frac{1}{2}S_{正方形ABCD} + S_{\triangle BCD} - S_{\triangle BCG} = \frac{1}{2} × BC × CD + \frac{1}{2}DE × CD - \frac{1}{2}CE^{2} = \frac{1}{2} × 6 × 6 + \frac{1}{2} × 6x - \frac{1}{2}(x^{2} + 36) = -\frac{1}{2}x^{2} + 3x = -\frac{1}{2}(x - 3)^{2} + \frac{9}{2}$.
$\because -\frac{1}{2} ＜ 0$,$\therefore$当$x = 3$时,$-\frac{1}{2}(x - 3)^{2} + \frac{9}{2}$最大,且最大值为$\frac{9}{2}$,即$\triangle DEG$面积的最大值为$\frac{9}{2}$.
2. 如图，矩形 $ABCD$ 中，$AB=2$，$BC=3$，$E$，$F$ 分别是边 $BC$，$CD$ 上的动点，且 $BE=CF$，将 $\triangle BCF$ 沿着 $BC$ 方向平移到 $\triangle EGH$，连接 $DH$，当 $DE=EH$ 时，$DH$ 的长是
$\frac{2\sqrt{5}}{3}$
；运动过程中，$\triangle DEH$ 的面积的最小值是$\frac{15}{8}$
.
答案：
2.$\frac{2\sqrt{5}}{3}$ $\frac{15}{8}$ 点拨:如答图,连接$FH$.$\because \triangle EGH \cong \triangle BCF$,$\therefore \angle DCB = \angle G = 90^{\circ}$,$FC = GH$,$BC = EG = 3$,$\therefore FC // GH$,$BE = CG$,$\therefore$四边形$FCGH$是矩形.$\because BE = CF$,$\therefore CG = CF$,$\therefore$四边形$CGHF$为正方形,$\therefore FH = CF$.
设$BE = x$,则$CF = FH = HG = x$,$\therefore EC = 3 - x$.
$\because DE = EH$,$\therefore (3 - x)^{2} + 2^{2} = 3^{2} + x^{2}$,
解得$x = \frac{2}{3}$,$\therefore CF = FH = \frac{2}{3}$,$\therefore DF = 2 - x = 2 - \frac{2}{3} = \frac{4}{3}$,
$\therefore DH = \sqrt{DF^{2} + FH^{2}} = \sqrt{(\frac{4}{3})^{2} + (\frac{2}{3})^{2}} = \frac{2\sqrt{5}}{3}$,
$\therefore S_{\triangle DEH} = S_{\triangle DEC} + S_{梯形DCGH} - S_{\triangle EHG} = \frac{1}{2}(3 - x) × 2 + \frac{1}{2} · (2 + x) · x - \frac{1}{2} × 3 · x = \frac{1}{2}x^{2} - \frac{3}{2}x + 3 = \frac{1}{2}(x - \frac{3}{2})^{2} + \frac{15}{8}$.
$\because \frac{1}{2} ＞ 0$,$\therefore$当$x = \frac{3}{2}$时,$S_{\triangle DEH}$有最小值,为$\frac{15}{8}$,即$\triangle DEH$的面积的最小值是$\frac{15}{8}$.
2.$\frac{2\sqrt{5}}{3}$ $\frac{15}{8}$ 点拨:如答图,连接$FH$.$\because \triangle EGH \cong \triangle BCF$,$\therefore \angle DCB = \angle G = 90^{\circ}$,$FC = GH$,$BC = EG = 3$,$\therefore FC // GH$,$BE = CG$,$\therefore$四边形$FCGH$是矩形.$\because BE = CF$,$\therefore CG = CF$,$\therefore$四边形$CGHF$为正方形,$\therefore FH = CF$.
设$BE = x$,则$CF = FH = HG = x$,$\therefore EC = 3 - x$.
$\because DE = EH$,$\therefore (3 - x)^{2} + 2^{2} = 3^{2} + x^{2}$,
解得$x = \frac{2}{3}$,$\therefore CF = FH = \frac{2}{3}$,$\therefore DF = 2 - x = 2 - \frac{2}{3} = \frac{4}{3}$,
$\therefore DH = \sqrt{DF^{2} + FH^{2}} = \sqrt{(\frac{4}{3})^{2} + (\frac{2}{3})^{2}} = \frac{2\sqrt{5}}{3}$,
$\therefore S_{\triangle DEH} = S_{\triangle DEC} + S_{梯形DCGH} - S_{\triangle EHG} = \frac{1}{2}(3 - x) × 2 + \frac{1}{2} · (2 + x) · x - \frac{1}{2} × 3 · x = \frac{1}{2}x^{2} - \frac{3}{2}x + 3 = \frac{1}{2}(x - \frac{3}{2})^{2} + \frac{15}{8}$.
$\because \frac{1}{2} ＞ 0$,$\therefore$当$x = \frac{3}{2}$时,$S_{\triangle DEH}$有最小值,为$\frac{15}{8}$,即$\triangle DEH$的面积的最小值是$\frac{15}{8}$.
3. 如图，$Rt\triangle ABC≌Rt\triangle DEF$，$\angle ACB=\angle DFE=30°$，$AB=DE=a$，且点 $B$，$C$，$F$，$E$ 在一条直线上.
(1) 如图①，沿着 $EB$ 方向平移 $\triangle DEF$，当点 $F$ 在线段 $BC$ 上时，两个三角形重合部分的面积最大值是
(2) 继续沿直线 $CF$ 平移 $\triangle DEF$，如图②，求图中阴影部分面积的最大值.
① ②
(1) 如图①，沿着 $EB$ 方向平移 $\triangle DEF$，当点 $F$ 在线段 $BC$ 上时，两个三角形重合部分的面积最大值是
$\frac{\sqrt{3}}{4}a^{2}$
；(2) 继续沿直线 $CF$ 平移 $\triangle DEF$，如图②，求图中阴影部分面积的最大值.
① ②
答案：
3.
(1)$\frac{\sqrt{3}}{4}a^{2}$ 点拨:当点$F$与点$B$重合时,两个三角形重合部分的面积取得最大值,其最大值为$\frac{1}{2} × \sqrt{3}a × \frac{1}{2}a = \frac{\sqrt{3}}{4}a^{2}$.
(2)解:如答图,设$AB$交$FD$于点$M$,$AC$交$DE$于点$N$,$AC$交$FD$于点$K$.在$Rt\triangle ABC$中,$\angle C = 30^{\circ}$,且$AB = a$,$\therefore BC = \sqrt{3}AB = \sqrt{3}a$.设$BM = x$,由对称性可知$EN = BM = x$,在$Rt\triangle ECN$中,$EC = \sqrt{3}EN = \sqrt{3}x$.在$\triangle AMK$中,$\angle A = \angle AMK = 60^{\circ}$,$\therefore \triangle AMK$为等边三角形,$\therefore AM = a - x$,
$\therefore S_{五边形KMBEN} = S_{\triangle ABC} - S_{\triangle AMK} - S_{\triangle CEN} = \frac{\sqrt{3}}{2}a^{2} - \frac{\sqrt{3}}{4}(a - x)^{2} - \frac{\sqrt{3}}{2}x^{2} = -\frac{3\sqrt{3}}{4}x^{2} + \frac{\sqrt{3}}{2}ax + \frac{\sqrt{3}}{4}a^{2} = -\frac{3\sqrt{3}}{4}(x^{2} - \frac{2}{3}ax - \frac{1}{3}a^{2}) = -\frac{3\sqrt{3}}{4}(x - \frac{1}{3}a)^{2} + \frac{\sqrt{3}}{3}a^{2}$.
$\because -\frac{3\sqrt{3}}{4} ＜ 0$,$\therefore$当$x = \frac{1}{3}a$时,阴影部分面积最大,最大值为$\frac{\sqrt{3}}{3}a^{2}$.
3.
(1)$\frac{\sqrt{3}}{4}a^{2}$ 点拨:当点$F$与点$B$重合时,两个三角形重合部分的面积取得最大值,其最大值为$\frac{1}{2} × \sqrt{3}a × \frac{1}{2}a = \frac{\sqrt{3}}{4}a^{2}$.
(2)解:如答图,设$AB$交$FD$于点$M$,$AC$交$DE$于点$N$,$AC$交$FD$于点$K$.在$Rt\triangle ABC$中,$\angle C = 30^{\circ}$,且$AB = a$,$\therefore BC = \sqrt{3}AB = \sqrt{3}a$.设$BM = x$,由对称性可知$EN = BM = x$,在$Rt\triangle ECN$中,$EC = \sqrt{3}EN = \sqrt{3}x$.在$\triangle AMK$中,$\angle A = \angle AMK = 60^{\circ}$,$\therefore \triangle AMK$为等边三角形,$\therefore AM = a - x$,
$\therefore S_{五边形KMBEN} = S_{\triangle ABC} - S_{\triangle AMK} - S_{\triangle CEN} = \frac{\sqrt{3}}{2}a^{2} - \frac{\sqrt{3}}{4}(a - x)^{2} - \frac{\sqrt{3}}{2}x^{2} = -\frac{3\sqrt{3}}{4}x^{2} + \frac{\sqrt{3}}{2}ax + \frac{\sqrt{3}}{4}a^{2} = -\frac{3\sqrt{3}}{4}(x^{2} - \frac{2}{3}ax - \frac{1}{3}a^{2}) = -\frac{3\sqrt{3}}{4}(x - \frac{1}{3}a)^{2} + \frac{\sqrt{3}}{3}a^{2}$.
$\because -\frac{3\sqrt{3}}{4} ＜ 0$,$\therefore$当$x = \frac{1}{3}a$时,阴影部分面积最大,最大值为$\frac{\sqrt{3}}{3}a^{2}$.
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