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3. 如图2-11,在$\triangle ABC$中,$AB = AC = 13$,$BC = 10$,$D是AB$的中点,过点$D作DE\perp AC$,垂足为$E$,则$DE$的长是______.
答案：
$\frac{60}{13}$ 解析：如答图2-2，作$AH\perp BC$，垂足为$H$，连接$CD$，
在$Rt\triangle ABH$中，$AH=\sqrt{13^{2}-5^{2}} = 12$，
$\therefore S_{\triangle ABC}=\frac{1}{2}×10×12 = 60$.
$\because D$为$AB$的中点，$\therefore S_{\triangle ADC}=S_{\triangle DBC}=30$，
$\therefore \frac{1}{2}\cdot AC\cdot DE = 30$，即$DE=\frac{60}{13}$.
$\frac{60}{13}$ 解析：如答图2-2，作$AH\perp BC$，垂足为$H$，连接$CD$，
在$Rt\triangle ABH$中，$AH=\sqrt{13^{2}-5^{2}} = 12$，
$\therefore S_{\triangle ABC}=\frac{1}{2}×10×12 = 60$.
$\because D$为$AB$的中点，$\therefore S_{\triangle ADC}=S_{\triangle DBC}=30$，
$\therefore \frac{1}{2}\cdot AC\cdot DE = 30$，即$DE=\frac{60}{13}$.
4. 如图2-12,小明在$A$时测得某树的影长为2m,$B$时又测得该树的影长为8m,若两次日照的光线互相垂直,则树的高度为______m.
答案：
解析：如答图2-3，$EH = 2$，$FH = 8$，$DH\perp EF$，$ED\perp DF$，
设$DH = x$，则由$DE^{2}+DF^{2}=EF^{2}$，
得$x^{2}+2^{2}+x^{2}+8^{2}=(2 + 8)^{2}$，解得$x = 4$.
解析：如答图2-3，$EH = 2$，$FH = 8$，$DH\perp EF$，$ED\perp DF$，
设$DH = x$，则由$DE^{2}+DF^{2}=EF^{2}$，
得$x^{2}+2^{2}+x^{2}+8^{2}=(2 + 8)^{2}$，解得$x = 4$.
5. 公园里有一块四边形$ABCD$的草地如图2-13所示,测得$BC = CD = 10m$,$\angle B = \angle C = 120^{\circ}$,$\angle A = 45^{\circ}$.请你求出这块草地的面积.
答案：
解：如答图2-4，连接$DB$，$\because DC = BC$，$\angle C = 120^{\circ}$，$\therefore \angle 1 = 30^{\circ}$，
$\therefore \angle 2 = 120^{\circ}-30^{\circ}=90^{\circ}$.作$CH\perp DB$，垂足为$H$，
在$Rt\triangle CHB$中，$CH = 5$，$HB = 5\sqrt{3}$，$AB = DB = 10\sqrt{3}$，
$\therefore S_{四边形ABCD}=S_{\triangle ABD}+S_{\triangle BDC}=\frac{1}{2}×(10\sqrt{3})^{2}+\frac{1}{2}×10\sqrt{3}×5=(150 + 25\sqrt{3})(m^{2})$.
解：如答图2-4，连接$DB$，$\because DC = BC$，$\angle C = 120^{\circ}$，$\therefore \angle 1 = 30^{\circ}$，
$\therefore \angle 2 = 120^{\circ}-30^{\circ}=90^{\circ}$.作$CH\perp DB$，垂足为$H$，
在$Rt\triangle CHB$中，$CH = 5$，$HB = 5\sqrt{3}$，$AB = DB = 10\sqrt{3}$，
$\therefore S_{四边形ABCD}=S_{\triangle ABD}+S_{\triangle BDC}=\frac{1}{2}×(10\sqrt{3})^{2}+\frac{1}{2}×10\sqrt{3}×5=(150 + 25\sqrt{3})(m^{2})$.
1. 如图2-14,有一张直角三角形纸片,两直角边$AC = 5cm$,$BC = 10cm$,将$\triangle ABC$折叠,使点$B与点A$重合,折痕为$DE$,则$CD$的长为( )
A.$\frac{25}{2}cm$
B.$\frac{15}{2}cm$
C.$\frac{25}{4}cm$
D.$\frac{15}{4}cm$
A.$\frac{25}{2}cm$
B.$\frac{15}{2}cm$
C.$\frac{25}{4}cm$
D.$\frac{15}{4}cm$
答案：
D 解析：如答图2-5，
$\because AE = EB$，$DE\perp AB$，$\therefore AD = DB$.
设$CD = x$，则$AD = BD = 10 - x$.
在$Rt\triangle ACD$中，$(10 - x)^{2}=x^{2}+5^{2}$，
解得$x=\frac{15}{4}$，$\therefore CD=\frac{15}{4}cm$.
D 解析：如答图2-5，
$\because AE = EB$，$DE\perp AB$，$\therefore AD = DB$.
设$CD = x$，则$AD = BD = 10 - x$.
在$Rt\triangle ACD$中，$(10 - x)^{2}=x^{2}+5^{2}$，
解得$x=\frac{15}{4}$，$\therefore CD=\frac{15}{4}cm$.
2. 在直线$l$上依次摆放着7个正方形(如图2-15).已知斜放置的3个正方形的面积分别是1,2,3,正放置的4个正方形的面积依次是$S_{1}$,$S_{2}$,$S_{3}$,$S_{4}$,试求$S_{1} + S_{2} + S_{3} + S_{4}$的值.
4
答案：
解：$\because \angle BAC+\angle ACB=\angle ACB+\angle ECD = 90^{\circ}$，
$\therefore \angle BAC=\angle ECD$，又$\because \angle ABC=\angle CDE = 90^{\circ}$，$AC = CE$，
$\therefore \triangle ABC\cong\triangle CDE$，
$\therefore AB = CD$，$BC = ED$，
$\therefore AC^{2}=3 = AB^{2}+BC^{2}=S_{3}+S_{4}$，即$S_{3}+S_{4}=3$.
同理，$2 = S_{2}+S_{3}$，$S_{1}+S_{2}=1$，$\therefore S_{1}+S_{2}+S_{3}+S_{4}=1 + 3 = 4$.
$\therefore \angle BAC=\angle ECD$，又$\because \angle ABC=\angle CDE = 90^{\circ}$，$AC = CE$，
$\therefore \triangle ABC\cong\triangle CDE$，
$\therefore AB = CD$，$BC = ED$，
$\therefore AC^{2}=3 = AB^{2}+BC^{2}=S_{3}+S_{4}$，即$S_{3}+S_{4}=3$.
同理，$2 = S_{2}+S_{3}$，$S_{1}+S_{2}=1$，$\therefore S_{1}+S_{2}+S_{3}+S_{4}=1 + 3 = 4$.
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