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1. 如图，在由边长为 1 的小正方形组成的网格中，点$A$，$B$，$C$，$D$为格点(小正方形的顶点)，$AB$，$CD$相交于点$P$，则$PC$的长为
$\frac{2\sqrt{5}}{7}$
.
答案：
1. $\frac{2\sqrt{5}}{7}$ 点拨：由勾股定理，得$CD = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$，如答图，由题意可知$E$是$CD$的中点， $\therefore EC = ED = \frac{1}{2}CD = \frac{\sqrt{5}}{2}$． 连接$BC,DF$，则$BC// DF$， $\therefore \angle BCE = \angle FDE$． 又$\because \angle BEC = \angle FED,CE = DE$，
∴△BEC≌△FED(ASA)， $\therefore BE = EF = \frac{3}{2}$． $\because AC// BE,\therefore \triangle APC \backsim \triangle BPE$， $\therefore \frac{PC}{PE} = \frac{AC}{BE} = \frac{4}{3}$，即$\frac{PC}{\frac{\sqrt{5}}{2}} = \frac{4}{3}$，解得$PC = \frac{2\sqrt{5}}{7}$．
1. $\frac{2\sqrt{5}}{7}$ 点拨：由勾股定理，得$CD = \sqrt{1^{2} + 2^{2}} = \sqrt{5}$，如答图，由题意可知$E$是$CD$的中点， $\therefore EC = ED = \frac{1}{2}CD = \frac{\sqrt{5}}{2}$． 连接$BC,DF$，则$BC// DF$， $\therefore \angle BCE = \angle FDE$． 又$\because \angle BEC = \angle FED,CE = DE$，
∴△BEC≌△FED(ASA)， $\therefore BE = EF = \frac{3}{2}$． $\because AC// BE,\therefore \triangle APC \backsim \triangle BPE$， $\therefore \frac{PC}{PE} = \frac{AC}{BE} = \frac{4}{3}$，即$\frac{PC}{\frac{\sqrt{5}}{2}} = \frac{4}{3}$，解得$PC = \frac{2\sqrt{5}}{7}$．
2. 如图，$P$是$\triangle ABC$内一点，过点$P$分别作直线平行于$\triangle ABC$的各边，所形成的三个小三角形(图中阴影部分)的面积分别是$1$，$9$和$49$，则$\triangle ABC$的面积是
121
.
答案：
2. 121 点拨：如答图，$\because DE// BC,FH// AC,GI// AB$， $\therefore \triangle DPF \backsim \triangle BHF,\triangle GHP \backsim \triangle BHF,\triangle PEI \backsim \triangle GCI$， $\triangle GHP \backsim \triangle GCI,\therefore \triangle DPF \backsim \triangle PEI \backsim \triangle GHP$． $\because \triangle DPF,\triangle PEI,\triangle GHP$的面积比为$1 : 9 : 49,\therefore$它们对应边的比为$1 : 3 : 7$．又$\because$四边形$BDPG$与四边形$CEPH$为平行四边形，$\therefore DP = BG,EP = CH$．设$DP$为$x$，则$PE = 3x,GH = 7x,\therefore BC = BG + GH + CH = DP + GH + PE = x + 7x + 3x = 11x,\therefore BC : DP = 11x : x = 11 : 1$，由相似三角形面积比等于相似比的平方，得$S_{\triangle ABC} : S_{\triangle FDP} = 121 : 1,\therefore \triangle ABC$的面积为121．
2. 121 点拨：如答图，$\because DE// BC,FH// AC,GI// AB$， $\therefore \triangle DPF \backsim \triangle BHF,\triangle GHP \backsim \triangle BHF,\triangle PEI \backsim \triangle GCI$， $\triangle GHP \backsim \triangle GCI,\therefore \triangle DPF \backsim \triangle PEI \backsim \triangle GHP$． $\because \triangle DPF,\triangle PEI,\triangle GHP$的面积比为$1 : 9 : 49,\therefore$它们对应边的比为$1 : 3 : 7$．又$\because$四边形$BDPG$与四边形$CEPH$为平行四边形，$\therefore DP = BG,EP = CH$．设$DP$为$x$，则$PE = 3x,GH = 7x,\therefore BC = BG + GH + CH = DP + GH + PE = x + 7x + 3x = 11x,\therefore BC : DP = 11x : x = 11 : 1$，由相似三角形面积比等于相似比的平方，得$S_{\triangle ABC} : S_{\triangle FDP} = 121 : 1,\therefore \triangle ABC$的面积为121．
3. 如图，点$A$，$B$在反比例函数$y = \frac{k}{x}$的图像上，射线$BA$交$y$轴于点$D$，且$AB = 2AD$，延长$BO$交反比例函数图像另一分支于点$C$，连接$AC$交$y$轴于点$E$，若$S_{\triangle BCE} = 6$，求$k$的值.
答案：
3. 解：设$B(x,y)$，则$C( - x, - y)$．$\because AB = 2AD,\therefore BD = 3AD$．如答图，过点$A$作$AF \perp y$轴于点$F$，过点$B$作$BM \perp y$轴于点$M$，过点$C$作$CN \perp y$轴于点$N$，则$AF// BM// CN$，$\therefore \triangle ADF \backsim \triangle BDM,\triangle AEF \backsim \triangle CEN$， $\therefore BM = 3AF$． 易知$\triangle BOM \cong \triangle CON$，$\therefore CN = BM = 3AF = x,OM = ON = y,\therefore EN = 3EF$．$\because$点$A,B$在反比例函数$y = \frac{k}{x}$的图像上，$\therefore A(\frac{1}{3}x,3y),\therefore OF = 3y,FN = 4y,\therefore EO = 2y$，
$\therefore S_{\triangle BCE} = \frac{1}{2}OE · (BM + CN) = \frac{1}{2} · 2y(x + x) = 6$， $\therefore 2xy = 6,\therefore k = xy = 3$．
3. 解：设$B(x,y)$，则$C( - x, - y)$．$\because AB = 2AD,\therefore BD = 3AD$．如答图，过点$A$作$AF \perp y$轴于点$F$，过点$B$作$BM \perp y$轴于点$M$，过点$C$作$CN \perp y$轴于点$N$，则$AF// BM// CN$，$\therefore \triangle ADF \backsim \triangle BDM,\triangle AEF \backsim \triangle CEN$， $\therefore BM = 3AF$． 易知$\triangle BOM \cong \triangle CON$，$\therefore CN = BM = 3AF = x,OM = ON = y,\therefore EN = 3EF$．$\because$点$A,B$在反比例函数$y = \frac{k}{x}$的图像上，$\therefore A(\frac{1}{3}x,3y),\therefore OF = 3y,FN = 4y,\therefore EO = 2y$，
$\therefore S_{\triangle BCE} = \frac{1}{2}OE · (BM + CN) = \frac{1}{2} · 2y(x + x) = 6$， $\therefore 2xy = 6,\therefore k = xy = 3$．
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