【答案】
分析:(1)根據(jù)正方形的面積可求出正方形的邊長,根據(jù)勾股定理可求出BG的長,易證Rt△BEH∽Rt△BCG,求出BH、EH的長,再根據(jù)相似三角形的性質(zhì)可求出△DOG∽△FOH,根據(jù)三角形邊長的比即可求出答案.
(2)過O作OL⊥CG,則△GOL∽△GBC,即可求出△DBO的面積.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/images0.png)
解:(1)∵正方形ABCD的面積為64cm
2,
∴BC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/0.png)
=8,
∵正方形CEFG的面積為36cm
2,
∴CG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/1.png)
=6,
∴BG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/2.png)
=10,
∵BC=8,CE=6,CG=6,BE=BC-CE=8-6=2,
∵EF∥CG,
∴Rt△BEH∽Rt△BCG,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/3.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/4.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/5.png)
,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/8.png)
,
∴BH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/9.png)
,EH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/10.png)
,
在△DOG與△FOH中,∠DOG=∠FOH,
∵EF∥CG,
∴∠HFO=∠FDC,
∴△DOG∽△FOH,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/12.png)
,HF=EF-EH=6-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/13.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/14.png)
,DC+CG=8+6=14,OG=BG-BH-OH=10-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/15.png)
-OH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/16.png)
-OH,
故
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/18.png)
,
∴OH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/19.png)
,BO=BH+OH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/20.png)
+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/22.png)
.
(2)過O作OL⊥CG,
∵△GOL∽△GBC,
∴OG=BG-BO=10-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/23.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/24.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/25.png)
=即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/26.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/27.png)
,
解得OL=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/28.png)
,
∴S
△DBO=S
△BDG-S
△DOG=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/29.png)
DG•BC-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/30.png)
DG•OL=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/31.png)
DG×(BC-OL)=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/32.png)
×14×(8-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/33.png)
)=7×
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/34.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/201310201209436955876116/SYS201310201209436955876026_DA/35.png)
.
點評:解答本題要充分利用正方形的特殊性質(zhì).注意在正方形中的特殊三角形的應(yīng)用,及勾股定理的應(yīng)用.