【答案】
分析:●觀察計算:分別代入計算即可得出
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/0.png)
與
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/1.png)
的大小關(guān)系;
●探究證明:
(1)由于OC是直徑AB的一半,則OC易得.通過證明△ACD∽△CBD,可求CD;
(2)分a=b,a≠b討論可得出
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/2.png)
與
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/3.png)
的大小關(guān)系;
●實踐應(yīng)用:通過前面的結(jié)論長方形為正方形時,周長最小.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/images4.png)
解:●觀察計算:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/4.png)
>
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/5.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/7.png)
.(2分)
●探究證明:
(1)∵AB=AD+BD=2OC,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/8.png)
(3分)
∵AB為⊙O直徑,
∴∠ACB=90°.
∵∠A+∠ACD=90°,∠ACD+∠BCD=90°,
∴∠A=∠BCD.
∴△ACD∽△CBD.(4分)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/9.png)
.
即CD
2=AD•BD=ab,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/10.png)
.(5分)
(2)當(dāng)a=b時,OC=CD,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/11.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/12.png)
;
a≠b時,OC>CD,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/13.png)
>
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/14.png)
.(6分)
●結(jié)論歸納:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/15.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/16.png)
.(7分)
●實踐應(yīng)用
設(shè)長方形一邊長為x米,則另一邊長為
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/17.png)
米,設(shè)鏡框周長為l米,則
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/18.png)
≥
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/19.png)
.(9分)
當(dāng)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/20.png)
,即x=1(米)時,鏡框周長最�。�
此時四邊形為正方形時,周長最小為4米.(10分)
點評:本題綜合考查了幾何不等式,相似三角形的判定與性質(zhì),通過計算和證明得出結(jié)論:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/21.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022164336199769124/SYS201310221643361997691025_DA/22.png)
是解題的關(guān)鍵.