10、(陕西)王师傅有两块板材边角料,其中30cm,下底为一块是边长为60cm的正方形板子;另一块是上底为30cm,下底为120cm,高为60cm的直角梯形板子(如图①).王师傅想将这两块板子裁成两块全等的矩形板材.他将两块板子叠放在一起,使梯形的两个直角顶点分别与正方形的两个顶点重合,两块板子的重叠部分为五边形ABCFE围成的区域(如图②).由于受材料纹理的限制,要求裁出的矩形要以点B为一个顶点.

   (1)求FC的长;

   (2)利用图②求出矩形顶点B所对的顶点到BC边的距离x(cm)为多少时,矩形的面积y(cm2)最大?最大面积是多少?

(3)若想使裁出的矩形为正方形,试求出面积最大的正方形的边长.

例2、[解题分析]⑴ 由和

得.

从而可得与的函数表达式是

解:(1)(或)······················································ 3分

(2),又

····································································································· 5分

,即······························· 6分

,,,

解得································································ 7分

梯形的中位线长为······································································· 8分

9:分析:(1)当三角形QAP为等腰三角形时,由于∠A为直角,只能是AQ=AP,建立等量关系,,即时,三角形QAP为等腰三角形;

(2)四边形QAPC的面积=ABCD的面积-三角形QDC的面积-三角形PBC的面积

==36,即当P、Q运动时,四边形QAPC的面积不变。

(3)显然有两种情况:△PAQ∽△ABC,△QAP∽△ABC,

由相似关系得或,解之得或

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