如图,在平面直角坐标系中,直线l为正比例函数y=x的图象,点A1的坐

如图,在平面直角坐标系中,直线l为正比例函数y=x的图象,点A1的坐标为(10),过点A1x轴的垂线交直线l于点D1,以A1D1为边作正方形A1B1C1D1;过点C1作直线l的垂线,垂足为A2,交x轴于点B2,以A2B2为边作正方形A2B2C2D2;过点C2x轴的垂线,垂足为A3,交直线l于点D3,以A3D3为边作正方形A3B3C3D3,按此规律操作下所得到的正方形AnBnCnDn的面积是     

 

答案

n1 

【分析】根据正比例函数的性质得到∠D1OA1=45°,分别求出正方形A1B1C1D1的面积、正方形A2B2C2D2的面积,总结规律解答.

【解答】解:∵直线l为正比例函数y=x的图象,

∴∠D1OA1=45°

D1A1=OA1=1

∴正方形A1B1C1D1的面积=1=11

由勾股定理得,OD1=D1A2=

A2B2=A2O=

∴正方形A2B2C2D2的面积==21

同理,A3D3=OA3=

∴正方形A3B3C3D3的面积==31

由规律可知,正方形AnBnCnDn的面积=n1

故答案为:(n1

【点评】本题考查的是正方形的性质、一次函数图象上点的坐标特征,根据一次函数解析式得到∠D1OA1=45°,正确找出规律是解题的关键.

 

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