2014天津中考数学试题第18题

要思路,解析(就是为什么这模做)急急急!!!
2025-03-15 20:41:43
推荐回答(2个)
回答(1):

以A为原点建立平面直角坐标系:则直线AB的函数关系式为y=(1/4)x
假设所做矩形为ABDE如草图2,作BM⊥x轴于点M,直线DE交y轴于N,
则△ABM∽△ANE         (AB/AN)=(AM/AE)
因为AB=√(17)      AE=(11/√(17))       AM=4
所以(√(17)/AN)=(4/(11/√(17)))           所以AN=(11/4)     即点N(0,-(11/4))
所以直线DE函数关系式为y=(1/4)x-(11/4)         于是x=4y+11
在网格内取整数y=-2时,x=3  ;       y=-3时,x=-1;
即DE经过点(3,-2)和(-1,-3)
在原图中过点(3,-2)和(-1,-3)作直线与以AB为边的正方形
一组对边分别交于点D、E
则矩形ABDE即为所求作的图形,即SABCD=11=(AC^2)+(BC^2)

回答(2):

题目是什么

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