如图是用4个全等的直角三角形与1个小正方形镶嵌而成的正方形图案，已知大正方形面积为49，小正方形面积为

Question

如图是用4个全等的直角三角形与1个小正方形镶嵌而成的正方形图案，已知大正方形面积为49，小正方形面积为9，若用x，y表示直角三角形的两直角边（x＞y）观察图片，指出下列错误的（ ）
A.x2+y2=49，B.x-y=23，C.2xy+9=49，D.x+y=13

Answer 1

A对 X^2+Y^2=AB^2=正方形面积=49

B错 X-Y=CD=23 则CD>AB , 而AB才等于9 显然不成立

D错 应为若等式成立 则其平方为（X+Y）^2=13^2

                                                     X^2+2XY+Y^2=169

                                                      2XY+49=169

                                                      2XY=120

                             而三角形面积为1/2XY^2 , 所以4个三角形面积为2XY 而四个三角形面积显然小于正方形面积小于49  所以120小于49不可能成立

C，有可能成立 

    因为2XY+9=49，所以XY=20  y=20/x    又X^2+Y^2=49  联立解方程 （20/X）^2+X^2=49

                                                                                                             400+Y^4=49Y^2

                                                                                                              Y^4-49Y^2+400=0

                                                                                                              因 49^2-4*1*400

                                                                                                                 =2401-1600=801>0 

                                                                                                                 故方程有解   

                                                                                                                        X=6.21699057

                                                                                                              另一根X=3.21699057舍去（因为X>y）

               所以C有可能成立，但并不一定。所以C是错误的

C选项可由如下图作出

做法由   dony_zheng_  提供 感谢 dony_zheng_  

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