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来源：学生作业帮 编辑：大师作文网作业帮 分类：数学作业 时间：2024/11/12 10:44:58
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12.设AD为X,CD为Y,求ABCD的面积,即求XY的值.
由ABCD周长为16可知,
X+Y=16/2=8
又四个正方形面积为68,其中上下两个相等,左右两个相等,
所以上面正方形面积+右边正方形面积=68/2=34
即X^2+Y^2=34
因为（X+Y）^2=64
（X+Y）^2-(X^2+Y^2)=2XY=64-34=30
所以XY=30/2=15
即ABCD的面积为15
由ABCD周长为16可知,
X+Y=16/2=8
又四个正方形面积为68,其中上下两个相等,左右两个相等,
所以上面正方形面积+右边正方形面积=68/2=34
即X^2+Y^2=34
因为（X+Y）^2=64
（X+Y）^2-(X^2+Y^2)=2XY=64-34=30
所以XY=30/2=15
即ABCD的面积为15
I need you help Thank you for helping me!
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