矩形纸片ABCD中,AB=3cm,BC=4cm,现将纸片折叠压平,使A与C重合,设折痕为EF,则重叠部分△AEF的面积等于_.
矩形纸片ABCD中,AB=3cm,BC=4cm,现将纸片折叠压平,使A与C重合,设折痕为EF,则重叠部分△AEF的面积等于______.
数学人气:384 ℃时间:2019-10-18 08:45:29
优质解答
设AE=x,由折叠可知,EC=x,BE=4-x,
在Rt△ABE中,AB
2+BE
2=AE
2,即3
2+(4-x)
2=x
2,
解得:x=
由折叠可知∠AEF=∠CEF,
∵AD∥BC,
∴∠CEF=∠AFE,
∴∠AEF=∠AFE,即AE=AF=
,
∴S
△AEF=
×AF×AB=
×
×3=
.
故答案为:
.
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