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32(m) (20+4
![](https://img.yulucn.com/upload/9/cd/9cd9833e3a076203a11b41ebdafb0430_thumb.jpg)
) m
![](https://img.yulucn.com/upload/d/53/d5315387c26778a4c3d87431c8d43603_thumb.jpg)
(m)
分三类情况讨论如下:
(1)如题图1所示,原来的花圃为Rt△ABC,其中BC=6 m,AC=8 m,∠ACB=90°.由勾股定理易知AB=10 m,将△ABC沿直线AC翻折180°后,得等腰三角形ABD,此时,AD=10 m,CD=6 m.故扩建后的等腰三角形花圃的周长为12+10+10=32(m).
(2)如题图2,因为BC=6 m,CD=4 m,所以BD=AB=10 m,在Rt△ACD中,由勾股定理得AD=
![](https://img.yulucn.com/upload/d/62/d6232f6b57b4bc08f0ad1c6343da6c12_thumb.jpg)
=4
![](https://img.yulucn.com/upload/9/cd/9cd9833e3a076203a11b41ebdafb0430_thumb.jpg)
,此时,扩建后的等腰三角形花圃的周长为4
![](https://img.yulucn.com/upload/9/cd/9cd9833e3a076203a11b41ebdafb0430_thumb.jpg)
+10+10=(20+4
![](https://img.yulucn.com/upload/9/cd/9cd9833e3a076203a11b41ebdafb0430_thumb.jpg)
) m.
(3)如题图3,设△ABD中DA=DB,
再设CD=x m,则DA=(x+6)m,
在Rt△ACD中,由勾股定理得x
2 +8
2 =(x+6)
2 ,
解得x=
![](https://img.yulucn.com/upload/a/c8/ac8d304aab3726f99a4e39cb1fedc9a6_thumb.jpg)
.
∴扩建后等腰三角形花圃的周长=10+2(x+6)=
![](https://img.yulucn.com/upload/d/53/d5315387c26778a4c3d87431c8d43603_thumb.jpg)
(m).
1年前
7