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如图所示,在四边形ABCD中,AM=MN=ND,BE=EF=FC,四边形ABEM,MEFN,NFCD的面积分别记为S1,S2和S3,则S2S1+S3值等于.
题目详情
如图所示,在四边形ABCD中,AM=MN=ND,BE=EF=FC,四边形ABEM,MEFN,NFCD的面积分别记为S 1 ,S 2 和S 3 ,则
![]() |
S 2 |
S 1 + S 3 |
![](https://www.zaojiaoba.cn/zhidao/pic/item/42a98226cffc1e17a64e6b804990f603728de98a.jpg)
S 2 |
S 1 + S 3 |
![](https://www.zaojiaoba.cn/zhidao/pic/item/42a98226cffc1e17a64e6b804990f603728de98a.jpg)
S 2 |
S 1 + S 3 |
![](https://www.zaojiaoba.cn/zhidao/pic/item/42a98226cffc1e17a64e6b804990f603728de98a.jpg)
S 2 |
S 1 + S 3 |
![](https://www.zaojiaoba.cn/zhidao/pic/item/42a98226cffc1e17a64e6b804990f603728de98a.jpg)
S 2 |
S 1 + S 3 |
![](https://www.zaojiaoba.cn/zhidao/pic/item/42a98226cffc1e17a64e6b804990f603728de98a.jpg)
S 2 |
S 1 + S 3 |
![](https://www.zaojiaoba.cn/zhidao/pic/item/42a98226cffc1e17a64e6b804990f603728de98a.jpg)
![](https://www.zaojiaoba.cn/zhidao/pic/item/42a98226cffc1e17a64e6b804990f603728de98a.jpg)
▼优质解答
答案和解析
如图3a,连接AE、EN和NC,易知
由S △AEM =S △MEN ,S △CNF =S △EFN ,
上面两个式子相加得S △AEM +S △CNF =S 2 (1)
并且四边形AECN的面积=2S 2 .
连接AC,如图3b,由三角形面积公式,
易知 S △ABE =
S △AEC , S △CDN =
S △CNA
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
.
如图3a,连接AE、EN和NC,易知
由S △AEM =S △MEN ,S △CNF =S △EFN ,
上面两个式子相加得S △AEM +S △CNF =S 2 (1)
并且四边形AECN的面积=2S 2 .
连接AC,如图3b,由三角形面积公式,
易知 S △ABE =
S △AEC , S △CDN =
S △CNA
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
.
如图3a,连接AE、EN和NC,易知
由S △AEM =S △MEN ,S △CNF =S △EFN ,
上面两个式子相加得S △AEM +S △CNF =S 2 (1)
并且四边形AECN的面积=2S 2 .
连接AC,如图3b,由三角形面积公式,
易知 S △ABE =
S △AEC , S △CDN =
S △CNA
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
.
如图3a,连接AE、EN和NC,易知
由S △AEM △AEM =S △MEN △MEN ,S △CNF △CNF =S △EFN △EFN ,
上面两个式子相加得S △AEM △AEM +S △CNF △CNF =S 2 2 (1)
并且四边形AECN的面积=2S 2 2 .
连接AC,如图3b,由三角形面积公式,
易知 S △ABE =
S △AEC , S △CDN =
S △CNA
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
. S △ABE =
S △AEC , S △CDN =
S △CNA
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
. △ABE =
1 2 1 1 1 2 2 2 S △AEC , S △CDN =
S △CNA
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
. △AEC , S △CDN =
S △CNA
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
. S △CDN =
S △CNA
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
. △CDN =
1 2 1 1 1 2 2 2 S △CNA
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
. △CNA
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
. S △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
. △ABE + S △CDN =
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
=
.
答:
=
. △CDN =
1 2 1 1 1 2 2 2
四边形AECN的面积=S 2 2 (2)
将(1)式和(2)相加,
得到S △AEM △AEM +S △CNF △CNF +S △ABE △ABE +S △CDN △CDN =2S 2 2 ,
既然S △AEM △AEM +S △ABE △ABE =S 1 1 ,S △CNF △CNF +S △ABE △ABE =S 3 3
因此S 1 1 +S 3 3 =2S 2 2 ,
=
.
答:
=
.
S 2 S 1 + S 3 S 2 S 2 S 2 2 S 1 + S 3 S 1 + S 3 S 1 + S 3 1 + S 3 3 =
1 2 1 1 1 2 2 2 .
答:
=
.
S 2 S 1 + S 3 S 2 S 2 S 2 2 S 1 + S 3 S 1 + S 3 S 1 + S 3 1 + S 3 3 =
1 2 1 1 1 2 2 2 .
![]() 如图3a,连接AE、EN和NC,易知 由S △AEM =S △MEN ,S △CNF =S △EFN , 上面两个式子相加得S △AEM +S △CNF =S 2 (1) 并且四边形AECN的面积=2S 2 . 连接AC,如图3b,由三角形面积公式, ![]() 易知 S △ABE =
上面两个式子相加得 S △ABE + S △CDN =
四边形AECN的面积=S 2 (2) 将(1)式和(2)相加, 得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 , 既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3 因此S 1 +S 3 =2S 2 ,
答:
|
![](http://hiphotos.baidu.com/zhidao/pic/item/48540923dd54564e5cea0388b0de9c82d0584ff0.jpg)
如图3a,连接AE、EN和NC,易知
由S △AEM =S △MEN ,S △CNF =S △EFN ,
上面两个式子相加得S △AEM +S △CNF =S 2 (1)
并且四边形AECN的面积=2S 2 .
连接AC,如图3b,由三角形面积公式,
![](http://hiphotos.baidu.com/zhidao/pic/item/a8773912b31bb051cdd55499357adab44bede08a.jpg)
易知 S △ABE =
1 |
2 |
1 |
2 |
上面两个式子相加得 S △ABE + S △CDN =
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
![](http://hiphotos.baidu.com/zhidao/pic/item/48540923dd54564e5cea0388b0de9c82d0584ff0.jpg)
如图3a,连接AE、EN和NC,易知
由S △AEM =S △MEN ,S △CNF =S △EFN ,
上面两个式子相加得S △AEM +S △CNF =S 2 (1)
并且四边形AECN的面积=2S 2 .
连接AC,如图3b,由三角形面积公式,
![](http://hiphotos.baidu.com/zhidao/pic/item/a8773912b31bb051cdd55499357adab44bede08a.jpg)
易知 S △ABE =
1 |
2 |
1 |
2 |
上面两个式子相加得 S △ABE + S △CDN =
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
![](http://hiphotos.baidu.com/zhidao/pic/item/48540923dd54564e5cea0388b0de9c82d0584ff0.jpg)
如图3a,连接AE、EN和NC,易知
由S △AEM =S △MEN ,S △CNF =S △EFN ,
上面两个式子相加得S △AEM +S △CNF =S 2 (1)
并且四边形AECN的面积=2S 2 .
连接AC,如图3b,由三角形面积公式,
![](http://hiphotos.baidu.com/zhidao/pic/item/a8773912b31bb051cdd55499357adab44bede08a.jpg)
易知 S △ABE =
1 |
2 |
1 |
2 |
上面两个式子相加得 S △ABE + S △CDN =
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
![](http://hiphotos.baidu.com/zhidao/pic/item/48540923dd54564e5cea0388b0de9c82d0584ff0.jpg)
如图3a,连接AE、EN和NC,易知
由S △AEM △AEM =S △MEN △MEN ,S △CNF △CNF =S △EFN △EFN ,
上面两个式子相加得S △AEM △AEM +S △CNF △CNF =S 2 2 (1)
并且四边形AECN的面积=2S 2 2 .
连接AC,如图3b,由三角形面积公式,
![](http://hiphotos.baidu.com/zhidao/pic/item/a8773912b31bb051cdd55499357adab44bede08a.jpg)
易知 S △ABE =
1 |
2 |
1 |
2 |
上面两个式子相加得 S △ABE + S △CDN =
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
1 |
2 |
1 |
2 |
上面两个式子相加得 S △ABE + S △CDN =
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
1 |
2 |
1 |
2 |
上面两个式子相加得 S △ABE + S △CDN =
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
1 |
2 |
上面两个式子相加得 S △ABE + S △CDN =
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
1 |
2 |
上面两个式子相加得 S △ABE + S △CDN =
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
1 |
2 |
上面两个式子相加得 S △ABE + S △CDN =
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
上面两个式子相加得 S △ABE + S △CDN =
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
1 |
2 |
四边形AECN的面积=S 2 (2)
将(1)式和(2)相加,
得到S △AEM +S △CNF +S △ABE +S △CDN =2S 2 ,
既然S △AEM +S △ABE =S 1 ,S △CNF +S △ABE =S 3
因此S 1 +S 3 =2S 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
1 |
2 |
四边形AECN的面积=S 2 2 (2)
将(1)式和(2)相加,
得到S △AEM △AEM +S △CNF △CNF +S △ABE △ABE +S △CDN △CDN =2S 2 2 ,
既然S △AEM △AEM +S △ABE △ABE =S 1 1 ,S △CNF △CNF +S △ABE △ABE =S 3 3
因此S 1 1 +S 3 3 =2S 2 2 ,
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
S 2 |
S 1 + S 3 |
1 |
2 |
答:
S 2 |
S 1 + S 3 |
1 |
2 |
S 2 |
S 1 + S 3 |
1 |
2 |
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