作业帮嘿!帮帮忙!帮忙解一些这道数学题!求解题过程!紧急!O(∩_∩)O谢谢!写出解题过程,谢谢啦!
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嘿!帮帮忙!帮忙解一些这道数学题!求解题过程!紧急!O(∩_∩)O谢谢!写出解题过程,谢谢啦!
嘿!帮帮忙!帮忙解一些这道数学题!求解题过程!紧急!O(∩_∩)O谢谢!
写出解题过程,谢谢啦!
嘿!帮帮忙!帮忙解一些这道数学题!求解题过程!紧急!O(∩_∩)O谢谢!写出解题过程,谢谢啦!
勾股定理,S1+S2=A,S2+S3=B.S3+S4=C,所以s1+s2+s3+s4=A+C=16
S1+S2=SA S2+S3=SB S3+S4=SC
所以S1+S2+S3+S4=SA+SC=16
由勾股定理得S1+S2=7，S3+S4=9，所以，S1+S2+S3+S4=7+9=16
设A与s(1),s(2)的交点分别为点D，E。
s(1) 的右下角点，A的下方角点，s(2)的左下角点分别为F,G,H.
|DF| = [s(1)]^(1/2),
|EH| = [s(2)]^(1/2).
|DG| = A^(1/2) = |GE|.
|DE|^2 = 2A.
则，s(1) + |FG|^2 = A = |GH|^2 + s(2),<...
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设A与s(1),s(2)的交点分别为点D，E。
s(1) 的右下角点，A的下方角点，s(2)的左下角点分别为F,G,H.
|DF| = [s(1)]^(1/2),
|EH| = [s(2)]^(1/2).
|DG| = A^(1/2) = |GE|.
|DE|^2 = 2A.
则，s(1) + |FG|^2 = A = |GH|^2 + s(2),
|FG| = [A - s(1)]^(1/2),
|GH| = [A - s(2)]^(1/2).
[|FG| + |GH|]^2 + [|EH| - |DF|]^2 = |DE|^2 = 2A
= { [A-s(1)]^(1/2) + [A-s(2)]^(1/2) }^2 + { [s(2)]^(1/2) - [s(1)]^(1/2) }^2
= A-s(1)+A-s(2)+2[A-s(1)]^(1/2)[A-s(2)]^(1/2) + s(2)+s(1)-2[s(1)s(2)]^(1/2)
= 2A + 2[A-s(1)]^(1/2)[A-s(2)]^(1/2) - 2[s(1)s(2)]^(1/2),
0 = [A-s(1)]^(1/2)[A-s(2)]^(1/2) - [s(1)s(2)]^(1/2),
[A-s(1)]^(1/2)[A-s(2)]^(1/2) = [s(1)s(2)]^(1/2),
[A-s(1)][A-s(2)] = s(1)s(2) = A^2 + s(1)s(2) - A[s(1)+s(2)],
0 = A^2 - A[s(1)+s(2)] = A[A-s(1)-s(2)],
0 = A-s(1)-s(2),
s(1)+s(2) = A = 7.
同理，s(3)+s(4)=C = 9,
s(1)+s(2)+s(3)+s(4) = 7+9=16.
收起
利用角边角可证A、B、C两边三角形全等，其余同理。
再利用勾股即可得A=S1+S2,S2+S3=B.S3+S4=C,得s1+s2+s3+s4=A+C=16