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help.

 Mar 30, 2018

Best Answer 

 #1
avatar+7612 
+3

AB  =  \(\sqrt{(3-1)^2+(5-3)^2}\,=\,\sqrt{2^2+2^2}\,=\,\sqrt{4+4}\,=\,\sqrt8\)

 

BC  =  \(\sqrt{(3-1)^2+(3--1)^2}\,=\,\sqrt{2^2+4^2}\,=\,\sqrt{4+16}\,=\,\sqrt{20}\)

 

CD  =  \(\sqrt{(5-3)^2+(3--1)^2}\,=\,\sqrt{2^2+4^2}\,=\,\sqrt{4+16}\,=\,\sqrt{20}\)

 

DA  =  \(\sqrt{(5-3)^2+(5-3)^2}\,=\,\sqrt{2^2+2^2}\,=\,\sqrt{4+4}\,=\,\sqrt8\)

 

 

AB + BC + CD + DA   =   √8 + √20 + √20 + √8   ≈  14.6

 Mar 31, 2018
 #1
avatar+7612 
+3
Best Answer

AB  =  \(\sqrt{(3-1)^2+(5-3)^2}\,=\,\sqrt{2^2+2^2}\,=\,\sqrt{4+4}\,=\,\sqrt8\)

 

BC  =  \(\sqrt{(3-1)^2+(3--1)^2}\,=\,\sqrt{2^2+4^2}\,=\,\sqrt{4+16}\,=\,\sqrt{20}\)

 

CD  =  \(\sqrt{(5-3)^2+(3--1)^2}\,=\,\sqrt{2^2+4^2}\,=\,\sqrt{4+16}\,=\,\sqrt{20}\)

 

DA  =  \(\sqrt{(5-3)^2+(5-3)^2}\,=\,\sqrt{2^2+2^2}\,=\,\sqrt{4+4}\,=\,\sqrt8\)

 

 

AB + BC + CD + DA   =   √8 + √20 + √20 + √8   ≈  14.6

hectictar Mar 31, 2018

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