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sangangsx Mar 30, 2018

+9466

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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

hectictar · Answer 1 · 2018-03-31T12:13:20

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