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Misc 2 Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0). Let Δ ABC where AD, BE, CF are medians Since median bisects the opposite side D is Midpoint of BC E is Midpoint of AC F Midpoint of AB Misc 2 Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and (6, 0, 0). Let Δ ABC where AD, BE, CF are medians Since median bisects the opposite side D is Midpoint of BC E is Midpoint of AC F Midpoint of AB E is midpoint of AC D = ((0 + 6)/2, (0 + 0)/2, (6 + 0)/2) D = (3, 0, 3) F is midpoint of AB D = ((0 + 0)/2, (0 + 4)/2, (6 + 0)/2) D = (0, 2, 3) Calculating AD AD = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Hence x1 = 0, y1 = 0, z1 = 6 x2 = 3, y2 = – 2, z2 = 0 AD = √((3−0)2+(2−0)2+(0−6)2) = √((3)2+(2)2+(−6)2) = √(9+4+36) = √49 = 7 Lenth of median AD = 7 Calculating BE BE = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Hence x1 = 0, y1 = 4, z1 = 0 x2 = 3, y2 = 0, z2 = 3 BE = √((3−0)2+(0−4)2+(3−0)2) = √((3)2+(−4)2+(3)2) = √(9+16+9) = √𝟑𝟒 Length of median BE = √34 Calculate CF CF = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Hence x1 = 6, y1 = 0, z1 = 0 x2 = 0, y2 = 2, z2 = 3 CF = √((0−6)2+(2−0)2+(3−0)2) = √((−6)2+(2)2+(3)2) = √(36+4+9) = √49 = 7

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo