In a triangle ABC,if the midpoints of sides A | vedclass

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(D) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.Given midpoints are $M_{AB} = (3,0,0)$,$M_{BC} =

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$400$

Vedclass · Answer

(D) Let the vertices of the triangle be $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$.Given midpoints are $M_{AB} = (3,0,0)$,$M_{BC} = (0,4,0)$,and $M_{CA} = (0,0,5)$.By the midpoint formula:$\frac{x_1+x_2}{2} = 3, \frac{y_1+y_2}{2} = 0, \frac{z_1+z_2}{2} = 0 \implies x_1+x_2=6, y_1+y_2=0, z_1+z_2=0$$\frac{x_2+x_3}{2} = 0, \frac{y_2+y_3}{2} = 4, \frac{z_2+z_3}{2} = 0 \implies x_2+x_3=0, y_2+y_3=8, z_2+z_3=0$$\frac{x_3+x_1}{2} = 0, \frac{y_3+y_1}{2} = 0, \frac{z_3+z_1}{2} = 5 \implies x_3+x_1=0, y_3+y_1=0, z_3+z_1=10$Solving these systems,we get $A(3, -4, 5)$,$B(3, 4, -5)$,and $C(-3, 4, 5)$.Now,calculate the squared lengths of the sides:$AB^2 = (3-3)^2 + (4 - (-4))^2 + (-5-5)^2 = 0^2 + 8^2 + (-10)^2 = 64 + 100 = 164$$BC^2 = (-3-3)^2 + (4-4)^2 + (5 - (-5))^2 = (-6)^2 + 0^2 + 10^2 = 36 + 100 = 136$$CA^2 = (3 - (-3))^2 + (-4-4)^2 + (5-5)^2 = 6^2 + (-8)^2 + 0^2 = 36 + 64 = 100$Therefore,$AB^2 + BC^2 + CA^2 = 164 + 136 + 100 = 400$.

In a triangle $ABC$,if the midpoints of sides $AB, BC, CA$ are $(3,0,0), (0,4,0), (0,0,5)$ respectively,then $AB^2 + BC^2 + CA^2 = $

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