In a triangle ABC, if the midpoints of sides | 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 the midpoints of $AB, BC, CA$ are $M_1(3,

Vedclass · Accepted Answer

$400$

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 the midpoints of $AB, BC, CA$ are $M_1(3,0,0)$, $M_2(0,4,0)$, and $M_3(0,0,5)$ respectively.Using the midpoint formula:$x_1+x_2=6, x_2+x_3=0, x_3+x_1=0$Solving these, we get $x_1=3, x_2=3, x_3=-3$.Similarly for $y$ coordinates:$y_1+y_2=0, y_2+y_3=8, y_3+y_1=0$Solving these, we get $y_1=-4, y_2=4, y_3=4$.Similarly for $z$ coordinates:$z_1+z_2=0, z_2+z_3=0, z_3+z_1=10$Solving these, we get $z_1=5, z_2=-5, z_3=5$.Thus, the vertices are $A(3, -4, 5)$, $B(3, 4, -5)$, and $C(-3, 4, 5)$.Now, calculate the squares of the side lengths:$AB^2 = (3-3)^2 + (4-(-4))^2 + (-5-5)^2 = 0^2 + 8^2 + (-10)^2 = 64 + 100 = 164$.$BC^2 = (-3-3)^2 + (4-4)^2 + (5-(-5))^2 = (-6)^2 + 0^2 + 10^2 = 36 + 100 = 136$.$CA^2 = (3-(-3))^2 + (-4-4)^2 + (5-5)^2 = 6^2 + (-8)^2 + 0^2 = 36 + 64 = 100$.Finally, $AB^2+BC^2+CA^2 = 164 + 136 + 100 = 400$.

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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