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Question: Answered & Verified by Expert
In a triangle $A B C$, if the mid points of sides $A B, B C$, CA are $(3,0,0),(0,4,0),(0,0,5)$ respectively, then $\mathrm{AB}^2+\mathrm{BC}^2+\mathrm{CA}^2=$
MathematicsThree Dimensional GeometryTS EAMCETTS EAMCET 2023 (14 May Shift 1)
Options:
  • A $50$
  • B $200$
  • C $300$
  • D $400$
Solution:
2260 Upvotes Verified Answer
The correct answer is: $400$


$x_1+x_2=6$
$$
\begin{aligned}
& x_2+x_3=0 \\
& x_1+x_3=0 \\
& \therefore x_1=x_2=3 \\
& x_1=-3
\end{aligned}
$$

Similarly
$$
\begin{aligned}
& y_1+y_2=0 \\
& y_2+y_3=8 \\
& y_1+y_3=0 \\
& \therefore y_2=y_3=4 \\
& y_1=-4 \\
& \text { and } z_1+z_2=0 \\
& z_2+z_3=0 \\
& z_3+z_1=10 \\
& z_1=z_3=5 \\
& z_2=-5 \\
& \therefore \mathrm{AB}^2+\mathrm{BC}^2+\mathrm{CA}^2
\end{aligned}
$$
$\begin{aligned} & =\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2+\left(z_1-z_2\right)^2 \\ & \quad+\left(x_2-x_3\right)^2+\left(y_2-y_3\right)^2+\left(z_2-z_3\right)^2 \\ & +\left(x_1-x_3\right)^2+\left(y_1-y_3\right)^2+\left(z_1-z_3\right)^2 \\ & =0+64+100+36+0+100+36+64+0=400\end{aligned}$

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