Q99 / 160 · TS EAMCET 2009 · Mathematics

The perimeter of the triangle with vertices at $(1,0,0),(0,1,0)$ and $(0,0,1)$ is

  • A
    3
  • B
    2
  • C
    $2 \sqrt{2}$
  • Correct answer
    D
    $3 \sqrt{2}$

Solution

Let $A=(1,0,0), B=(0,1,0)$ and $C=(0,0,1)$ Now, $\begin{aligned} \text { Now, } A B & =\sqrt{(0-1)^2+(1-0)^2+0^2}=\sqrt{2} \\ B C & =\sqrt{0^2+(0-1)^2+(1-0)^2}=\sqrt{2} \\ \text { and } \quad C A & =\sqrt{(1-0)^2+0^2+(0-1)^2}=\sqrt{2} \end{aligned}$ $\begin{aligned} \therefore \quad \text { Perimeter of triangle } & =A B+B C+C A \\ = & \sqrt{2}+\sqrt{2}+\sqrt{2}=3 \sqrt{2} \end{aligned}$

Hence, the correct answer is option: 'D'.

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