Q167 / 270 · NDA 2012 (Phase 1) · Mathematics

What is the perimeter of the triangle with vertices $\mathrm{A}(-4,2), \mathrm{B}(0,-1)$ and $\mathrm{C}(3,3)$ ?

  • A
    $7+3 \sqrt{2}$
  • Correct answer
    B
    $10+5 \sqrt{2}$
  • C
    $11+6 \sqrt{2}$
  • D
    $5+\sqrt{2}$

Solution

$\begin{aligned} & \mathrm{AB}=\sqrt{(0+4)^2}+ \\ &+(-1-2)^2 \\ &=\sqrt{16+9}=5 \\ & \mathrm{BC}=\sqrt{9+16}=5 \\ & \mathrm{CA}=\sqrt{49+(1)^2}=\sqrt{50} \\ &=5 \sqrt{2} \\ &=10+5 \sqrt{2} \\ &=5(2+\sqrt{2})\end{aligned}$

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

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