Question:
The perimeter of the triangle formed by the points (0, 0), (0, 1) and (0, 1) is
(a) $1 \pm \sqrt{2}$
(b) $\sqrt{2}+1$
(c) 3
(d) $2+\sqrt{2}$
Solution:
We have a triangle
whose co-ordinates are A (0, 0); B (1, 0); C (0, 1). So clearly the triangle is right angled triangle, right angled at A. So,
$\mathrm{AB}=1$ unit
$\mathrm{AC}=1$ unit
Now apply Pythagoras theorem to get the hypotenuse,
$\mathrm{BC}=\sqrt{\mathrm{AB}^{2}+\mathrm{AC}^{2}}$
$=\sqrt{2}$
So the perimeter of the triangle is,
$=\mathrm{AB}+\mathrm{BC}+\mathrm{AC}$
$=1+1+\sqrt{2}$
$=2+\sqrt{2}$
Therefore the answer is (d)