Question

The perimeter of the triangle whose vertices are (0,0),(1,0) and (0,1) is

Answer 1

$1 + 1 + \sqrt{2 }$

$2 + \sqrt{2}$

Answer 2

The perimeter of the triangle = $2 + \sqrt{2}$ units

Given: Vertices (0,0), (1,0), and (0,1)

To find: The perimeter of the triangle.

Solution:

Let us assume the vertices as

  A = (0,0)

  B = (1,0)

  C = (0,1)

∴ OA = 1 unit

  OB = 1 unit

To find AB we will use the Pythagoras theorem,

i.e, $AB^{2} = OA^{2} + OB^{2}$

     $AB^{2} = 1^{2} + 1^{2}\\ AB^{2} = 1 + 1\\ AB = \sqrt{2}$

∴ The perimeter of the triangle = OA + OB + AB

                                                    = 1 +1+$\sqrt{2}$

                                                    = $2 + \sqrt{2}$ units.

Perimeter = $2 + \sqrt{2}$ units