Question

Find perimeter of a triangle on a coordinate plane with coordinates A(-3,6), B(-3,2) and C(3,2) rounded off to 2 decimals places.

Answer 1

Hint:In the given question, we need to calculate the perimeter of a triangle whose coordinates of vertices are given to us. In such types of questions, we have to first find the lengths of sides of the triangle using distance formula and then sum up the distances to find the perimeter of the required triangle.

Complete step by step answer:
Perimeter of a triangle $ = $ sum of all sides.
So, we need to find the lengths of sides, namely AB, BC and CA.
Using distance formula,
Distance between two points whose coordinates are given as\[({x_1},{y_1})\] and \[({x_2},{y_2})\] is given as $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $.
So, distance between A and B = $AB = \sqrt {{{( - 3 - ( - 3))}^2} + {{(2 - 6)}^2}} $
 $ = AB = \sqrt {0 + 16} $
$ = AB = 4$units
Distance between B and C = $BC = \sqrt {{{(3 - ( - 3))}^2} + {{(2 - 2)}^2}} $
 $ = BC = \sqrt {36 + 0} $
 $ = BC = 6$units
Distance between C and A = $CA = \sqrt {{{( - 3 - 3)}^2} + {{(6 - 2)}^2}} $
$ = CA = \sqrt {36 + 16} $
$ = CA = \sqrt {52} $
$ = CA = 2\sqrt {13} $
$ = CA = 2 \times 3.6055$
$ = CA = 7.21{\text{ }}$units (rounded off up to two decimal places)
So, perimeter of triangle ABC $ = AB + BC + CA$
Perimeter of triangle ABC $ = \left( {4 + 6 + 7.21} \right)$ units
Perimeter of triangle ABC $ = 17.21{\text{ }}$units.
So, the perimeter of the triangle on a coordinate plane with coordinates A(-3,6), B(-3,2) and C(3,2) rounded off to 2 decimals places is $17.21{\text{ }}$ units .

Note: The given triangle in the problem is a right angled triangle as side AB is along the x axis and side BC is along the y axis. So, the side CA can also be calculated using the Pythagoras theorem and then the perimeter of triangle ABC can be evaluated by adding up all the sides.