Question

find the perimeter of the triangle whose vertices are -2,4 4,6 6,3​

Answer 1

Given :

• Co-ordinates of A = ( - 2 , 4 )

• Co-ordinates of B = ( 4 , 6 )

• Co-ordinates of C = ( 6 , 3 )

To Find :

• Perimeter of triangle

Solution :

First we have to find distance of AB , BC , AC by using distance formula

$\boxed{ \boxed{ \sf Distance = \sqrt{ {(x_2 - x_1)}^{2} +{(y_2 - y_1)}^{2}}}}$

Distance of AB

$\sf \implies AB = \sqrt{ {(4 + 2)}^{2}+ {(6 - 4)}^{2}} \\ \\ \sf \implies AB = \sqrt{ {6}^{2} + {2}^{2}} \\ \\ \sf \implies AB = \sqrt{36 + 4} \\ \\ \sf \implies AB = \sqrt{40}$

Distance of BC

$\sf \implies BC = \sqrt{ {(6 - 4)}^{2} + {(3 - 6)}^{2} } \\ \\\sf \implies BC = \sqrt{ {2}^{2} + { (- 3)}^{2} } \\ \\\sf \implies BC = \sqrt{4 + 9} \\ \\\sf \implies BC = \sqrt{13}$

Distance of AC

$\sf \implies AC = \sqrt{ {(6 + 2)}^{2} + {(3 - 4)}^{2}} \\ \\\sf \implies AC = \sqrt{ {8}^{2} + ( - {1)}^{2} } \\ \\\sf \implies AC = \sqrt{64 + 1} \\ \\\sf \implies AC = \sqrt{65}$

Perimeter of triangle

$\sf Perimeter = AB + BC + AC \\ \\ \sf Perimeter = \sqrt{40} + \sqrt{65} + \sqrt{13} \\ \\ \sf \large Perimeter = 18 \: unit$

Answer 2

Given: A triangle formed by the coordinates (-2, 4), (4, 6) and (6, 3).

To find: The perimeter.

Answer:

To do so, we'll have to find the length of each side [distance between two points.]

Distance formula:

$\tt Distance\ =\ \sqrt{\Bigg(x_2\ -\ x_1\Bigg)^2\ +\ \Bigg(y_2\ -\ y_1\Bigg)^2}$

Let's first find the distance between the points (-2, 4) and (4, 6).

From those points,

$\tt x_1\ =\ -2\\\\x_2\ =\ 4\\\\y_1\ =\ 4\\\\y_2\ =\ 6$

Using them in the formula,

$\tt Distance\ =\ \sqrt{\Bigg(4\ +\ 2\Bigg)^2\ +\ \Bigg(6\ -\ 4\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(6\Bigg)^2\ +\ \Bigg(2\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(36\ +\ 4\Bigg)}\\\\\\\\\bf Distance\ =\ \sqrt{40}\ =\ 6.32\ units\ [approx.]$

Now, let's find the distance between the points (4, 6) and (6, 3).

From those points,

$\tt x_1\ =\ 4\\\\x_2\ =\ 6\\\\y_1\ =\ 6\\\\y_2\ =\ 3$

Using them in the formula,

$\tt Distance\ = \sqrt{\Bigg(6\ -\ 4\Bigg)^2\ +\ \Bigg(3\ -\ 6\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(2\Bigg)^2\ +\ \Bigg(-3\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(4\ +\ 9\Bigg)^2}\\\\\\\\\bf Distance\ =\ \sqrt{13}\ =\ 3.6\ units\ [approx.]$

Now, the distance between the points (6, 3) and (-2, 4).

From those points,

$\tt x_1\ =\ 6\\\\x_2\ =\ -2\\\\y_1\ =\ 3\\\\y_2\ =\ 4$

Using them in the formula,

$\tt Distance\ =\ \sqrt{\Bigg(-2\ -\ 6\Bigg)^2\ +\ \Bigg(4\ -\ 3\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(-8\Bigg)^2\ +\ \Bigg(1\Bigg)^2}\\\\\\\\Distance\ =\ \sqrt{\Bigg(64\ +\ 1\Bigg)}\\\\\\\\\bf Distance\ =\ \sqrt{65}\ =\ 8.06\ units\ [approx.]$

Now, to find the perimeter.

Perimeter = Sum of the lengths of all sides

Perimeter = 6.32 + 3.6 + 8.06

Perimeter = 17.98

Therefore, the perimeter of the triangle formed by the points (-2, 4), (4, 6) and (6, 3) = 17.98 units.