Find the perimeter of triangle ABC, with coordinates A(-3, 0), B(0, 4), and C(3, 0).

Solution:

We know that the perimeter of a triangle is the sum of the lengths of all sides.

The length of sides can be found using the distance formula

\(d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

It is given that

A = (-3, 0), B = (0, 4) and C = (3, 0)

\(\\|AB|=\sqrt{(0-(-3))^{2}+(4-0)^{2}}=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5 \\ \\|AC|=\sqrt{(3-(-3))^{2}+(0-0)^{2}}=\sqrt{6^{2}+0^{2}}=\sqrt{36}=6 \\ \\|BC|=\sqrt{(3-0)^{2}+(0-4)^{2}}=\sqrt{3^{2}+(-4)^{2}}=\sqrt{9+16}=\sqrt{25}=5\)

So the perimeter is

P = |AB| + |AC| + |BC|

Substituting the values

P = 5 + 6 + 5

P = 16

Therefore, the perimeter of triangle ABC is 16.

Find the perimeter of triangle ABC, with coordinates A(-3, 0), B(0, 4), and C(3, 0).

Summary:

The perimeter of triangle ABC, with coordinates A(-3, 0), B(0, 4), and C(3, 0) is 16.