Triangle ABC is an isosceles triangle in which AB = AC. What is the perimeter of △ABC?

Solution:

Using the distance formula we can find the length of each line

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

Length of AB, \(AB=\sqrt{(-1- (-1))^{2}+(1-6)^{2}}\)

So we get,

\(AB=\sqrt{(0)^{2}+(-5)^{2}}\)

AB = √25 = 5 units

Length of AC is same as length of AB

AC = 5 units

Length of BC, \(BC=\sqrt{(2- (-1))^{2}+(2-1)^{2}}\)

So we get,

\(BC=\sqrt{(3)^{2}+(1)^{2}}\)

BC = √10 units

We know that

Perimeter = Sum of all lengths of the triangle

= AB + AC + BC

= 5 + 5 + √10

= 10 + √10units

Therefore, the perimeter of △ABC is 10 + √10units.

Summary:

Triangle ABC is an isosceles triangle in which AB = AC. The perimeter of △ABC is 10 + √10units.