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