Question

Find the perimeter of (triangle) ABC with vertices A(-2, 3), B(3,-3), and C(-2, -3).​

Answer 1

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Answer 2

Given,

A triangle with vertices  A(-2, 3), B(3,-3), and C(-2, -3).​

To find,

The perimeter of the triangle.

Solution,

The perimeter of a triangle is the sum of all three sides of the triangle.

The three sides of the triangle can be found by calculating the distance between points AB, AC, and CA with the help of the distance formula.

The distance formula is -

$\sqrt{(x2-x1)^{2}+((y2-y1)^{2} }$.

We have A(-2, 3), B(3,-3), and C(-2, -3)

⇒   Side AB = $\sqrt{(3-(-2))^{2}+((-3-(3))^{2} }$

⇒   Side AB = $\sqrt{(5)^{2}+(-6)^{2} }$

⇒   Side AB = $\sqrt{25+36}=\sqrt{61}$

⇒   Side BC= $\sqrt{(-2-(3))^{2}+((-3-(-3))^{2} }$

⇒   Side BC = $\sqrt{(-5^{2}+(0)^{2} }=\sqrt{25}$

⇒   Side BC = 5

⇒   Side CA = $\sqrt{(-2-(-2))^{2}+((-3-(3))^{2} }$

⇒   Side CA =$\sqrt{(0)^{2}+(-6)^{2} }=\sqrt{36}$

⇒   Side CA = 6.

Hence, perimeter = Side AB + Side BC + Side CA = 5 + 6 + √61

perimeter = 11 + √61 or 18.81.

Therefore, The perimeter of the triangle with vertices  A(-2, 3), B(3,-3), and C(-2, -3)  is (11 + √61) or 18.81.