Question

The vertices of ABC are A(-2,2), B(6,2), and C(0,8). The perimeter of ABC

Answer 1

The perimeter of triangle ABC = 22.78 units

Explanation:

Formula:-

The length of line segment with end points (x₁, y₁) and (x₂, y₂) is given by,

Length = √ [(x₂ - x₁)² + (y₂ - y₁)²]

To find the each side of triangle

We have A(-2,2), B(6,2), and C(0,8).

AB = √[(6 - -2)² + (2 - 2)²] = √64 = 8

BC = √[(0 - 6)² + (8 - 2)²] = √(36 + 36) = 8.48

AC = √[(0 - -2)² + (8 - 2)²] = √(4 + 36) = 6.3

To find the perimeter of triangle ABC

Perimeter = AB + BC + AC

= 8 + 8.48 + 6.3 = 22.78 units

Answer 2

The perimeter of the triangle ABC is 22.8 units

Explanation:

* Lets study the information in the problem

- There is Δ ABC with vertices:

A (-2 , 2) , B (6 , 2) , C (0 , 8)

- The perimeter of the triangle is the sum of the length of its

three sides

* We must to find the lengths of AB , BC and CD

- The rule to find the distance between 2 points (x1 , y1) and (x2 , y2) is

√[(x2 - x1)² + (y2 - y1)²]

* Lets find the lengths of the three sides

- Length of AB

∵ A = (-2 , 2) and B = (6 , 2)

∴ AB = √[(6 - -2)² + (2 - 2)²] = √8² = 8 units

- Length of BC

∵ B = (6 , 2) and C = (0 , 8)

∴ BC = √[(0 - 6)² + (8 - 2)²] = √[6² + 6²] = √72 = 6√2 units

- Length of AC

∵ A = (-2 , 2) and C = (0 , 8)

∴ AC = √[(0 - -2)² + (8 - 2)²] = √[2² + 6²] = √40 = 2√10 units

* Now lets find the perimeter of the triangle

∵ The perimeter = AB + BD + AC

∴ The perimeter = 8 + 6√2 + 2√10 = 22.8 units

* The perimeter of the triangle ABC is 22.8 units