Answer:
The perimeter of the triangle ABC is 22.8 units
Step-by-step 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
