What is the perimeter of the triangle with vertices A(- 4, 2), B(0, - 1) and C(3, 3)?

Options

A. `7+3sqrt(2)`
B. `10+5sqrt(2)`
C. `11+6sqrt(2)`
D. `5+sqrt(2)`

Answer

To find the perimeter of the triangle with vertices A(-4, 2), B(0, -1), and C(3, 3), we will follow these steps:

### Step 1: Identify the vertices
The vertices of the triangle are:
- A(-4, 2)
- B(0, -1)
- C(3, 3)

### Step 2: Calculate the lengths of the sides using the distance formula
The distance formula between two points (x1, y1) and (x2, y2) is given by:
\[ d = \sqrt{(x2 - x1)^2 + (y2 - y1)^2} \]

#### Step 2.1: Calculate AB
Using points A(-4, 2) and B(0, -1):
\[ AB = \sqrt{(0 - (-4))^2 + (-1 - 2)^2} = \sqrt{(0 + 4)^2 + (-3)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

#### Step 2.2: Calculate BC
Using points B(0, -1) and C(3, 3):
\[ BC = \sqrt{(3 - 0)^2 + (3 - (-1))^2} = \sqrt{3^2 + (3 + 1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

#### Step 2.3: Calculate AC
Using points A(-4, 2) and C(3, 3):
\[ AC = \sqrt{(3 - (-4))^2 + (3 - 2)^2} = \sqrt{(3 + 4)^2 + (3 - 2)^2} = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \]

### Step 3: Calculate the perimeter
The perimeter \( P \) of the triangle is the sum of the lengths of its sides:
\[ P = AB + BC + AC = 5 + 5 + 5\sqrt{2} = 10 + 5\sqrt{2} \]

### Final Answer
The perimeter of the triangle is \( 10 + 5\sqrt{2} \).

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