Answer :
Sure! Let's go through the steps to find the area and perimeter of a rectangle given its vertices [tex]\((-3,-5)\)[/tex], [tex]\((1,-5)\)[/tex], [tex]\((1,0)\)[/tex], and [tex]\((-3,0)\)[/tex].
### Step 1: Identify the side lengths
- Length: To find the length, we can examine the horizontal distance between [tex]\((-3, -5)\)[/tex] and [tex]\((1, -5)\)[/tex]. The distance between the x-coordinates of these points is [tex]\(1 - (-3) = 1 + 3 = 4\)[/tex].
- Width: To find the width, we examine the vertical distance between [tex]\((1, -5)\)[/tex] and [tex]\((1, 0)\)[/tex]. The distance between the y-coordinates of these points is [tex]\(0 - (-5) = 0 + 5 = 5\)[/tex].
### Step 2: Calculate the area
The area [tex]\(A\)[/tex] of a rectangle is given by:
[tex]\[ A = \text{length} \times \text{width} \][/tex]
Substituting the length and width found:
[tex]\[ A = 4 \times 5 = 20 \][/tex]
### Step 3: Calculate the perimeter
The perimeter [tex]\(P\)[/tex] of a rectangle is given by:
[tex]\[ P = 2 \times (\text{length} + \text{width}) \][/tex]
Substituting the length and width found:
[tex]\[ P = 2 \times (4 + 5) = 2 \times 9 = 18 \][/tex]
### Final Result
- Length: 4 units
- Width: 5 units
- Area: 20 square units
- Perimeter: 18 units
### Step 1: Identify the side lengths
- Length: To find the length, we can examine the horizontal distance between [tex]\((-3, -5)\)[/tex] and [tex]\((1, -5)\)[/tex]. The distance between the x-coordinates of these points is [tex]\(1 - (-3) = 1 + 3 = 4\)[/tex].
- Width: To find the width, we examine the vertical distance between [tex]\((1, -5)\)[/tex] and [tex]\((1, 0)\)[/tex]. The distance between the y-coordinates of these points is [tex]\(0 - (-5) = 0 + 5 = 5\)[/tex].
### Step 2: Calculate the area
The area [tex]\(A\)[/tex] of a rectangle is given by:
[tex]\[ A = \text{length} \times \text{width} \][/tex]
Substituting the length and width found:
[tex]\[ A = 4 \times 5 = 20 \][/tex]
### Step 3: Calculate the perimeter
The perimeter [tex]\(P\)[/tex] of a rectangle is given by:
[tex]\[ P = 2 \times (\text{length} + \text{width}) \][/tex]
Substituting the length and width found:
[tex]\[ P = 2 \times (4 + 5) = 2 \times 9 = 18 \][/tex]
### Final Result
- Length: 4 units
- Width: 5 units
- Area: 20 square units
- Perimeter: 18 units