On the same graph paper, draw the straight lines represented by equations :
x= 5 , x+ 5= 0 , y + 3=0 and y = 3
Also find the area and perimeter of the rectangle formed by the intersections of these lines.

Answer

To solve the problem step by step, we will first plot the lines represented by the given equations and then find the area and perimeter of the rectangle formed by their intersections.

### Step 1: Plot the lines on the graph paper

1. **Draw the line for x = 5**:
- This is a vertical line that passes through the point (5, y) for all values of y.
- On the graph, draw a vertical line at x = 5.

2. **Draw the line for x + 5 = 0**:
- Rearranging gives us x = -5.
- This is another vertical line that passes through the point (-5, y) for all values of y.
- On the graph, draw a vertical line at x = -5.

3. **Draw the line for y + 3 = 0**:
- Rearranging gives us y = -3.
- This is a horizontal line that passes through the point (x, -3) for all values of x.
- On the graph, draw a horizontal line at y = -3.

4. **Draw the line for y = 3**:
- This is a horizontal line that passes through the point (x, 3) for all values of x.
- On the graph, draw a horizontal line at y = 3.

### Step 2: Identify the intersections of the lines

The lines will intersect at four points, which will be the vertices of the rectangle formed by these lines:

- Intersection of x = 5 and y = 3: (5, 3)
- Intersection of x = 5 and y = -3: (5, -3)
- Intersection of x = -5 and y = 3: (-5, 3)
- Intersection of x = -5 and y = -3: (-5, -3)

### Step 3: Calculate the lengths of the sides of the rectangle

1. **Length of the rectangle (horizontal distance)**:
- The distance between the points (-5, 3) and (5, 3) is:
\[ \text{Length} = 5 - (-5) = 5 + 5 = 10 \text{ units} \]

2. **Width of the rectangle (vertical distance)**:
- The distance between the points (-5, 3) and (-5, -3) is:
\[ \text{Width} = 3 - (-3) = 3 + 3 = 6 \text{ units} \]

### Step 4: Calculate the area and perimeter of the rectangle

1. **Area of the rectangle**:
\[ \text{Area} = \text{Length} \times \text{Width} = 10 \times 6 = 60 \text{ square units} \]

2. **Perimeter of the rectangle**:
\[ \text{Perimeter} = 2 \times (\text{Length} + \text{Width}) = 2 \times (10 + 6) = 2 \times 16 = 32 \text{ units} \]

### Final Results
- **Area**: 60 square units
- **Perimeter**: 32 units