The length of a rectangle is five times its width.
If the area of the rectangle is 245cm squared, find its perimeter.

Let's solve this step by step:

1. **Given information:**
   - Let the width of the rectangle be \( w \).
   - The length of the rectangle is 5 times the width, so the length is \( 5w \).
   - The area of the rectangle is 245 cm².

2. **Formula for the area of a rectangle:**
   The area \( A \) is given by:
   \[
   A = \text{length} \times \text{width}
   \]
   Substituting the values:
   \[
   245 = 5w \times w
   \]
   This simplifies to:
   \[
   245 = 5w^2
   \]

3. **Solve for \( w \):**
   Divide both sides by 5:
   \[
   w^2 = \frac{245}{5} = 49
   \]
   Now, take the square root of both sides:
   \[
   w = \sqrt{49} = 7
   \]

   So, the width of the rectangle is \( 7 \) cm.

4. **Find the length:**
   Since the length is 5 times the width:
   \[
   \text{Length} = 5w = 5 \times 7 = 35 \, \text{cm}
   \]

5. **Find the perimeter:**
   The formula for the perimeter \( P \) of a rectangle is:
   \[
   P = 2 \times (\text{length} + \text{width})
   \]
   Substituting the values:
   \[
   P = 2 \times (35 + 7) = 2 \times 42 = 84 \, \text{cm}
   \]

Thus, the perimeter of the rectangle is **84 cm**.

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Would you like further details or clarifications?

Here are 5 related questions for practice:
1. If the width of the rectangle was 10 cm, what would be the length and perimeter?
2. What would the perimeter be if the area was doubled while keeping the same ratio between length and width?
3. How would the problem change if the length was 3 times the width instead of 5 times?
4. Can you find the diagonal of the rectangle using the width and length?
5. How does the perimeter formula change for a square?

**Tip:** When solving geometric problems, drawing a diagram can help visualize relationships between different dimensions.

The length of a rectangle is five times its width. If the area of the rectangle is 245cm squared, find its perimeter.

Learn how to find the perimeter of a rectangle where the length is five times the width, and the area is 245 cm². This problem involves key geometric and algebraic concepts suitable for students in Grades 6-8.

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