A rectangle has the width being 7 cm less than the length,if the area is 198cm²,find the perimeter of the rectangle

Let the length of the rectangle be \( L \) cm. Since the width is 7 cm less than the length, the width \( W \) can be expressed as:

\[ W = L - 7 \]

The area of the rectangle is given by:

\[ \text{Area} = L \times W \]

Substituting the width into the area formula:

\[ 198 = L \times (L - 7) \]

This simplifies to the quadratic equation:

\[ L^2 - 7L - 198 = 0 \]

To solve this quadratic equation, we can use the quadratic formula:

\[ L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -7 \), and \( c = -198 \). Plugging in these values:

\[ L = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-198)}}{2 \cdot 1} \]

\[ L = \frac{7 \pm \sqrt{49 + 792}}{2} \]

\[ L = \frac{7 \pm \sqrt{841}}{2} \]

\[ L = \frac{7 \pm 29}{2} \]

This gives us two possible solutions:

\[ L = \frac{7 + 29}{2} = 18 \]
\[ L = \frac{7 - 29}{2} = -11 \]

Since the length cannot be negative, we have:

\[ L = 18 \]

Now, we find the width:

\[ W = L - 7 = 18 - 7 = 11 \]

Next, we calculate the perimeter \( P \) of the rectangle:

\[ P = 2L + 2W \]

Substituting the values of \( L \) and \( W \):

\[ P = 2 \times 18 + 2 \times 11 \]

\[ P = 36 + 22 \]

\[ P = 58 \]

Thus, the perimeter of the rectangle is:

\[ \boxed{58 \text{ cm}} \]

Would you like more details or have any questions?

Here are 8 related questions:

1. What are the properties of a rectangle?
2. How is the area of a rectangle calculated?
3. Can the width of a rectangle ever be greater than its length?
4. How do you solve quadratic equations?
5. What is the significance of the discriminant in the quadratic formula?
6. Can a rectangle have negative dimensions?
7. What are other methods to solve quadratic equations besides the quadratic formula?
8. How is the perimeter of a rectangle related to its length and width?

**Tip:** Always double-check your solutions for potential errors, especially when solving quadratic equations and substituting back into original formulas.

Discover how to find the perimeter of a rectangle with a given area of 198cm². This problem involves geometry, quadratic equations, and perimeter calculations. Suitable for high school students learning about geometric formulas and quadratic equations.

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