3. The length of a rectangle is three more than twice the width. a) Draw a diagram to represent the length of sides of the rectangle. b) Determine the dimensions that will give a total area of $$ 27 \mathrm{~m}^{2} $$ [3]

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### Problem 3

**The length of a rectangle is three more than twice the width.**

---

#### **a) Diagram of the Rectangle**

While I can't provide a visual diagram here, I can describe how to represent the rectangle with labeled sides based on the given information.

1. **Label the Width:**
   - Let’s denote the width of the rectangle as $$ W $$ meters.

2. **Label the Length:**
   - According to the problem, the length $$ L $$ is three more than twice the width. Mathematically, this can be expressed as:
     $$
     L = 2W + 3
     $$

3. **Visual Representation:**

```
+---------------------------+
|                           |
|                           |
|                           |  L = 2W + 3
|                           |
|                           |
+---------------------------+
           W
```

- **Width (W):** The shorter side of the rectangle.
- **Length (L):** The longer side, which is three more than twice the width.

---

#### **b) Determining the Dimensions for an Area of $$ 27 \, \text{m}² $$**

**Given:**
- Area ($$ A $$) of the rectangle is $$ 27 \, \text{m}² $$.
- Relationship between length and width: $$ L = 2W + 3 $$.

**Steps to Find Dimensions:**

1. **Write the Area Formula:**
   $$
   A = L \times W
   $$
   Substituting the given area:
   $$
   27 = (2W + 3) \times W
   $$

2. **Formulate the Equation:**
   $$
   27 = 2W^2 + 3W
   $$
   Rearranging the terms to form a quadratic equation:
   $$
   2W^2 + 3W - 27 = 0
   $$

3. **Solve the Quadratic Equation:**
   Use the quadratic formula:
   $$
   W = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$
   Where $$ a = 2 $$, $$ b = 3 $$, and $$ c = -27 $$.

$$
   W = \frac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-27)}}{2 \times 2}
   $$
   $$
   W = \frac{-3 \pm \sqrt{9 + 216}}{4}
   $$
   $$
   W = \frac{-3 \pm \sqrt{225}}{4}
   $$
   $$
   W = \frac{-3 \pm 15}{4}
   $$

4. **Calculate the Potential Solutions:**
   - **First Solution:**
     $$
     W = \frac{-3 + 15}{4} = \frac{12}{4} = 3 \, \text{m}
     $$
   - **Second Solution:**
     $$
     W = \frac{-3 - 15}{4} = \frac{-18}{4} = -4.5 \, \text{m}
     $$
     *(Since width cannot be negative, we discard this solution.)*

5. **Determine the Length:**
   $$
   L = 2W + 3 = 2(3) + 3 = 6 + 3 = 9 \, \text{m}
   $$

**Conclusion:**

- **Width ($$ W $$)** = **3 meters**
- **Length ($$ L $$)** = **9 meters**

These dimensions will yield an area of $$ 27 \, \text{m}² $$.

---

#### **Verification**

To ensure the dimensions are correct, calculate the area using the found values:

$$
A = L \times W = 9 \, \text{m} \times 3 \, \text{m} = 27 \, \text{m}²
$$

The calculation confirms that the dimensions satisfy the area requirement.
**Problem 3** **a) Diagram of the Rectangle** - **Width (W):** The shorter side. - **Length (L):** Three more than twice the width ($$ L = 2W + 3 $$). **b) Dimensions for an Area of $$ 27 \, \text{m}² $$** - **Width (W):** 3 meters - **Length (L):** 9 meters These dimensions give an area of $$ 27 \, \text{m}² $$.

3. The length of a rectangle is three more than twice the width. a) Draw a diagram to represent the length of sides of the rectangle. b) Determine the dimensions that will give a total area of \( 27 \mathrm{~m}^{2} \) [3]

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