2. For each area, what are the dimensions of the rectangle with the minimum perimeter? What is that perimeter? (2 marks each) \( \begin{array}{ll}\text { a) } 20 \mathrm{ft}^{2} & \text { b) } 81 \mathrm{~m}^{2} \\ \text { c) } 144 \mathrm{~cm}^{2} & \text { d) } 60 \mathrm{in}^{2}\end{array} \)

To find the dimensions of the rectangle with the minimum perimeter for a given area, we can use the fact that for a fixed area, a square has the minimum perimeter. Therefore, we can find the side length of a square that has the same area as the given rectangle. 1. **Area = 20 ft²** - For a square, \( A = s^2 \) - \( s^2 = 20 \) - \( s = \sqrt{20} \approx 4.47 \) ft - Dimensions: \( 4.47 \, \text{ft} \times 4.47 \, \text{ft} \) - Perimeter: \( P = 4s = 4 \times 4.47 \approx 17.88 \, \text{ft} \) 2. **Area = 81 m²** - For a square, \( A = s^2 \) - \( s^2 = 81 \) - \( s = \sqrt{81} = 9 \) m - Dimensions: \( 9 \, \text{m} \times 9 \, \text{m} \) - Perimeter: \( P = 4s = 4 \times 9 = 36 \, \text{m} \) 3. **Area = 144 cm²** - For a square, \( A = s^2 \) - \( s^2 = 144 \) - \( s = \sqrt{144} = 12 \) cm - Dimensions: \( 12 \, \text{cm} \times 12 \, \text{cm} \) - Perimeter: \( P = 4s = 4 \times 12 = 48 \, \text{cm} \) 4. **Area = 60 in²** - For a square, \( A = s^2 \) - \( s^2 = 60 \) - \( s = \sqrt{60} \approx 7.75 \) in - Dimensions: \( 7.75 \, \text{in} \times 7.75 \, \text{in} \) - Perimeter: \( P = 4s = 4 \times 7.75 \approx 31 \, \text{in} \) ### Summary of Results: - **a)** Dimensions: \( 4.47 \, \text{ft} \times 4.47 \, \text{ft} \), Perimeter: \( 17.88 \, \text{ft} \) - **b)** Dimensions: \( 9 \, \text{m} \times 9 \, \text{m} \), Perimeter: \( 36 \, \text{m} \) - **c)** Dimensions: \( 12 \, \text{cm} \times 12 \, \text{cm} \), Perimeter: \( 48 \, \text{cm} \) - **d)** Dimensions: \( 7.75 \, \text{in} \times 7.75 \, \text{in} \), Perimeter: \( 31 \, \text{in} \)