The length of a rectangle, \( I \mathrm{~cm} \), is 46.2 cm rounded to 1 decimal place. The breadth of the rectangle, b cm, is 21.2 cm rounded to 1 decimal place. a Find the range of possible values of \( I \) and \( b \). b Find the smallest possible area.

**a)** Since the values are given rounded to 1 decimal place, the true value lies within 0.05 below the rounded value up to (but not including) 0.05 above it. Thus, - For the length \( I \), which is 46.2 cm rounded to 1 decimal place, we have \[ 46.15 \le I < 46.25 \] - For the breadth \( b \), which is 21.2 cm rounded to 1 decimal place, we have \[ 21.15 \le b < 21.25 \] **b)** The area of the rectangle is given by \[ A = I \times b. \] To obtain the smallest possible area, we take the smallest possible values of \( I \) and \( b \) in their respective ranges. That is, \[ I_{\text{min}} = 46.15 \quad \text{and} \quad b_{\text{min}} = 21.15. \] Thus, the smallest area \( A_{\text{min}} \) is \[ A_{\text{min}} = 46.15 \times 21.15. \] Calculating this product: \[ 46.15 \times 21.15 = \frac{4615 \times 2115}{10000}. \] Compute the numerator: \[ 4615 \times 2115 = 4615 \times (2000 + 115) = (4615 \times 2000) + (4615 \times 115). \] \[ 4615 \times 2000 = 9\,230\,000, \] \[ 4615 \times 115 = 4615 \times 100 + 4615 \times 15 = 461\,500 + 69\,225 = 530\,725. \] Thus, \[ 4615 \times 2115 = 9\,230\,000 + 530\,725 = 9\,760\,725. \] So, \[ A_{\text{min}} = \frac{9\,760\,725}{10\,000} = 976.0725. \] Therefore, the smallest possible area is approximately \[ 976.07 \text{ cm}^2. \]