(a) The actual area of a reservoir is \( 3.875 \mathrm{~km}^{2} \). The scale of Map \( A \) is \( 1: 25000 \). Find the area of the reservoir, in \( \mathrm{cm}^{2} \), on Map \( A \).

Let the actual area of the reservoir be \[ A_{\text{actual}} = 3.875\,\mathrm{km}^2. \] The map has a scale of \[ 1:25000, \] which means that a length on the map corresponds to a real length that is \(25000\) times as long. Since the area scales as the square of the linear scale factor, the area scaling factor is \[ \left(\frac{1}{25000}\right)^2. \] Before applying the area scaling factor, we need to convert the actual area from \(\mathrm{km}^2\) to \(\mathrm{cm}^2\). We know that \[ 1\,\mathrm{km} = 100000\,\mathrm{cm}, \] so \[ 1\,\mathrm{km}^2 = \left(100000\,\mathrm{cm}\right)^2 = 10^{10}\,\mathrm{cm}^2. \] Thus, the actual area in \(\mathrm{cm}^2\) is \[ A_{\text{actual}} = 3.875 \times 10^{10}\,\mathrm{cm}^2. \] Now, applying the area scale factor, the area on the map is \[ A_{\text{map}} = A_{\text{actual}} \times \left(\frac{1}{25000}\right)^2. \] First, compute the square of the scale denominator: \[ 25000^2 = 6.25 \times 10^8. \] Then, \[ A_{\text{map}} = \frac{3.875 \times 10^{10}}{6.25 \times 10^8}. \] Simplify the division: \[ A_{\text{map}} = \left(\frac{3.875}{6.25}\right) \times 10^{10-8} = \left(\frac{3.875}{6.25}\right) \times 10^{2}. \] Calculating \(\frac{3.875}{6.25}\): \[ \frac{3.875}{6.25} = 0.62. \] Thus, \[ A_{\text{map}} = 0.62 \times 10^{2} = 62\,\mathrm{cm}^2. \] So, the area of the reservoir on Map \(A\) is \[ 62\,\mathrm{cm}^2. \]