(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 $$ A $$.

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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  
$$
62\,\mathrm{cm}^2.
$$
The area of the reservoir on Map $$A$$ is $$62\,\mathrm{cm}^{2}$$.

(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 \( A \).

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