Pregunta
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(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 \( A \).

Ask by Stewart Gibbs. in Singapore
Mar 16,2025

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

Solución

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 \[ 62\,\mathrm{cm}^2. \]

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To find the area of the reservoir on Map \( A \), we start with the actual area of the reservoir, which is \( 3.875 \, \text{km}^2 \). To convert this area from square kilometers to square centimeters, we use the conversion factor that \( 1 \, \text{km}^2 = 10^8 \, \text{cm}^2 \). So, \[ 3.875 \, \text{km}^2 = 3.875 \times 10^8 \, \text{cm}^2 = 387500000 \, \text{cm}^2. \] Next, we apply the scale of the map, which is \( 1:25000 \). This means that \( 1 \, \text{cm} \) on the map represents \( 25000 \, \text{cm} \) in reality. To find the area on the map, we square the scale factor: \[ \left(\frac{1}{25000}\right)^2 = \frac{1}{625000000} \, \text{cm}^2/\text{cm}^2. \] Now, we multiply the actual area in square centimeters by the scale factor: \[ \text{Area on Map A} = 387500000 \, \text{cm}^2 \times \frac{1}{625000000} = \frac{387500000}{625000000} \, \text{cm}^2. \] Calculating this gives: \[ \text{Area on Map A} = 0.620 \, \text{cm}^2. \] Therefore, the area of the reservoir on Map \( A \) is approximately \( 0.620 \, \text{cm}^2 \).

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