To find the distance between the library (L) and the community swimming pool (P), let's visualize the scenario based on the given information.
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Placement of Points:
- Let the courthouse (C) be at the origin (0, 0).
- The library (L) is directly south of the courthouse, so we can place it at (0, -24) because it is 24 kilometers south.
- The community swimming pool (P) is directly east of the courthouse to make calculations easier. The distance from the courthouse to the swimming pool is 30 kilometers, which means the swimming pool can be placed at (30, 0).
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Distance Calculation: We now need to calculate the distance between the library (0, -24) and the community swimming pool (30, 0).
We can use the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] where \((x_1, y_1)\) are the coordinates of the library and \((x_2, y_2)\) are the coordinates of the swimming pool.
Here: \[ (x_1, y_1) = (0, -24) \] \[ (x_2, y_2) = (30, 0) \]
Plugging these values into the distance formula, we get: \[ d = \sqrt{(30 - 0)^2 + (0 - (-24))^2} \] \[ d = \sqrt{30^2 + (0 + 24)^2} \] \[ d = \sqrt{30^2 + 24^2} \] \[ d = \sqrt{900 + 576} \] \[ d = \sqrt{1476} \]
We can simplify \(\sqrt{1476}\): \[ \sqrt{1476} = \sqrt{4 \times 369} = 2\sqrt{369} \]
Finding the approximate value of \(\sqrt{369}\): \[ \sqrt{369} \approx 19.2 \implies d \approx 2 \times 19.2 = 38.4 \]
Thus, the distance between the library and the community swimming pool is approximately \(38.4\) kilometers.