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Calculating Area on Topographic Maps

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Area of a Square Formula Area = Length x Width

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Calculating the area of an even boundary A map with a scale of 1:50 000. The length of this map is 5cm and the width is 3cm. Calculate the area represented by the map. Convert the length and width into what they represent in reality before using the formula*. Scale 1:50 000 - therefore each cm on the map represents 0.5 km in reality Length: 5cm x 0.5 km = 2.5km Width: 3cm x 0.5 km = 1.5 km Area = Length x Width = 2.5km x 1.5km = 3.75 sq. km

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Calculating the area of an uneven boundary Calculate the area of Klinger Lake.

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How to do it… Step 1: Work out the area of one grid square on the map. Scale of this map: 1:50 000, therefore each cm represents 0.5km Length: 2cm x 0.5km = 1.0km Width: 2cm X 0.5 km = 1.0km Area of One Grid Square = 1.0km x 1.0km = 1.0 sq. km Step 2: Determine the number of squares that the object occupies. Klinger Lake occupies a total of about 0.5 of a grid square. Step 3: Multiply the number of grid squares that the object occupies by the area of one grid square. Hence, 0.5 of a square x 1.0 sq. km = 0.5 sq. km which is the area of Klinger Lake.

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Another way to do it… Calculate the area of the built up portion of the city (i.e. the pink shaded area).

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Working with Percentages… Step 1: Work out the area of one grid square on the map. It is the same as the previous map: 1.0 sq.km. Step 2: Determine the number of squares that the object occupies. In this situation, one could go from grid square to grid square determing how much the built up area occupies each square. i.e. 0.1 of a grid square (5866) + 0.4 (5966) + 0.7 (5967) + 0.7 (6067) + 0.03 (5868) + 0.5 (5968) + 1.0 (6068) + 0.25 (6168) + 0.2 (5869) + 0.6 (5969) + 1.0 (6069) + 0.85 (6169) + 0.2 (6269) + 0.8 (6070) + 0.85 (6170) + 0.2 (6270) + 0.2 (6071) + 0.4 (6171) = 8.98 squares Step 3: Multiply the number of grid squares that the object occupies by the area of one grid square. Hence, 8.98 squares x 1.0 sq.km = 8.98 sq. km which is the area of downtown St. John's.

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