Sewall's Point, Florida

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Sewall's Point is a town located on the peninsula of the same name in Martin County, Florida, United States. The population was 1,946 at the 2000 census. As of 2004, the population recorded by the U.S. Census Bureau is 2,053 [1]. Both the town and the peninsula are named for Capt. Henry Edwin Sewall (August 22, 1848-August 1, 1925).[3]

Contents

Geography

Sewall's Point is located at 27°11′42″N 80°11′52″W / 27.195°N 80.19778°W / 27.195; -80.19778 (27.195057, -80.197731).[4] It is bordered by water on the south, east, and west due to its peninsula shape of the entire town. On the south and est] is the St. Lucie River and to the east is the Indian River Lagoon. On the north it is bordered by unincorporated Jensen Beach

According to the United States Census Bureau, the town has a total area of 4.1 square miles (10.7 km²), of which, 1.2 square miles (3.2 km²) of it is land and 2.9 square miles (7.5 km²) of it (70.05%) is water.

Demographics

As of the census[1] of 2000, there were 1,946 people, 758 households, and 607 families residing in the town. The population density was 1,578.1 people per square mile (610.9/km²). There were 828 housing units at an average density of 671.5/sq mi (259.9/km²). The racial makeup of the town was 98.72% White, 0.41% African American, 0.41% Asian, 0.10% from other races, and 0.36% from two or more races. Hispanic or Latino of any race were 1.39% of the population.

There were 758 households out of which 33.4% had children under the age of 18 living with them, 73.4% were married couples living together, 4.4% had a female householder with no husband present, and 19.8% were non-families. 15.3% of all households were made up of individuals and 6.5% had someone living alone who was 65 years of age or older. The average household size was 2.57 and the average family size was 2.85.

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