Horseshoe Bay, Texas

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Horseshoe Bay is a city in Burnet and Llano Counties in the U.S. state of Texas. Prior to its incorporation in September 2005[3], it was a census-designated place (CDP). The CDP population was 3,337 at the 2000 census.

Contents

Geography

Horseshoe Bay is located at 30°32′41″N 98°22′06″W / 30.544814°N 98.368428°W / 30.544814; -98.368428 (30.544814, -98.368428).[4] It is located in both Burnet County and Llano County on the southern shore of Lake Lyndon B. Johnson. This is about 23 miles southeast of Llano and 42 miles northwest of Austin.

According to the United States Census Bureau, the CDP had a total area of 26.7 square miles (69.1 km2), of which, 23.4 square miles (60.5 km2) of it is land and 3.3 square miles (8.6 km2) of it (12.48%) is water.

Demographics

As of the census[1] of 2000, there were 3,337 people, 1,623 households, and 1,186 families residing in the CDP. The population density was 142.8 people per square mile (55.1/km2). There were 2,773 housing units at an average density of 118.7/sq mi (45.8/km2). The racial makeup of the CDP was 96.22% White, 0.33% African American, 0.24% Native American, 0.78% Asian, 1.47% from other races, and 0.96% from two or more races. Hispanic or Latino of any race were 4.29% of the population.

There were 1,623 households out of which 11.6% had children under the age of 18 living with them, 68.5% were married couples living together, 2.8% had a female householder with no husband present, and 26.9% were non-families. 24.0% of all households were made up of individuals and 12.1% had someone living alone who was 65 years of age or older. The average household size was 2.06 and the average family size was 2.37.

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