Jackson Township, Dallas County, Arkansas
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Jackson Township | |
|---|---|
  Jackson Township Location in Arkansas  Jackson Township Jackson Township (the United States) | |
| Coordinates: 33°50′48″N 92°31′59″W / 33.846733°N 92.532990°W[1] | |
| Country | United States |
| State | Arkansas |
| County | Dallas |
| Area | |
• Total | 35.844 sq mi (92.84 km2) |
| • Land | 35.839 sq mi (92.82 km2) |
| • Water | 0.005 sq mi (0.013 km2) |
| Elevation | 292 ft (89 m) |
| Population | |
• Total | 190 |
| • Density | 5.3/sq mi (2.0/km2) |
| Time zone | UTC-6 (CST) |
| • Summer (DST) | UTC-5 (CDT) |
| FIPS code[2][3] | 05-91860 |
| GNIS ID[2][3] | 66488 |
Jackson Township is a township in Dallas County, Arkansas, United States.[4] Its total population was 190 as of the 2010 United States census,[1][5] a decrease of 18.10 percent from 232 at the 2000 census.[6][7]
According to the 2010 Census, Jackson Township is located at 33°50′48″N 92°31′59″W / 33.846733°N 92.532990°W (33.846733, -92.532990).[1] It has a total area of 35.844 square miles (92.84 square kilometers); of which 35.839 sq mi (92.82 km2) is land and 0.005 sq mi (0.013 km2) is water (0.01%).[1] As per the USGS National Elevation Dataset, the elevation is 292 feet (89 m).[2]
