Fife Lake[1] is a hamlet located between Coronach and Rockglen within Rural Municipality of Poplar Valley No. 12 in south-central Saskatchewan, Canada near the border with the United States. Approximately 40 people inhabited the village of Fife Lake in 2006. It is about 11 kilometres (6.8 mi) east of Rockin Beach Park and Fife Lake.[2]
History
editPrior to January 27, 2005, Fife Lake was incorporated as a village, and was restructured as a hamlet under the jurisdiction of the RM of Poplar Valley on that date.[3]
Demographics
editIn the 2021 Census of Population conducted by Statistics Canada, Fife Lake had a population of 25 living in 17 of its 23 total private dwellings, a change of 0% from its 2016 population of 25. With a land area of 0.36 km2 (0.14 sq mi), it had a population density of 69.4/km2 (179.9/sq mi) in 2021.[4]
See also
editReferences
edit- ^ "Fife Lake". Canadian Geographical Names Database. Government of Canada. Retrieved January 7, 2023.
- ^ "Rockin Beach Park". Tourism Saskatchewan. Government of Saskatchewan. Retrieved January 7, 2023.
- ^ "Restructured Villages". Saskatchewan Ministry of Municipal Affairs. Archived from the original on March 25, 2008. Retrieved February 10, 2008.
- ^ "Population and dwelling counts: Canada and designated places". Statistics Canada. February 9, 2022. Retrieved August 31, 2022.
49°11′39″N 105°43′52″W / 49.19417°N 105.73111°W
 | This article about a location in the Census Division No. 3 of Saskatchewan is a stub. You can help Wikipedia by expanding it. |