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Lakes / South Dakota / Walworth County / Horseshoe Lake

Horseshoe Lake, Walworth County, South Dakota

66.22 acres
Surface Area

Where It Sits, and How Deep It Goes

Locator - South Dakota

Sioux FallsPierreHorseshoe LakeWalworth Co.

45.55 degrees N, 99.72 degrees W - Walworth County. State and county outlines drawn from Census boundary data; the lake marker is placed by its USGS National Hydrography Dataset coordinates, not by eye. Schematic - not a survey map.

How Horseshoe Lake Stacks Up

Lakes Nearby

LakeCountyDistanceSurface Area
Bowdle-Hosmer LakeEdmunds3.2 mi129.73 acres
East Flat LakeCampbell8.3 mi151.72 acres
Heupel LakeMcPherson8.5 mi45.71 acres
Flat LakeCampbell9.7 mi113.17 acres
Spring LakeWalworth9.7 mi677.81 acres
Feinstein LakeMcPherson10 mi152.46 acres

Straight-line distance between lake centers.

Facts and Figures

Waterbody ID147898020
CountyWalworth
Coordinates45.5478, -99.7166

Questions About Horseshoe Lake

How Big Is Horseshoe Lake?

Horseshoe Lake covers 66.22 acres in Walworth County, South Dakota.

What Is the Largest Lake in Walworth County?

Swan Lake is the largest lake in Walworth County, at 2,467.35 acres.

Largest Lakes in South Dakota

#LakeCountySurface AreaMax Depth
1Lake OaheDewey309,912.57 acres-
2Lake Francis CaseCharles Mix82,166.92 acres-
3Lake SharpeHughes58,419.67 acres-
4Lewis and Clark LakeBon Homme25,141.99 acres-
5Lake ThompsonKingsbury13,887.93 acres-

All 533 South Dakota lakes ranked by surface area

Sources: USGS National Hydrography Dataset (The National Map) lake data (ID 147898020). Verified October 30, 2026.