Alexander Island, Florida, United States
About
Alexander Island is a island located in Volusia county Florida in the United States of America. It is at 28.8513798 latitude and -81.3522885 longitude. Alexander Island is at an elevation of 0 ft (0 m). The USGS map that contains this location is titled "Sanford".
Location Details
| Type of Place: | Island |
| Latitude: | 28.8513798 |
| Longitude: | -81.3522885 |
| Elevation: | 0 ft / 0 m |
| County: | Volusia |
| USGS Map Name: | Sanford |
