### Solution: Traverse Area

#### Problem (2)

The owner of the parcel shown below hired a surveyor to determine the parcel area.

The surveyor was unable to run a simple loop traverse through the four property corners due to various obstructions. Instead, she created and measured a crossing-loop traverse (shown in blue). In addition to the parcel corners, additional points were needed to work the traverse around the obstructions.

Reducing the field measurements resulted in a 1/17,300 traverse precision.The traverse was adjusted by the Compass Rule and coordinates were computed.

Point |
North (ft) |
East (ft) |

A | 600.000 | 1,000.000 |

B | 569.612 | 1,293.788 |

C | 254.542 | 1,209.288 |

D | 463.631 | 766.846 |

E | 470.303 | 397.969 |

F | 97.543 | 566.876 |

What is the area and its expected error?

#### Answer

Arrange the parcel corner coordinates in order going around the parcel. Repeat the first corner at the end of the list. Use **only** points which which are parcel corners.

Cross-multiply and sum.

This solution starts at point A and travels clockwise around the parcel.

Point |
North (ft) |
East (ft) |
||

A | 600.000 | 1,000.000 | 254,542 | |

C | 254.542 | 1,209.288 | 117,960 | 725,573 |

F | 97.543 | 566.876 | 266,603 | 144,294 |

E | 470.303 | 397.969 | 238,781 | 38,819 |

A | 600.000 | 1,000.000 | 470,303 | |

sums: |
877,890 | 1,378,989 |

If you use sig fig and assume an additional digit was carried computing coordinates, then area should be statewd to 4-5 sig fig. If we use 5 sig fig, area would be 250,550 ft^{2} and the error would be a compatible ±20 ft^{2}.

**Area = 250,550 ft ^{2} ±20 ft^{2}**