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December 18, 2014

December 18, 2014

Posted by **sarah** on Sunday, March 7, 2010 at 8:27pm.

- math -
**MathMate**, Sunday, March 7, 2010 at 8:44pmThe shortest distance from the SW corner of the lot to the NE corner of the lot is along two straight lines which pass through the SE or NW corner of the building.

If the path passes through the SE corner of the building, the x-displacement is 250+50=300 ft, and the y-displacement is 250-50=200 ft. Find the distance by Pythagoras theorem. Since by symmetry, the other leg has the same length, the total distance is 2√(200²+300²)

- math -
**Ms. Sue**, Sunday, March 7, 2010 at 8:45pmThe diagonal (c) across the lot:

500^2 + 500^2 = c^2

250,000 + 250,000 = 500,000^2

c = 707.11 feet

Less the diagonal through the building:

100^2 + 100^2 = c^2

10,000 + 10,000 = 20,000^2

c = 141.43 feet

707.11 - 141.42 = 565.69 feet

565.69 + 200 = 765.69 = 766 feet

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