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