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)
a city block 500ft by 500ft is an unobstructed paved lot, except for a small office building 100ft by 100ft, centered in the middle of the block. What's the shortest distance from the SW corner to the NE corner, going through the paved lot and along (but not through) the building (to the nearest foot)?
2 answers
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
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
1 answer