Solution to the Problem:
Here is the solution:
The shortest route is: A, B, D, G, F, E, C.
CFDB must be a rectangle because of the last three clues listed.
Hence, CF is 5 miles long and parallel to BD.
Since CE is 3 miles and EF is 2 miles, E must be a point on line segment CF.
CG is 7 miles, CE is 3 miles, and EG is 4 miles, so CEFG is a straight line.
G must be 2 miles southeast of corner F of the rectangle.
AD is 7 miles, AB is 2 miles, and BD is 5 miles, so ABD is a straight line, and A
is 2 miles northwest of corner B of the rectangle.
So, the shortest route that travels through every stop must be A, B, D, G, F, E, C,
or the reverse A, C, E, F, G, D, B.
But since the bus stops at B before C, the correct answer is A, B, D, G, F, E, C.
Dr. Kishan gets extra credit for sending in the shortest distance, given by
d=2+5+2√5+2+2+3+2√5=(14+4√5) units.