How far does the spider crawl?

Question

An unfurnished room has a rectangular floor 26 feet long and 12 feet wide with walls 7ft 6in high, and a spider has to crawl along surfaces from one corner on the floor of the room to the diagonally opposite corner on the ceiling on the other side of the room. What is the length of its shortest route?

Answer 1

If spider gotta kraw, him go kros diagonal av floor...12 ft bi 26 ft

dist=sqrt(12^2+26^2)=sqrt(144+676)=sqrt(820)=28.6356421265527

Then add hite av wall...7.5 ft...dist=36.135642126552696

more or less

Comments

The shortest distance is 32ft 6in. The spider has a choice of routes: here's one:
It crawls across the floor to a point 16 feet along the adjacent longer side (20 feet), then it crawls up the wall from that point to the corner (12ft 6in). Geometrical solution consists of opening out the room, folding the sides and ceiling flat and drawing a straight line between the point on the floor and the point on the ceiling. Your solution is a crooked path.

Answer 2

Given rectangular floor 26 feet long

And 12 feet wide with walls 7ft 6in high,

So dist = √(12^2+26^2)= 28.6356

Length of its shortest route= 28.6356+7.6 = 36.13

Geometry Help
 

Comments

Thanks for your answer, but the shortest distance is 32.5ft.
Convert the 3-D room to 2-D by folding the long wall down so that it's level with the floor, making a rectangle 19.5' by 26'. The shortest distance is a straight line running along the diagonal across the floor meeting the long edge and running from this point to the corner (20'+12.5').
The length of the diagonal is sqrt((12+7.5)^2+26^2)=32.5ft. The same result is reached if the spider climbs at an angle across the long wall first and then crawls across the ceiling to the corner. But the distance is longer if the spider crawls across the short wall and then crosses the ceiling (sqrt((26+7.5)^2+12^2=35.58ft), because this is the diagonal of a different rectangle.
Crossing the floor diagonally then climbing to the corner (or climbing the wall vertically and crossing the ceiling diagonally) is in fact a crooked path, as can be seen in the 2-D model.