The length and width of the floor of a room are 10 m. ?

Question

The length and width of the floor of a room are 10 m. The height of the room is 3 m. A spider, on the floor in a corner, sees a fly on the ceiling in the corner diagonally opposite. If the fly does not move, what is the shortest distance the spider can travel to reach the fly, to the nearest tenth of a metre?

Answer 1

If the spider walks diagonally across the room to the opposite corner, that would be √(10²+10²) = 14.142 meters. Up the wall is another 3 meters. The spider would have to travel 17.1 meters ( to the nearest tenth).

Now, if it was one of those flying spiders, Attus volans, it'd have to fly [ √(14.142²+3²) = √(208.996164 = ] 14.5 meters.

Answer 2

Stan S is right ---you have to apply Pythagoras Theorem twice to get the distance which is 14.45 m, say 14.5m rounded off.

Answer 3

Assumption: spider can't fly. The ability to string webs doesn't help, as it still has to crawl along the surface first.

There are two paths the spider can crawl, up the wall and diagonally across the ceiling or diagonally across the floor and up the wall. Both are the same distance.

Diagonally across is √(10² + 10²) = 14.14 meters

Up the wall is 3 meters

answer is 17.14 meters, or 17.1

.

Answer 4

10 root 3 metre is the shortest distance the spider has to travel!

Answer 5

Well, if the spider has to crawl along the wall I believe it would come out to about 17.4 meters.  Because the spider would have to crawl up the wall which is 3 meters, and then he would have to crawl diagonally across the ceiling (or floor).  You can find this distance using the Pythagorean Theorem.  If you are looking for the straight line distence between the two then you would have to figure out how long the line would be if you drew one between the fly and the spider.  Again, using the Pythagorean Theorem I believe it would come out to 10.4 meters.  Let me know if I'm right.

Answer 6

The spider should move in such a way that there is no wasted travel distance (ie. as directly towards the fly as possible).

It should proceed diagonally across the floor towards the fly's corner and then straight up the wall. The diagonal on the floor creates a 45-45-90 triangle, with the hypotenuse being the spiders path. The hypotenuse of a 45-45-90 triangle is sqrt(2) = 1.41421352 times the length of its side, so the path length along the floor is 10*1.41 = 14.1 m long. Finally the spider climbs the 3 m vertical distance to the fly.

Total path length = 14.1 + 3 = 17.1 meters