0
$\begingroup$

A spider is sitting exactly in the middle of one of the smallest walls in a living room, whilst a fly is resting by the side of the window of the opposite wall, 1.5 m above the ground and o.5 m from the adjacent wall. The room is 5m long, 4m wide and 2.5m high. Work out the distance between spider and fly. (P.s: according to the answer sheet, the answer should be 5.23 m correct to 2dp, but i dont know how to get this answer)

The room and the location of the fly in the spider:

enter image description here

$\endgroup$
9
  • $\begingroup$ Weird thing (or not) that there is a window...and I guess the distance is meant going on walls/floor/ceiling, not in straight line, which seems to be reasonable easy, with straight triangles...? $\endgroup$
    – DonAntonio
    Apr 13, 2019 at 9:23
  • $\begingroup$ I think it must be air distance, as I get this is 5.226 mts.... $\endgroup$
    – DonAntonio
    Apr 13, 2019 at 9:30
  • $\begingroup$ @DonAntonio It certainly is about the path along the room's inner surface – a spider can't fly! :) $\endgroup$
    – CiaPan
    Apr 13, 2019 at 10:00
  • $\begingroup$ This is an old problem, presented with different numeric parameters (room's dimensions). The most clear way to solve it is unfolding the cuboid room into a flat net and considering a straight way (line segment) between the two arthropods. Be aware, however, that you can draw the net in several different ways, which cause the 'straight' way to go through different edges, so you eventually have to choose the globally shortest path from several locally shortest. $\endgroup$
    – CiaPan
    Apr 13, 2019 at 10:01
  • $\begingroup$ @CiaPan Who said the spider can fly? Who even said anything about the spider wanting to hunt the fly?! $\endgroup$
    – DonAntonio
    Apr 13, 2019 at 10:08

1 Answer 1

0
$\begingroup$

spider and fly$A$ is the fly, $B$ the spider. Let $AKLC$ be a rectangle parallel to the floor, and draw $CD$ vertically and $BD$ horizontally. Since $LC=AK=.5m$, and $B$ is at the wall's center, then $$BD=2-.5=1.5m$$ and $$CD=1.5-1.25=.25m$$Therefore by Pythagoras$$CB=\sqrt {1.5^2+.25^2}=1.52m$$Then drawing $AB$, since $\triangle ACB$ is right, again by Pythagoras$$AB=\sqrt{5^2+1.52^2}=5.23m$$the distance between spider and fly.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .