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:
On
$A$ 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.