A room is 30 X 12 X 12. a spider is ont the middle of the samller wall, 1 feet from the top, and a fly is ont he middle of the opposite wall 1 feet from the bottom. what is the min distance reqd for the spider to crawl to the fly.

2 Answers on this Question

1. length is 30 breadth is 12 and height is 12
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   1
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   1 .
   1 .
   1 10 .
   1
   1 .
   1
   1
   1 30 .
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   1
   consider right angled triangle
   so the distance= square root of (30*30+10*10)

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2. h=30; b=12; h=12;
   …………..
   | ” |
   | |
   | | 12
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   | ” |
   |…………..|
   30
   Apply Geometry to get the minimum distance.
   Mid-point [A] of side wall is at 6.
   Spider [B] is at 5 from Mid-point of side wall (or) 11 from bottom (or) 1 form top.
   Center point [C] of box is 15 away form the mid-point of side wall.
   Then ABC is a right angled triangle.
   Sides AB=5; AC=15;
   Angle at A=90 degrees
   BC=((AB^2)+(AC^2))^1/2
   BC=15.811
   Similarly,
   Mid-point[D] of ooposite sidewall is at 6.
   Fly [E] is at 5 from Mid-point of opposite side wall (or) 11 from top (or) 1 form bottom.
   Center point [C] of box is 15 away form the mid-point of opposite side wall.
   Then DEC is a right angled triangle.
   Sides DE=5; DC=15;
   Angle at D=90 degrees
   CE=((DE^2)+(DC^2))^1/2
   CE=15.811
   Shortest distance from B(Spider) to E(Fly)
   BE = BC + CE
   BE = 31.622

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