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Q. 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.
Read Solution (Total 2)
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- h=30; b=12; h=12;
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| " |
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| | 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
- 12 years agoHelpfull: Yes(10) No(2)
- very simple geometry, use pythagores theorem
distance b/w them=sq root(30^2+10^2) =31.622 - 9 years agoHelpfull: Yes(3) No(3)
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