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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.
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Answer Posted / hitesh
Answer is 40.
1. We need to "flatten" the cuboid room(or open it up as a plane sheet i.e. how all 6 faces of cuboid can be opened up).
2. The question is asked about the shortest distance spider
can take to reach Fly by "CRAWLING".
As Sarathchand answered the shortest distance as 31.622, that means the spider must have the ability to "fly" to reach that Fly which unfortunately not mentioned in the porblem.
3. So, if we flatten the cuboid and join those 2 points(say A (spider and B Fly), you can get the distance as 40 by using right angle triangle property(for triangle ABC for 'flattened' or(Opened Up) cuboid.
Sqrt(32*32+24*24)=40.
In which base AC will be = 32 i.e.(1+30+1) where 1 is dist. from top to spider, 30 is length and 1 is dist. from bottom to Fly.
And height BC will be = 24 i.e. (6+12+6) where 6 is centre distance(Mid point) and 12 is one side.
When we flatten the cuboid it'll look like this(can't post the picture, no option available. I tried my best, I hope u can understand.)
____________
|____________|__ 12
|____________|B_| 12 (mid point 6) 6+
__|____________|1 12 12+
|_A|____________|C 12 (mid point 6) 6= 24 (=BC)
1 30
1+ 30+ 1= 32 (=AC)
| Is This Answer Correct ? | 1 Yes | 0 No |
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modulo(373,7)+round(5.8)+truncat(7.2)-round(3.4) = ? i exactly don't remember the truncate function.but the function was where we skip the '.' part.
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