FoolForMath
Another super easy problem,
If an ant wants to crawl over the rectangular block of dimensions \( 6\times5\times4 \) from one vertex to a diagonally opposite vertex, what is the shortest distance it would need to travel?
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binary3i
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\[\sqrt{125}\]
FoolForMath
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No.
rebeccaskell94
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Is it a flying ant?
FoolForMath
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No, no :)
apoorvk
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4 + sqrt(61)
ninhi5
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sqr 74
Callisto
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No......
FoolForMath
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Read the question carefully Parth.
ninhi5
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sqr 117
FoolForMath
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Bingo ninhi5! \(\sqrt{117} \) is the right answer.
ninhi5
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hooray
Arnab09
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15
apoorvk
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The ant needs to travel one side, and one diagonal.
We have three cases:
side '4' + diagonal of (6 and 5) = 4 + sqrt61= 11.something
side '6' + diagonal of (4 and 5) = 6 + sqrt41= 12.something
side '5' + diagonal of (4 and 6) = 5 + sqrt52 =12.something
hence shortest = 4 + sqrt61
apoorvk
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Now where am I wrong?
SmoothMath
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|dw:1338100316725:dw|\[\sqrt{x^2 + 5^2} + \sqrt{(6-x)^2 + 4^2}\]
By the pythagorean theorem, the sum of those two diagonals is:
To minimize this distance, derive and set equal to 0.
ninhi5
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dont make it over complicated :)
apoorvk
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Oh damn! I completely forgot that !! :/ Damn my soul
FoolForMath
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There are only 3 possible roots, consistent to Smooth's diagram.
binary3i
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|dw:1338145571548:dw|
Ishaan94
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11.7? lol
Ishaan94
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Oh No :(
FoolForMath
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@ninhi5: Can you post the explanation?
binary3i
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\[\sqrt{117}\]
ninhi5
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i just use pythogorean theorem
ninhi5
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My new question guys
rebeccaskell94
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can antonia travel around the outside of the box?