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Mathematics
OpenStudy (anonymous):

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?

OpenStudy (binary3i):

\[\sqrt{125}\]

OpenStudy (anonymous):

No.

OpenStudy (anonymous):

Is it a flying ant?

OpenStudy (anonymous):

No, no :)

OpenStudy (apoorvk):

4 + sqrt(61)

OpenStudy (anonymous):

sqr 74

OpenStudy (anonymous):

No No, http://www.nooooooooooooooo.com/

OpenStudy (callisto):

No......

OpenStudy (anonymous):

Read the question carefully Parth.

OpenStudy (anonymous):

sqr 117

OpenStudy (anonymous):

Bingo ninhi5! \(\sqrt{117} \) is the right answer.

OpenStudy (anonymous):

hooray

OpenStudy (anonymous):

15

OpenStudy (apoorvk):

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

OpenStudy (apoorvk):

Now where am I wrong?

OpenStudy (anonymous):

|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.

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