Question

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?

Answer 1

\[\sqrt{125}\]

Answer 2

No.

Answer 3

Is it a flying ant?

Answer 4

No, no :)

Answer 5

4 + sqrt(61)

Answer 6

sqr 74

Answer 7

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

Answer 8

No......

Answer 9

Read the question carefully Parth.

Answer 10

sqr 117

Answer 11

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

Answer 12

hooray

Answer 13

15

Answer 14

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

Answer 15

Now where am I wrong?

Answer 16

\[\sqrt{x^2 + 5^2} + \sqrt{(6-x)^2 + 4^2}\]
By the pythagorean theorem, the sum of those two diagonals is:
To minimize this distance, derive and set equal to 0.