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23 Aug 2018, 04:16
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
74% (02:07) correct
26%
(02:16)
wrong
based on 50
sessions
History
Date
Time
Result
Not Attempted Yet
Herman is trying to box a gift for his cousin Edward. The gift included a rod that is 26 inches long and the box Herman is using measures 8 inches long by 6 inches wide. If the rod fits exactly in the box when one end of the rod is placed in the lower right corner and the other end is placed in the upper left corner, how tall is the box?
A. 12
B. 13
C. 24
D. 27
E. 104
A. 12
B. 13
C. 24
D. 27
E. 104
23 Aug 2018, 04:40
Kudos
Bookmarks
Bunuel
The box in which Herman will fit the gift(which has a rod - 26 inches long) will be a cuboid.
We are given two dimensions of the cuboid - Length: 8 inches | Width: 6 inches. Since, the
maximum distance between any two points in a cuboid is the length of the diagonal, we can
find out the height of the cuboid using the formula of the diagonal in a cuboid
Formula used: Diagonal of cuboid = \(\sqrt{Length^2 + Width^2 + Height^2}\)
Substituting values, we get \(26 = \sqrt{8^2 + 6^2 + x^2}\) (Here, x - height of cuboid)
Squaring on both sides, we will get \(26^2 = (\sqrt{64 + 36 + x^2})^2\) -> \(x^2 = 576(676 - 100)\)
Therefore, the height of the box in which the rod fits exactly is \(\sqrt{576} = \sqrt{24^2}\) = 24 (Option C)
23 Aug 2018, 07:20
Kudos
Bookmarks
Bunuel
Since the rod directly fits into the box then the diagonal of the box = 26
Now, \(26^2 = 8^2 + 6^2 + width^2\)
Or, \(676 = 64 + 36 + width^2\)
Or, \(Width = \sqrt{676 - 100}\)
Or, \(Width = 24\), Answer must be (C)
