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02 Mar 2012, 15:53
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
81% (01:09) correct
19%
(01:27)
wrong
based on 501
sessions
History
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A rectangular box has dimensions of 8 feet, 8 feet, and z feet. In terms of z, what is the greatest possible (straight-line) distance, in feet, between any two points on the box?
(A) 8 + z
(B) \(8\sqrt{2} + z\)
(C) \(8z\sqrt{2}\)
(D) \(\sqrt{64 + z^2}\)
(E) \(\sqrt{128 + z^2}\)
(A) 8 + z
(B) \(8\sqrt{2} + z\)
(C) \(8z\sqrt{2}\)
(D) \(\sqrt{64 + z^2}\)
(E) \(\sqrt{128 + z^2}\)
Guy - I know that this will be solved using the right angle triangle, but struggling. For me the longest distance will be the diagonal of the box, but I don't get an answer "E". Can you please help?
02 Mar 2012, 21:50
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enigma123
You are right the longest distance will be the diagonal of a rectangular box. Look at the diagram below:
Square of the diagonal of the face (base) is \(d^2=a^2+b^2\) and the square of the diagonal of a rectangular box is \(D^2=d^2+c^2=(a^2+b^2)+c^2\) --> \(D=\sqrt{a^2+b^2+c^2}\).
Applying this to our question, we get: \(D=\sqrt{8^2+8^2+z^2}=\sqrt{128+z^2}\).
Answer: E.
Hope it's clear.
Attachment:
Diagonal.jpg [ 17.98 KiB | Viewed 38624 times ]
General Discussion
02 Mar 2012, 20:20
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enigma123
The longest distance is the diagonal of the box. Formula for the diagonal of a rectangular is \(\sqrt{a^2 + b^2 + c^2}\).
Just use the information from the text and insert it in the formula:
\(\sqrt{8^2 + 8^2 + z^2} = \sqrt{64 + 64 + z^2} = \sqrt{128 + z^2}\)
E, correct.
