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# A rectangular box is 10 inches wide, 10 inches long, and 5

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A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]  26 Dec 2012, 06:23
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A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A) 15
(B) 20
(C) 25
(D) $$10\sqrt{2}$$
(E) $$10\sqrt{3}$$
[Reveal] Spoiler: OA
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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]  26 Dec 2012, 06:26
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Walkabout wrote:
A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box?

(A) 15
(B) 20
(C) 25
(D) $$10\sqrt{2}$$
(E) $$10\sqrt{3}$$

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: $$D=\sqrt{10^2+10^2+5^2}=15$$.

Answer: A.

Similar question to practice: a-rectangular-box-has-dimensions-of-8-feet-8-feet-and-z-128483.html
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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]  06 May 2014, 11:50
One can also apply the 3-D or Deluxe Pythagorean theorem directly, which is D^2 = L^2 + W^2 + H^2, to get the value directly.
Re: A rectangular box is 10 inches wide, 10 inches long, and 5   [#permalink] 06 May 2014, 11:50
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# A rectangular box is 10 inches wide, 10 inches long, and 5

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