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

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Manager
Joined: 02 Dec 2012
Posts: 177
A rectangular box is 10 inches wide, 10 inches long, and 5  [#permalink]

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26 Dec 2012, 06:23
4
15
00:00

Difficulty:

15% (low)

Question Stats:

77% (00:49) correct 23% (01:08) wrong based on 1015 sessions

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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) $$10\sqrt{3}$$
Math Expert
Joined: 02 Sep 2009
Posts: 52437
Re: A rectangular box is 10 inches wide, 10 inches long, and 5  [#permalink]

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26 Dec 2012, 06:26
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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) $$10\sqrt{3}$$

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

Similar question to practice: a-rectangular-box-has-dimensions-of-8-feet-8-feet-and-z-128483.html
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##### General Discussion
Manager
Joined: 12 Feb 2011
Posts: 83
Re: A rectangular box is 10 inches wide, 10 inches long, and 5  [#permalink]

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06 May 2014, 11:50
1
1
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.
Target Test Prep Representative
Affiliations: Target Test Prep
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Posts: 2830
Re: A rectangular box is 10 inches wide, 10 inches long, and 5  [#permalink]

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07 Jul 2016, 09:19
1
1
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) $$10\sqrt{3}$$

To solve this problem we must remember that given any rectangular solid, the longest line segment that can be drawn within the solid will be one that goes from a corner of the solid, through the center of the solid, to the opposite corner, or in other words, the space diagonal of the solid.

The space diagonal can be calculated using the extended Pythagorean theorem:

diagonal^2 = length^2 + width^2 + height^2

Using the values from the given information we have:

d^2 = 10^2 + 10^2 + 5^2

d^2 = 100 + 100 + 25

d^2 = 225

√d^2 = √225

d = 15
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A rectangular box is 10 inches wide, 10 inches long, and 5  [#permalink]

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16 Jul 2016, 23:51
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 greatest distance between any two points in a CUBE/CUBOID is given by the formula
$$d=\sqrt{l^2+b^2+h^2}$$

$$d=\sqrt{100+100+25}$$; $${given===> l=10; b=10; h=5}$$

$$d=\sqrt{225}= 15$$

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Re: A rectangular box is 10 inches wide, 10 inches long, and 5  [#permalink]

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01 Oct 2018, 01:26
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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 &nbs [#permalink] 01 Oct 2018, 01:26
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