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A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
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26 Dec 2012, 07:23
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A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straightline) distance, in inches, between any two points on the box? (A) 15 (B) 20 (C) 25 (D) \(10\sqrt{2}\) (E) \(10\sqrt{3}\)
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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
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26 Dec 2012, 07:26




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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
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06 May 2014, 12:50
One can also apply the 3D or Deluxe Pythagorean theorem directly, which is D^2 = L^2 + W^2 + H^2, to get the value directly.



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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
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07 Jul 2016, 10:19
Walkabout wrote: A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straightline) 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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17 Jul 2016, 00:51
Walkabout wrote: A rectangular box is 10 inches wide, 10 inches long, and 5 inches high. What is the greatest possible (straightline) 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 \(d=\sqrt{l^2+b^2+h^2}\) \(d=\sqrt{100+100+25}\); \({given===> l=10; b=10; h=5}\) \(d=\sqrt{225}= 15\) Answer is A
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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
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22 Sep 2016, 08:33
Hi I wonder if there is a faster way to solve this problem?
I read that if you find yourself using the Pythagorean theorem you missed a shortcut. Naturally I thought about triangles involving multiples of the following sides: 3,4,5 5,12,13 and 8,15,17. However none of these would suffice. Any other shortcuts?



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Re: A rectangular box is 10 inches wide, 10 inches long, and 5 [#permalink]
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26 Feb 2018, 12:19
Dienekes wrote: One can also apply the 3D or Deluxe Pythagorean theorem directly, which is D^2 = L^2 + W^2 + H^2, to get the value directly. Deluxe ? are you talking about some high end Pythagorean theorem ? what does deluxe mean in this context ?




Re: A rectangular box is 10 inches wide, 10 inches long, and 5
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26 Feb 2018, 12:19






