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A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all

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A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all  [#permalink]

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New post 25 Feb 2015, 03:02
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A
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C
D
E

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Re: A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all  [#permalink]

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New post 25 Feb 2015, 07:42
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In a inscribed rectangle in a sphere we will have a line joining the opposite vertices as diameter.

with help of Pythagoras theorem 3, 4 give diagonal as 5>> with 5 and 12 we get 13 , 13 is the diameter of the sphere. answer> A
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Re: A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all  [#permalink]

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New post 25 Feb 2015, 08:19
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I think A is the correct answer, as d²=3²+4²+12² is the diagonal of the rectangular which is also the diameter of the sphere
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Re: A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all  [#permalink]

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New post 25 Feb 2015, 17:24
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Hi All,

When dealing with a rectangular solid, there is a formula for calculating the distance from one corner to "opposite opposite" corner (meaning "through the box"). This also happens to be the LONGEST straight-line distance from any point on the solid to any other point on the solid:

Longest Diagonal = SqRt(Length^2 + Width^2 + Height^2).

Here, we have dimensions of 3, 4 and 12. Placing each of those numbers in any of the dimensions gives us...

SqRt(3^2 + 4^2 + 12^2) =
SqRt(9 + 16 + 144) =
SqRt(169) =
13

This particular concept does not show up on the GMAT too often (you probably won't see it on Test Day). However, if you do, then this formula will come in handy.

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A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all  [#permalink]

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New post 25 Feb 2015, 19:37
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Bunuel wrote:
A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all eight of its vertices are on the sphere. What is the diameter of the sphere?

(A) 13
(B) 15
(C) 18
(D) 19
(E) 20


Kudos for a correct solution.


A rectangular solid inscribed in a sphere will have longest diagonal equal to diameter of the circle.

Hence the diameter = \(\sqrt{3^2 + 4^2 + 12 ^2}\)=13

Hence, A.
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Re: A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all  [#permalink]

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New post 02 Mar 2015, 05:26
Bunuel wrote:
A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all eight of its vertices are on the sphere. What is the diameter of the sphere?

(A) 13
(B) 15
(C) 18
(D) 19
(E) 20


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

First of all, what’s important to appreciate — and this is something that does appear with some frequency on the GMAT math section — the most famous formula in mathematics, the Pythagorean theorem, extends seamlessly to three-dimensions. If you have a solid with a length L, a width W, and a height H, then the “space diagonal”, the line form one vertex to the catty-corner opposite vertex, has a length R, which satisfies the equation: Y^2 = L^2 + W^2 +H^2.

Here, we can easily find the space diagonal of the 3 x 4 x 12 solid. We get
Y^2 = 3^2 + 4^2 + 12^2
Y^2 = 9 + 16 + 144 = 169
\(Y = \sqrt{169} = 13\)

So the space diagonal of the rectangular solid is 13. It may stretch your visualizing abilities a bit, but this space diagonal must be equal to the diameter of the sphere — the line from one vertex to the catty-corner opposite vertex must pass through the center of the sphere, and a line segment from one point on a sphere to another that passes through the center is, by definition, a diameter.

So diameter = space diagonal = Y = 13.

Answer = A
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Re: A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all  [#permalink]

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New post 02 Mar 2015, 05:30
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Re: A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all  [#permalink]

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Re: A rectangular solid, 3 x 4 x 12, is inscribed in a sphere, so that all   [#permalink] 25 Nov 2018, 05:59
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