Bunuel
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