A cylinder, which has a diameter of 27 and a height of 30, contains two lead spheres with radii 6 and 9, with the large sphere sitting on the bottom of the cylinder. Water is poured into the cylinder so that it just manages to submerge both the spheres. The volume of the water required is?
I'm happy to help.
First of all, this should not be posted in this particular forum, the "Ask GMAT Experts" forum. This space is for general, non-subject questions about strategy, logistics, etc. A math question should be posted in the math subject forum. This question, though, it not at all a GMAT question. It's well beyond what the GMAT would ask --- a very challenging geometry question. I don't know where you found it, but this is not something you need to know.
Nevertheless, because I love math, here's a solution:
two spheres in a cylinder.JPG [ 37.86 KiB | Viewed 218 times ]
Just by the symmetry of the situation, I know that the largest circle on each sphere will be tangent and coplanar. In the diagram, all the points shown (A, B, C, D, E, F, and G) all must be coplanar. That allows us to construct this diagram.
AC = BC = 9 cm, the radius of the larger sphere
EG = EF = 6 cm, the radius of the smaller sphere
CE = 15 = the line between the two centers of the sphere. I have extended CD and DE. Originally, we know neither of those red lengths.
If we add all the horizontal lengths, AC + CD + EF, that should equal 27 cm.
9 + CD + 6 = 27
CD = 12
Then, in right triangle CDE, we can use the Pythagorean Theorem. We have hypotenuse CE = 15, and leg CD = 12, so this is just the (3, 4, 5) triangle multiplied by 3 ---- the remaining side, DE = 3*3 = 9.
Now, add vertical lengths ---- BC = 9, ED = 9 and EG = 6, so point G is 24 cm above the base of the cylinder.
So, the volume = (pi)(r^2)*h = (pi)(13.5^2)*24 = 4374pi
I don't know how any normal person would get that final number without a calculator --- that's another incredibly un-GMAT-like aspect to this question. Nevertheless, that's the answer.
Does all this make sense?
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