Edthesock wrote:A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
The question is correct, and I don't remember what the available answers are because I did this question a while ago, but I will show how to get the solution:
If the Cube has a sphere inscribed in it, and the cube has an edge of 10 units, then the radius of the sphere is 5.
that being said, if we draw an imaginary line from the center of the sphere towards a cube's vertex, doesn't matter which one obviously, then the portion of that line inside the sphere is 5. The total line, as in from the center of the sphere to the vertex, minus 5 is the answer. So how to find the full length of that imaginary line?
Imagine you cut the cube with the sphere inscribed in such a way that we are left with a smaller cube that has an edge of 5 instead of 10. One vertex of this smaller cube will be the center of the sphere inscribed in the larger cube, and opposite that vertex in the smaller cube is the larger cube's vertex that we were drawing an imaginary line to. In other words, the diagonal of this smaller 5 x 5 x 5 cube is the imaginary line discussed in the last paragraph.
Remember, imaginary line - 5, (where 5 is the sphere radius), will be our answer. We know our new smaller cube is 5 x 5 x 5. The diagonal of a surface of this cube, using standard Pythagorean theorem is:
root[5^2 + 5^2] = root[50]
Getting back to the imaginary line from sphere's center to the cube's vertex... draw that line as the hypotenuse of a triangle that uses one of the bottom edges of the small cube (which has a measure of 5 units) and the other is our newly formed small cube surface diagonal line that measures 5 x root[2].
the imaginary line's length is:
= root [(root[50])^2 + 5^2] = root[50 + 25] = root[75] = root [25 x 3]
and finally, = 5 x root[3]
that is the length of the full imaginary line, but if you recall, we need to remove 5 from this, since that is the portion of the line inscribed inside the sphere.
Therefore,
5 x root[3] - 5 = 5(root[3] -1)
is the shortest distance from the cubes' vertex to the outer shell of the sphere.
this was a tough question for me to describe, sorry if it's not entirely comprehensible....
Good luck all!
you are absolutely right Edthesock...
I calculated wrongly in the beginning as 5(sqrt(2)-1)
but that is the distane between center the square (one of the faces of the cube) to the vertex...
To calculate the distance from center of the sphere to the vertex, we can use pythagoras theorem, where
Hypotenuse is the line from center to vertex (say X)
(1/2) times diagonal of the face of the cube (square with side 10, so this is sqrt(50) )
radius of the sphere (5)
we get
X^2 = (sqrt(50))^2 + 5^2
X^2 = 50+25
x = 5 sqrt (3)
now we got the distance from center to vertex.
subtracting 5 (radius) from this distance will give the distance between vertex and surface of the sphere.
ie 5(sqrt (3) - 1)
I tried to draw this, but it is difficult. if anyone couldn understand please let me know i will draw this and upload.
