A sphere is inscribed in a cube with an edge of 10. What is

Wicked Question and great simple explanation Bunuel!

Also if it were not sphere and just a two dimensional square, the shortest distance would be S/2(\sqrt{2} - 1) = 10/2(\sqrt{2} - 1) = 5(\sqrt{2} - 1)

Bunuel.. i cant understand the question ? can u elaborate it further..


And ya instead of sphere if it wud be a square inscribed in a cube then wat wud b the answer?

i dint understood why diameter is 10?
how smallest area is 1/2(diameter-diagonal)
question was dreaded for me

bunuel help

Consider the cross-section as shown below:The diameter = The edge.

As for your second question, check here: a-sphere-is-inscribed-in-a-cube-with-an-edge-of-10-what-is-127461.html#p1097531 The shortest distance from one of the vertices of the cube to the surface of the sphere is 1/2(diagonal of the cube - diameter of the circle). Diagonal of the cube - diameter of the circle, is the length of two little black arrows shown here:


Hope it's clear.

hi bunuel
thanks for an explanation.
why diagonal-diameter divided by 2

Diagonal - diameter is the length of two little black arrows we need one...

thanks bunuel for patiently providin g solution
you rock!!
too much and too little study is fatal thats what happening to me i have missed such a small stuff.

Posted from my mobile device

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Hope it helps.

If you visualize the problem in your head, you realize that what you want is 1/2 the diagonal of the cube- the radius of the circle.

We know the radius of the circle is 5, because the circle touches the sided of the cube, which has a total length of 10.

If you memorized the diagonal of a cube, which I found helpful to do for my test, then you would know it is side*(sqrt(3)), but we want half of that so it is 5(sqrt(3)).

So, the distance of the vertice to the sphere is 5(sqrt(3))-5. That is not an answer, but we can see that D is the same thing, it just divided out the 5.

I did not get how the diameter is 100 in this step
distance: x=Diagonal−Diameter2=10∗√3−102=5(√3−1)

Could you plz explain?

Hi SonaliT,

Since the sphere is inscribed in the cube, it's diameter is the SAME as the edge of the cube. They are BOTH 10 (not 100).

The calculation that you referred to should be written as....

10(Root 3) - 10 = 10(Root 3 -1)

This calculation is TWICE the length that we're looking for (one on both "sides" of the sphere). Since the question asks for the shortest distance from any of the vertices on the cube to the sphere, we have to divide this entire calculation by 2....

10(Root 3 - 1)/2 = 5(Root 3 - 1)

GMAT assassins aren't born, they're made,
Rich

Here is the official explanation.

Answer: Option D (Just replace edge length as 10 instead of 20)

Check solution as attached

Do we know that sphere is touching the cube from inside?

Hi lakshya14,

The prompt tells us that the sphere is 'inscribed' in the cube, which means that the sphere is inside the cube and touching all 6 sides of the cube.

GMAT assassins aren't born, they're made,
Rich

Keep in mind that the edge of the cube is equal to the diameter of the sphere inscribed in said cube.

In other words, 10 = edge of the cube = diameter of the sphere.

If we draw the diagonal of the cube, we have that:

Diagonal of the cube = sphere diameter + two equal distances (from the edge to the sphere x 2)

Thus we can express:

Cube diagonal = x + sphere diameter + x

Where x is the distance from the sphere to one of the edges.

And x is the shortest distance between an edge and the sphere, which is what the exercise asks for.

Let us remember that the diagonal of a cube joins two opposite edges of said cube.

So we have to:

Diagonal of the cube = x + 10 +x

Now we must establish the value of the diagonal of a cube of edge 10.

The diagonal of a cube can be constructed:

Let us consider an edge of the cube, if we take the base of the cube, a square of side 20, and obtain its diagonal, we will have that the edge of the cube and the diagonal of the base of the cube form an angle of 90 degrees, and if we draw the diagonal of the cube, to form a triangle with the edge and the diagonal of the base, we will have a right triangle, with leg 1 edge of the cube, leg 2 diagonal of the base of the cube (square of side 10) and the diagonal hypotenuse of the cube.

The diagonal of a square of side 10 is 10√2 (applying Pythagoras it is an isosceles right triangle, whose equal sides are 10, half of a square).

So we have a new rectangle:

Leg 1 (cube edge) = 10
Leg 2 (diagonal of the base of the cube) = 10√2
Hypotenuse = diagonal of the cube.

So applying Pythagoras we have:

10exp2 + (10√2)exp2 = (diagonal cube)exp2

Solving we have:

10exp2 + 10exp2 x 2 = (diagonal cube)exp2

3 x 10exp2 = (diagonal cube)exp2

If we apply square root to both sides, we have:

10√3 = diagonal cube

Now if you know the direct relationship of a cube with edge a, where the diagonal of one of its faces is a√2 and the diagonal of the cube is a√3. It allows you to work much faster. I suggest you, learn the above relationship.

Going back to the situation:

cube diagonal = x + cube edge + x

where x is the requested distance.

10√3 = 2x + 10

Then x = (10√3 -10)/2

x = 5√3 -5

x = 5(√3 -1)

Answer D