The question states: A cube has edge length a cm. What is the radius length, in terms of a of a sphere that just contains the cube?
I visualized it with a 2d diagram, where the radius spans from the center of the cube (and sphere) to the edge of the cube, where one side of the cube is the chord length (a cm).
I then took the right angled triangle, with a base length of a/2 cm (as it spanned from the center of the cube (and sphere) to the end of the cube) and the height as a/2 (as it is equal to half the chord length, with is a).
Then ,using Pythagoras' theorem to find the radius (and the hypotenuse in this triangle), I calculated that
c² = (a/2)² + (a/2)² = a²/4 + a²/4
c² = 2a²/4
c² =a²/2
c = ± a/√2
l > 0
c = a/√2
I then rationalized the denominator to get an answer of (a√2)/2
The book says it is (a√3)/2, but I don't understand how my answer is wrong. Could someone please tell me what I did wrong, or if I'm right... Thanks.
+130514 I forgot to add how I arrived at that formula......
Draw a diagonal across the bottom of the cube.....this will be the hypotenuse of a right triangle whose legs are both 'a.'
Then, the length of this will be a√2
Now.....another right triangle is formed by this diagonal and the edge of the cube extending from one end of the diagonal to a top vertex. These are the two legs. The hypotenuse is the distance from the bottom vertex on the other end of this diagonal to this top vertex. This is what we're looking for. It is given by √[a^2 + ( a√2)^2] = √[a^2 + 2a^2] = √[3a^2] = a√3 .......!!!!
+130514
Notice that the center of the cube and the center of the sphere will coincide. Now....let's forget the sphere for a moment.
Let us take some cube of edge length, a.
A diagonal drawn from a bottom vertex of the cube to an opposite top vertex will go through the center of the cube and have a length given by √[a^2 + a^2 + a^2] = √[3a^2] = a√3. And 1/2 of this length will be from the center of the cube to either the top vertex or the bottom vertex........whatever your preference. And this equals a√3/2.
But.....from the center of the cube to either vertex will be where that vertex touches the sphere. And since the center of the cube and the center of the sphere coincide, this distance will be = to the radius of the sphere. Thus.....the radius = a√3/2....
Does that make sense???
+130514 I forgot to add how I arrived at that formula......
Draw a diagonal across the bottom of the cube.....this will be the hypotenuse of a right triangle whose legs are both 'a.'
Then, the length of this will be a√2
Now.....another right triangle is formed by this diagonal and the edge of the cube extending from one end of the diagonal to a top vertex. These are the two legs. The hypotenuse is the distance from the bottom vertex on the other end of this diagonal to this top vertex. This is what we're looking for. It is given by √[a^2 + ( a√2)^2] = √[a^2 + 2a^2] = √[3a^2] = a√3 .......!!!!
