a We are given a cube with edges that are 6 inches long. There is a sphere circumscribed about the cube.
We are asked to find the diagonal of the cube and the radius of the sphere. First, we will find the length of a diagonal of the square base. Let's analyze square ABCD.
Therefore, the diagonal of the base is d=6sqrt(2) inches. Now, let's analyze a cross section of the given cube and the sphere with a plane through D, B, H, and F.
Let's use the Pythagorean Theorem for â–ł BDF. Then, we will solve the equation for r.
Therefore, the radius of the sphere is r=3sqrt(3) inches. Since the diagonal of the cube is twice the radius, we get that it is equal to 2* 3sqrt(3)=6sqrt(3) inches.
b We are asked to find the volume of the space between the sphere and the cube to the nearest tenth.
It is equal to the difference of the volume of the cube and the sphere.
\begin{gathered}
V_\text{space}={\color{#FD9000}{V_\text{sphere}}}-{\color{#A800DD}{V_\text{cube}}}
\end{gathered}
Since the cube has edges 6 inches long, its volume is {\color{#A800DD}{V_\text{cube}}}=6^3={\color{#A800DD}{216}} cubic inches. From Part A we know that the radius of the sphere is r=6sqrt(3) inches. Now, let's use the formula for the volume of a sphere.
Now, let's find the volume of the space.
\begin{gathered}
V_\text{space}={\color{#FD9000}{V_\text{sphere}}}-{\color{#A800DD}{V_\text{cube}}} \\ \Downarrow \\
V_\text{space}={\color{#FD9000}{587.7}}-{\color{#A800DD}{216}}=371.6
\end{gathered}
Finally, we find that the volume of the space between the sphere and the cube is about 371.6 cubic inches.
