The figure above is a cube with edges of length 9. Points C and D lie

The figure above is a cube with edges of length 9. Points C and D lie on diagonal AB such that points A, C, D, and B are equally spaced. As shown, a right circular cylindrical hole is cut out of the cube so that segment CD is a diameter of the top of the hole. What is the volume of the resulting figure?

Side length of the cube = 9
Diagonal of a cube = Side \(\sqrt{2}\) = 9\(\sqrt{2}\)

The points A,C,D and B are equally placed , so consider the distance between any 2 points as x, so 3x = diagonal of the cube
3x = 9\(\sqrt{2}\)
x = \(\frac{9\sqrt{2} }{ 3}\) = 3\(\sqrt{2}\)
Diameter of the cylinder = 3\(\sqrt{2}\)
Radius of the cylinder = \(\frac{3\sqrt{2} }{ 2}\)

Area of cylinder = \(\pi* r^2 * h\) = \(\pi * (3\sqrt{2})^2*9\) = \(\frac{81 \pi }{ 2}\)

Area of cube = \(Side^3\) = \(9^3\) = 729

Volume = 729 - \(\frac{81 \pi }{ 2}\)

Ans: B

cub-vol: 9^3=729
a-c-d-b=9V2, c-d=9V2/3=3V2
r=3V2/2, h=9
cyl-vol: pi.r^2*h=81/2.pi

(B)