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Cube and Sphere example

Question:

ABCD-A1B1C1D1 is a cube with side length a. If all vertices of the cube lie on a sphere and the surface of the sphere is 27 pi, what is the value of a?

example of cube and sphere

Solution:

Because all vertices of the cube lie on a sphere, so the center of the sphere is the middle point of the longest diagonal of the cube. Connect A1C. The line segment A1C is one of the longest diagonal. In order to find A1C, we connect AC first.

solution of the example of cube and sphere

Because ABCD-A1B1C1D1 is a cube, so the base ABCD is a square. The angle ABC is 90 degrees, so triangle ABC is a right triangle. AB = BC = a (a is unknown).

In right triangle ABC,
AC2 = AB2 + BC2
= a2 + a2 = 2a2

Because ABCD-A1B1C1D1 is a cube, so edge A1A is perpendicular to the plane ABCD. Because AC lies on the plane ABCD, so A1A is perpendicular to AC, so the angle A1AC is 90 degrees. So, the triangle A1AC is a right triangle.

In right triangle A1AC,
A1C2 = A1A2 + AC2
= a2 + 2 a2
= 3 a2
So, A1C = square root of 3 times a.

The radius of the sphere = half of A1C = half of square root of 3 times a. Let R be the radius of the sphere, R = (square root of 3) a/2.

Surface of the sphere SA:
SA = 4 pi R2
= 4 pi [(square root of 3) a/2]2
= 4 pi × 3 × a2 ÷ 4
= 3 pi a2

We are given that the surface of the sphere is 27 pi, so we get the equation:

3 pi a2 = 27 pi.

Both side of the equation divide by pi and both side of the equation divide by 3, we get

a2 = 9
a = 3

Therefore, the side length of the cube is 3. Please watch the video for more details.