Question

A sphere is inscribed in a cube. Explain the relationship between the surface areas of the two solid figures.

Answer 1

Answer: The relationship between the surface areas of the two solid figures is that the surface area of cube is 1.1 times the surface of cube.

Step-by-step explanation:

Since we have given that

A sphere is inscribed in a cube.

Let the radius of sphere be 'r'.

Let the side of cube would be the diameter of sphere i.e. 2r.

So, Surface area of sphere would be

$4\pi r^2$

And Surface area of cube would be

$6a^2\\\\=6(2r)^2\\\\=6\times 4r^2\\\\=24r^2$

$\dfrac{\text{Surface area of cube}}{\text{Surface area of sphere}}=\dfrac{24r^2}{4\times \dfrac{22}{4}r^2}=\dfrac{24}{22}=\dfrac{12}{11}=1.09=1.1$

So, the relationship between the surface areas of the two solid figures is that the surface area of cube is 1.1 times the surface of cube.

Answer 2

Sphere has a radius of "r"

cube has a side of "2r"

SA of sphere is 4 *pi *r^2
Sa of cube is 6* (2r)^2