Cube is a Solid geometrical shape which six Square faces, means a Cube has 8 vertices and 12 edges, and three edges meet together at each vertex. It is also called regular hexadron.

Here we will discuss how to find the diagonal of a cube--> Means we have to find the Face diagonal of cube and Space diagonal of cube whose each side is of length 'a'.

Let’s draw a cube, assume that each side of this cube is 'a';

We know that each cube is made of 6 faces and each face has 4 edges. So first of all derive the face diagonal, from figure consider the face BCDE, and draw a line from D to B due to this BCDE is divided into two triangle, triangle BCD and triangle BDE, and DB is called diagonal, and we have to find out the length of this diagonal. We can see that both Triangles are right angled so we can apply pythagoras theorem on both triangles, from we can find DB, consider the triangle BCD, by applying theorem in this triangle,

DB

Length of DC = CB = a,

So DB

DB = face diagonal = 2 a ------equation 1,

Now to find the space diagonal draw a line from a to d and apply Pythagoras theorem on Right Triangle ADB,

So, AB

AB

= √3.a.

Here we will discuss how to find the diagonal of a cube--> Means we have to find the Face diagonal of cube and Space diagonal of cube whose each side is of length 'a'.

Let’s draw a cube, assume that each side of this cube is 'a';

We know that each cube is made of 6 faces and each face has 4 edges. So first of all derive the face diagonal, from figure consider the face BCDE, and draw a line from D to B due to this BCDE is divided into two triangle, triangle BCD and triangle BDE, and DB is called diagonal, and we have to find out the length of this diagonal. We can see that both Triangles are right angled so we can apply pythagoras theorem on both triangles, from we can find DB, consider the triangle BCD, by applying theorem in this triangle,

DB

^{2}= DC^{2}+ CB^{2},Length of DC = CB = a,

So DB

^{2}= a^{2}+ a^{2}=> DB^{2}= 2a^{2},DB = face diagonal = 2 a ------equation 1,

Now to find the space diagonal draw a line from a to d and apply Pythagoras theorem on Right Triangle ADB,

So, AB

^{2}= AD^{2}+ DB^{2}, value of AD = a and DB = √(2) a according to …...........equation 2,AB

^{2}= a^{2}+ (√2a)^{2},= √3.a.