Question

Show that the face diagonal of a cube is equal to `sqrt2  a`, where a is the edge length.

Answer 1

From the image, assume the cube has vertices at A, B, C, and D on the front face of the cube, with edge length a.

The face diagonal we need to find is the diagonal connecting A to C on the face ABCD, which is a square.

Apply the pythagorean theorem:

Since the face ABCD is a square, the face diagonal is the hypotenuse of a right triangle formed by the two edges, AB and AD, both of length a.

For a square with side length a, the diagonal AC forms the hypotenuse of a right-angled triangle, with the two sides being the edges of the square.

By the Pythagorean theorem:

(Face diagonal)² = AB² + AD²

AC² = AB² + AD²

AC = `sqrt(AD^2 + CD^2)` 

= `sqrt(a^2 + a^2)`    ...(∵ AD = CD = a)

= `sqrt(2  a^2)` 

= `sqrt 2  a`