$7^3$ Followers Problem!

Using Pythagorean Theorem

let y be the side of the cube, x be the diagonal of a face of the cube and a the diagonal of the cube.

y^2+y^2=x^2

2(y^2)=x^2

x^2+y^2=a^2

(2(y^2))+(y^2)=a^2

3(y^2)=a^2

a=y(sqrt(3))

a=7(sqrt(3))

Looking through the choices, only (sqrt((1078/3)x3)) is equal to (7(sqrt(3)))

so it's the answer

(sqrt((1078/3)x3))

Using Algebra and Phytagorean Theorem...

Let " $a$ " be the length of the side, substituting $7$

Green = $a\sqrt{2}$ Purple = $\sqrt{(a\sqrt{2})^2+a^2}=\sqrt{2a^2 + a^2}=\sqrt{3a^2}$

Substitute " $a$ " with 7

$\sqrt{3(7)^2}=\sqrt{3(49)}=\sqrt{\frac{1078}{22} * 3}$

The space diagonal of a cube with side length a is given by a√3 Answer is 7√3 from option A