Hi Arwaz,

The only information we are getting from the diagonals of the cube is the length of the sides of the triangle. We are not drawing any triangle inside the cube. We are being told that the diagonal lengths DF, AG and CE are the lengths of the sides of a triangle. So we just have to draw an equilateral triangle with sides having length same as the length of a diagonal.
Now if the side of cube is a, then side of the diagonal will be $$\sqrt{\ 3}$$a which is the length of the side of an equilateral triangle. We know that the area of an equilateral triangle is $$\dfrac{\sqrt{\ 3}}{4}s^2$$ , so the area becomes

Area = $$\frac{\sqrt{\ 3}}{4}\left(\sqrt{\ 3}a\right)^2\ =\ \frac{3\sqrt{\ 3}}{4}a^2$$

and we know that,    product of sides of triangle = 4 * R * Area of the triangle

==>   R =  $$\dfrac{product\ of\ sides\ of\ the\ triangle}{4\times\left(area\ of\ triangle\right)\ }\ =\ \dfrac{\sqrt{\ 3\ }a\times\ \sqrt{\ 3}a\times\ \sqrt{\ 3}a}{4\times\ \dfrac{3\sqrt{\ 3}}{4}\times\ a^2}\ =\ \dfrac{3\sqrt{\ 3}a^3}{3\sqrt{\ 3}a^2}\ =\ a$$

So R is the length of the side of the cube which is a.
Hope this clears your doubt.

What step of the solution did you not understand Lenin? Please go through the concepts again and attempt with a fresh mind.

Hi Vinay,
In an equilateral triangle, Circumradius is two-thirds of the height whereas in-radius is one-third of the height of the triangle.