Consider the cube given below: $NR,MS$ and $PT$ are the diagonals of the cube whose volume is $1728\ cm^{3}$. If a triangle is constructed whose three sides are same as the measure of the sides $NR,MS$ and $PT$, then what will be the radius of the circle circumscribing that triangle?

Given the cube's volume is 1728 cubic cm, its side length is the cube root of 1728, which is 12 cm. The lengths of NR, MS, and PT are space diagonals of the cube, each measuring 12 * sqrt(3) cm, so the constructed triangle is equilateral. When an equilateral triangle is circumscribed by a circle, the circumradius formula is side divided by sqrt(3). However, applying the general formula for the circumradius of any triangle, R = (a * b * c) / (4 * Area), the area of this equilateral triangle is (sqrt(3)/4) * (12 * sqrt(3))^2 = 108 * sqrt(3). The radius is therefore (12 * sqrt(3))^3 / (4 * 108 * sqrt(3)), which simplifies to 24/sqrt(5) cm.

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