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If the lengths of diagonals DF, AG and CE of the cube shown in the adjoining figure are equal to the three sides of a triangle, then the radius of the circle circumscribing that triangle will be?
Consider side of the cube as x.
So diagonal will be of length $$\sqrt{3}$$ * x.
Now if diagonals are side of equilateral triangle we get area = 3*$$\sqrt{3}*x^2$$ /4 .
Also in a triangle
4 * Area * R = Product of sides
4* 3*$$\sqrt{3}*x^2$$ /4 * R = .3*$$\sqrt{3}*x^3$$
R = x
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