The area of an equilateral triangle is $$6\sqrt{3}$$ times the area of a rhombus whose one side measures 13 cm and one diagonal is 10 cm. The length of side of the triangle, in cm, is:

Since, diagonals of a rhombus bisect each other at right angles, we get a right angled triangle with side of rhombus as the hypotenuse and the two half diagonals as base and perpendicular, let other diagonal be $$2x$$ cm

=> $$x=\sqrt{(13)^2-(5)^2}=\sqrt{169-25}=\sqrt{144}=12$$ cm

=> Other diagonal = $$24$$ cm

Thus, area of rhombus = $$\frac{1}{2}\times24\times10=120$$ $$cm^2$$

=> Area of triangle with side $$s$$ = $$\frac{\sqrt3}{4} s^2=6\sqrt{3}\times120$$

=> $$s^2=120\times24$$

=> $$s=24\sqrt5$$ $$cm^2$$

=> Ans - (C)