The surface area of circle of radius 3 cm and cube of edge length x cm being equal implies [tex]x = \sqrt{k\pi}[/tex] integer (k = 6).
Suppose that radius of the considered sphere is of 'r' units.
Then, its surface area S would be:
[tex]S = 4\pi r^2 \: \rm unit^2[/tex]
Given that:
Finding the surface area of the considered cube and sphere:
Thus, we get:
[tex]6x^2 = 36\pi\\\\\text{Dividing both the sides by 6}\\\\x^2 = 6\pi\\\\\text{Taking root of both the sides}\\\\x = \sqrt{6\pi} \: \rm cm[/tex]
(took positive value out of the root since x cannot be negative as it represents length of the edge of a cube which cannot be a negative quantity).
Thus, the surface area of circle of radius 3 cm and cube of edge length x cm being equal implies [tex]x = \sqrt{k\pi}[/tex] integer (k = 6).
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