Question 265045
I'm assuming that the circular base of the conical hole inscribed in the square side of the cube is an exact fit, that is the circle is tangent to the sides of the square.
And that the depth of the conical hole does not protrude beyond the other side of the cube.
So, given the above, you will have a cone (conical hole) whose circular base has a diameter (D) equal to the length of the side of the cube (D = 6cm.) and the height (h) of the cone (really the depth of the conical hole) is also equal to the length of the side of the cube (h = 6cm.).
The surface area (S) of a right circular cone, such as you have here, is given by:
{{{S = (pi)R*sqrt(R^2+h^2)}}} Where: R = radius of base of cone = 3cm. and h = height of cone = 6cm. Making the appropriate substitutions...
{{{S = (3.14)(3)sqrt(9+36)}}}
{{{S = (9.42)*sqrt(45)}}}
{{{highlight(S = 63.2)}}}sq.cm.