Question 153186
By "8-inch globe", I assume that you mean the diameter of the globe (sphere) is 8 inche!?
And by the roses' average width of 2" I assume that the diameter of the rose averages 2 inches!
So, let's find the surface area of the 8" globe (sphere).
The surface area of a sphere is given by:
{{{A[g] = pi*D^2}}} ...and the area of a rose, which approximates a circle, is given by:
{{{A[r] = pi*r^2}}} or {{{A[r] = pi*(D/2)^2}}} = {{{pi*D^2/4}}}
So, for the 8" globe, we have:
{{{A[g] = pi*(8)^2}}}
{{{A[g] = 64*pi}}}sq.in....and for the roses, we have:
{{{A[r] = pi*(2)^2/4}}}
{{{A[r] = pi}}}sq.in.
Now you want to know many times will the area of the rose ({{{pi}}}sq.in.) go into the area of the globe ({{{64*pi}}}sq.in.)? So we divide the globe area by the rose area.
{{{64*cross(pi)/cross(pi) = 64}}}
You would need about 64 2"-roses to cover the surface of an 8"-globe.